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\r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EA complex fraction is a fraction where the numerator, denominator, or both contain a fraction. 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\r\n\u003Cdiv\u003EIndirect variation is a relationship between two variables in which the product remains constant. When one variable increases the other decreases in proportion, so that the product does not change. If we say, b is inversely proportional to a, the equation is of the form b = k\/a (where k is a constant).\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to: \u003C\/div\u003E \r\n\u003Cdiv\u003E- Explain the concept of variation.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Recognize indirect variation.\u003C\/div\u003E","CONT_DESC_AR":"Indirect variation is a relationship between two variables in which the product is a constant. When one variable increases the other decreases in proportion so that the product is unchanged. 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In this expression, a is the real part and b is the imaginary part of the complex number.\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to:\u003C\/div\u003E \r\n\u003Cdiv\u003E- Define complex numbers.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Find the conjugate and the multiplicative inverse of a complex number.\u003C\/div\u003E","CONT_DESC_AR":"A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, satisfying the equation i^2 = \u0026amp;minus;1.\u0026lt;br \/\u0026gt;\nIn this expression, a is the real part and b is the imaginary part of the complex number.\u0026lt;br \/\u0026gt;\n\u0026amp;nbsp;\u0026lt;br \/\u0026gt;\n\u0026lt;strong\u0026gt;Learning Objectives\u0026lt;\/strong\u0026gt;\u0026lt;br \/\u0026gt;\n\u0026lt;br \/\u0026gt;\nAt the end of this simulation, you will able to:\u0026lt;br \/\u0026gt;\n\u0026amp;bull; explain the concept of complex numbers\u0026lt;br \/\u0026gt;\n\u0026amp;bull; find the conjugate and multiplicative inverse of a complex number","BACKING_FILE":null,"FILE_UID":null,"SCORM_COURSE_ID":null,"CONT_SRC":"","MOD_FILES":null,"FOLDER_NAME":null,"CONTTYPE_ID":"9","ANDROID_PKG":"com.umety.vr.hs300012","TOPIC_ID":"hs300012","IS_PUBLISH":"Y","IS_PUBLIC":"Y","CONT_PRICE":null,"PUBLIC_IMG":"thumb_HS300012.jpg","PUBLIC_BANNER_IMG":"HS300012.jpg","PUBLIC_VIDEO":"pvideo_hs300012.mp4","PUBLIC_VIDEO_URL":"https:\/\/youtu.be\/t2ti-mqDNXU","DIST":null,"SHOW_ON_HOME":"N","CONTROLLER_REQUIRED":"Y","DOMAIN":"3","CONCEPT":"0","STATUS":"A","EXPIRY_DAYS":null,"CREATED_ON":"0000-00-00 00:00:00","CREATED_BY":"1","UPDATED_ON":"0000-00-00 00:00:00","UPDATED_BY":"2","CONT_ORDER":"0","X_ROTATION":null,"Y_ROTATION":null,"Z_ROTATION":null,"BG_COLOR":"0x000000","X_POSITION":null,"Y_POSITION":null,"Z_POSITION":null,"TEMP_DESC":"\u0026lt;div\u0026gt;Overview:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;A complex number is defined as the combination of real and an imaginary number that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, satisfying the equation i^2 = \u22121. 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(synthetic)","CONT_DESC":"\u003Ch3\u003EOverview:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003ESynthetic division is a shortcut method of dividing a polynomial by another polynomial. To divide a polynomial by using synthetic division, we divide a linear expression by the leading coefficient (first number) that must be a 1. For example, you can use synthetic division to divide by x + 3 or x \u2013 6, but you cannot use synthetic division to divide by x\u00b2 + 2 or 3x\u00b2 \u2013 x + 7.\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to:\u003C\/div\u003E \r\n\u003Cdiv\u003E- Solve the division of polynomials using the synthetic method.\u003C\/div\u003E","CONT_DESC_AR":"Synthetic division is shorthand, or a shortcut, method of polynomial division in the special case of dividing by a linear factor, and it only works in this case.\u0026lt;br \/\u0026gt;\n\u0026amp;nbsp;\u0026lt;br \/\u0026gt;\n\u0026lt;strong\u0026gt;Learning Objective\u0026lt;\/strong\u0026gt;\u0026lt;br \/\u0026gt;\n\u0026lt;br \/\u0026gt;\nIn this topic you will be able to:\u0026lt;br \/\u0026gt;\n\u0026amp;bull; solve division of polynomials by the synthetic method","BACKING_FILE":"ss300059.apk","FILE_UID":null,"SCORM_COURSE_ID":null,"CONT_SRC":"","MOD_FILES":null,"FOLDER_NAME":null,"CONTTYPE_ID":"9","ANDROID_PKG":"com.umety.vr.ss300059","TOPIC_ID":"ss300059","IS_PUBLISH":"Y","IS_PUBLIC":"Y","CONT_PRICE":null,"PUBLIC_IMG":"thumb_SS300059.jpg","PUBLIC_BANNER_IMG":"SS300059.jpg","PUBLIC_VIDEO":"pvideo_ss300059.mp4","PUBLIC_VIDEO_URL":"https:\/\/youtu.be\/ODtQToJDKFQ","DIST":null,"SHOW_ON_HOME":"N","CONTROLLER_REQUIRED":"Y","DOMAIN":"3","CONCEPT":"0","STATUS":"A","EXPIRY_DAYS":null,"CREATED_ON":"0000-00-00 00:00:00","CREATED_BY":"1","UPDATED_ON":"0000-00-00 00:00:00","UPDATED_BY":"2","CONT_ORDER":"0","X_ROTATION":null,"Y_ROTATION":null,"Z_ROTATION":null,"BG_COLOR":"0x000000","X_POSITION":null,"Y_POSITION":null,"Z_POSITION":null,"TEMP_DESC":"\u0026lt;div\u0026gt;Overview:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;Synthetic division is a shortcut method of dividing a polynomial by another polynomial. To divide a polynomial by using synthetic division, we divide a linear expression by the leading coefficient (first number) that must be a 1. For example, you can use synthetic division to divide by x + 3 or x \u2013 6, but you cannot use synthetic division to divide by x\u00b2 + 2 or 3x\u00b2 \u2013 x + 7.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;Learning Objectives:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;div\u0026gt;After completing this module, you will be able to:\u0026amp;nbsp;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Solve the division of polynomials using the synthetic method.\u0026lt;\/div\u0026gt;\u0026lt;\/div\u0026gt;","IS_ANALYTICS":"Y","VR_ENABLE":"Y","VR_SESSION_ENABLE":"Y","YOUTUBE_URL":null,"CONT_TYPE":"VR Module","CAT_NAME":"\u0642\u0633\u0645\u0629 \u0643\u062b\u064a\u0631\u0627\u062a \u0627\u0644\u062d\u062f\u0648\u062f (\u0627\u0644\u062a\u0631\u0643\u064a\u0628\u064a\u0629)","ADMSUBJECT_ID":"733","ADMCOURSE_ID":"132","DISPLAY_NAME":"UAE - Grade 10 - 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Trigonometry","CONT_DESC":"\u003Ch3\u003EOverview:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EThe sides and angles of a triangle are related to each other. The relationships that define the connection between the angles and the sides of a right triangle are expressed in terms of six trigonometric ratios :Sine (sin), cosine (cos), tangent (tan), cosecant (cosec), secant (sec), and cotangent (cot). By using these trigonometric ratios , we can find the unknown values like height of a building or distance of any object from any building or tower etc if angle of elevation (angle that eye of observer makes with the top of a building) or angle of depression (the angle that is formed between the eye of the observer who is standing on any top of the building with the object that is placed on the ground) are known.\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to:\u003C\/div\u003E \r\n\u003Cdiv\u003E- Explain the applications of trigonometry in various fields of life.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Identify the angle of elevation and the angle of depression.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Determine \u0026#039;height\u0026#039; and \u0026#039;distance\u0026#039; without actually measuring them.\u003C\/div\u003E","CONT_DESC_AR":"Use of trigonometry, angle of elevation and depression. 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In plain English, this definition means: the domain is the set of all possible x-values which will make the function work, and will output real y-values.\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to:\u003C\/div\u003E \r\n\u003Cdiv\u003E- Identify and find the domain of a function.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Identify and find the range of a function.\u003C\/div\u003E","CONT_DESC_AR":"The domain of a function is the complete set of possible values of the independent variable.\u0026lt;br \/\u0026gt;\nIn plain English, this definition means: The domain is the set of all possible x-values which will make the function \u0026amp;quot;work\u0026amp;quot;, and will output real y-values.\u0026lt;br \/\u0026gt;\n\u0026amp;nbsp;\u0026lt;br \/\u0026gt;\n\u0026lt;strong\u0026gt;Learning Objective\u0026lt;\/strong\u0026gt;\u0026lt;br \/\u0026gt;\n\u0026lt;br \/\u0026gt;\nIn this topic you will be able to:\u0026lt;br \/\u0026gt;\n\u0026amp;bull; find the domain and range of a function graphically","BACKING_FILE":null,"FILE_UID":null,"SCORM_COURSE_ID":null,"CONT_SRC":"","MOD_FILES":null,"FOLDER_NAME":null,"CONTTYPE_ID":"9","ANDROID_PKG":"com.umety.vr.hs300051","TOPIC_ID":"hs300051","IS_PUBLISH":"Y","IS_PUBLIC":"Y","CONT_PRICE":null,"PUBLIC_IMG":"thumb_HS300051.jpg","PUBLIC_BANNER_IMG":"HS300051.jpg","PUBLIC_VIDEO":"pvideo_hs300051.mp4","PUBLIC_VIDEO_URL":"https:\/\/youtu.be\/0x50NsVhW4w","DIST":null,"SHOW_ON_HOME":"N","CONTROLLER_REQUIRED":"Y","DOMAIN":"3","CONCEPT":"0","STATUS":"A","EXPIRY_DAYS":null,"CREATED_ON":"0000-00-00 00:00:00","CREATED_BY":"1","UPDATED_ON":"0000-00-00 00:00:00","UPDATED_BY":"2","CONT_ORDER":"0","X_ROTATION":null,"Y_ROTATION":null,"Z_ROTATION":null,"BG_COLOR":"0x000000","X_POSITION":null,"Y_POSITION":null,"Z_POSITION":null,"TEMP_DESC":"\u0026lt;div\u0026gt;Overview:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;The domain of a function is the complete set of possible values of the independent variable. 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Variables","CONT_SLUG":"solving-system-of-equations-in-two-variables","CONT_TITLE_AR":"Solving System of Equations in Two Variables","CONT_DESC":"\u003Ch3\u003EOverview:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EA linear equation in two variable is defined as a first-degree equation that can be written in the form ax + by = c, where both a and b are not equal to zero and a, b, and c are real numbers. The graph of any equation of this form is a straight line.To solve the system of equations graphically, first of all we graph both the lines and then find the point of intersection. If the lines intersect at only one point, it is known as unique solutions. If the line coincides with each other, then they have infinitely many solutions and if the two lines are parallel, then no solution exists for those equations.\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to:\u003C\/div\u003E \r\n\u003Cdiv\u003E- Solve a linear equation in two variables using the graphical method.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Differentiate and construct a unique solution.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Differentiate and construct infinitely many solutions.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Differentiate and construct no solution.\u003C\/div\u003E","CONT_DESC_AR":"Solving systems of equations with two variables.\u003C\/br\u003E\r\nA system of a linear equation is comprised of two or more equations, used to find a common solution to the equations.\u003C\/br\u003E\u003C\/br\u003E\r\n\u003Cstrong\u003ELearning Objectives\u003C\/strong\u003E\u003C\/br\u003E\u003C\/br\u003E\r\nAt the end of this simulation, you will be able to:\u003C\/br\u003E\r\n\u2022 solve linear equation in two variable using graphical method\u003C\/br\u003E\r\n\u2022 differentiate and construct unique solution\u003C\/br\u003E\r\n\u2022 differentiate and construct infinitely many solution\u003C\/br\u003E\r\n\u2022 differentiate and construct no solution","BACKING_FILE":null,"FILE_UID":null,"SCORM_COURSE_ID":null,"CONT_SRC":"","MOD_FILES":null,"FOLDER_NAME":null,"CONTTYPE_ID":"9","ANDROID_PKG":"com.umety.vr.hs300076","TOPIC_ID":"hs300076","IS_PUBLISH":"Y","IS_PUBLIC":"Y","CONT_PRICE":null,"PUBLIC_IMG":"thumb_HS300076.jpg","PUBLIC_BANNER_IMG":"hs300076.jpg","PUBLIC_VIDEO":"pvideo_hs300076.mp4","PUBLIC_VIDEO_URL":"https:\/\/youtu.be\/tc7Z4gGoOwU","DIST":null,"SHOW_ON_HOME":"N","CONTROLLER_REQUIRED":"Y","DOMAIN":"3","CONCEPT":"0","STATUS":"A","EXPIRY_DAYS":null,"CREATED_ON":"0000-00-00 00:00:00","CREATED_BY":"1","UPDATED_ON":"0000-00-00 00:00:00","UPDATED_BY":"2","CONT_ORDER":"0","X_ROTATION":null,"Y_ROTATION":null,"Z_ROTATION":null,"BG_COLOR":"0x000000","X_POSITION":null,"Y_POSITION":null,"Z_POSITION":null,"TEMP_DESC":"\u0026lt;div\u0026gt;Overview:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;A linear equation in two variable is defined as a first-degree equation that can be written in the form ax + by = c, where both a and b are not equal to zero and \u0026amp;nbsp;a, b, and c are real numbers. The graph of any equation of this form is a straight line.To solve the system of equations graphically, first of all we graph both the lines and then find the point of intersection. If the lines intersect at only one point, it is known as unique solutions. If the line coincides with each other, then they have infinitely many solutions and if the two lines are parallel, then no solution exists for those equations.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;Learning Objectives:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;div\u0026gt;After completing this module, you will be able to:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Solve a linear equation in two variables using the graphical method.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Differentiate and construct a unique solution.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Differentiate and construct infinitely many solutions.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Differentiate and construct no solution.\u0026lt;\/div\u0026gt;\u0026lt;\/div\u0026gt;","IS_ANALYTICS":"Y","VR_ENABLE":"Y","VR_SESSION_ENABLE":"Y","YOUTUBE_URL":null,"CONT_TYPE":"VR 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origin.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Find the equation of a circle with any arbitrary origin.\u003C\/div\u003E","CONT_DESC_AR":"The center-radius form of the circle equation is in the format (x-h)\u0026lt;sup\u0026gt;2\u0026lt;\/sup\u0026gt; + (y-k)\u0026lt;sup\u0026gt;2\u0026lt;\/sup\u0026gt; = r\u0026lt;sup\u0026gt;2\u0026lt;\/sup\u0026gt;, with center \u0026amp;nbsp;(h, k) and the radius \u0026amp;quot;r\u0026amp;quot;.\u0026lt;br \/\u0026gt;\n\u0026amp;nbsp;\u0026lt;br \/\u0026gt;\n\u0026lt;strong\u0026gt;Learning Objectives\u0026lt;\/strong\u0026gt;\u0026lt;br \/\u0026gt;\n\u0026lt;br \/\u0026gt;\nAt the end of this simulation, you will be able to:\u0026lt;br \/\u0026gt;\n- define a circle\u0026lt;br \/\u0026gt;\n- find the equation of a circle with the centre at origin\u0026lt;br \/\u0026gt;\n- find the equation of a circle with any arbitary 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107%; font-family: Roboto; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;\u0026quot;\u0026gt;(x \u2013 h)\u0026lt;sup\u0026gt;2\u0026lt;\/sup\u0026gt; + (y \u2013 k)\u0026lt;sup\u0026gt;2\u0026lt;\/sup\u0026gt; = r\u0026lt;sup\u0026gt;2\u0026lt;\/sup\u0026gt;\u0026lt;\/span\u0026gt;, with center\u0026amp;nbsp; (h, k) and the radius r.\u0026lt;br\u0026gt;\u0026lt;\/p\u0026gt;\u0026lt;p\u0026gt;Learning Objectives:\u0026lt;br\u0026gt;\u0026lt;\/p\u0026gt;\u0026lt;div\u0026gt;After completing this module, you will be able to:\u0026lt;div\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Define a circle.\u0026lt;div\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Find the equation of a circle with its center at the origin.\u0026lt;div\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Find the equation of a circle with any arbitrary origin.\u0026lt;div\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;\/div\u0026gt;","IS_ANALYTICS":"Y","VR_ENABLE":"Y","VR_SESSION_ENABLE":"Y","YOUTUBE_URL":null,"CONT_TYPE":"VR Module","CAT_NAME":"\u0645\u0639\u0627\u062f\u0644\u0629 \u0627\u0644\u062f\u0627\u0626\u0631\u0629","ADMSUBJECT_ID":"733","ADMCOURSE_ID":"132","DISPLAY_NAME":"UAE - Grade 10 - Mathematics","DISPLAY_NAME_AR":"UAE - Grade 10 - Mathematics","SUBJECT_NAME":"Mathematics","SUBJECT_NAME_AR":"Mathematics","SUBJECT_DESC":"\u062a\u064f\u0639\u0631\u0651\u0650\u0641 \u0627\u0644\u0648\u062d\u062f\u0627\u062a \u0627\u0644\u062d\u0633\u0627\u0628\u064a\u0629 \u0644\u0644\u0635\u0641 \u0627\u0644\u0639\u0627\u0634\u0631 \u0627\u0644\u0623\u0639\u062f\u0627\u062f \u0627\u0644\u062d\u0642\u064a\u0642\u064a\u0629 \u0648\u0627\u0644\u0645\u0639\u0627\u062f\u0644\u0627\u062a \u0627\u0644\u062a\u0631\u0628\u064a\u0639\u064a\u0629 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The circumference of a circle is defined as the outer boundary of circle. The formula for calculating circumference is 2\u03c0r and for calculating area is \u03c0r\u00b2.\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAt the end of this module, you will be able to:\u003C\/div\u003E \r\n\u003Cdiv\u003E- Define a circle.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Identify and define the radius and diameter of a circle.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Calculate the circumference and the area of a circle.\u003C\/div\u003E","CONT_DESC_AR":"Circle : - A round closed figure whose boundary consists of points equidistant from a fixed point.\u0026lt;br \/\u0026gt;\nRadius : - Line connecting the centre to the outer boundary of circle.\u0026lt;br \/\u0026gt;\nDiameter : - Twice of the radius is diameter of the circle.\u0026lt;br \/\u0026gt;\nCircumference: - Outer boundary of circle.\u0026lt;br \/\u0026gt;\nFormula to calculate circumference is 2\u0026amp;pi;r\u0026lt;br \/\u0026gt;\nArea \u0026amp;nbsp;: - Formula for finding area of circle is \u0026amp;pi;r\u0026lt;sup\u0026gt;2\u0026lt;\/sup\u0026gt;\u0026lt;br \/\u0026gt;\n\u0026amp;nbsp;\u0026lt;br \/\u0026gt;\n\u0026lt;strong\u0026gt;Learning Objectives\u0026lt;\/strong\u0026gt;\u0026lt;br \/\u0026gt;\n\u0026lt;br \/\u0026gt;\nAt the end of this simulation, you will be able to:\u0026lt;br \/\u0026gt;\n\u0026amp;bull; define a circle\u0026lt;br \/\u0026gt;\n\u0026amp;bull; identify the radius and diameter of a circle\u0026lt;br \/\u0026gt;\n\u0026amp;bull; calculate the circumference and area of a circle","BACKING_FILE":null,"FILE_UID":null,"SCORM_COURSE_ID":null,"CONT_SRC":"","MOD_FILES":null,"FOLDER_NAME":null,"CONTTYPE_ID":"9","ANDROID_PKG":"com.umety.vr.ms300042","TOPIC_ID":"ms300042","IS_PUBLISH":"Y","IS_PUBLIC":"Y","CONT_PRICE":null,"PUBLIC_IMG":"thumb_MS300042.jpg","PUBLIC_BANNER_IMG":"MS300042.jpg","PUBLIC_VIDEO":"pvideo_ms300042.mp4","PUBLIC_VIDEO_URL":"https:\/\/youtu.be\/xKAEF2qfW3g","DIST":null,"SHOW_ON_HOME":"N","CONTROLLER_REQUIRED":"Y","DOMAIN":"3","CONCEPT":"0","STATUS":"A","EXPIRY_DAYS":null,"CREATED_ON":"0000-00-00 00:00:00","CREATED_BY":"1","UPDATED_ON":"0000-00-00 00:00:00","UPDATED_BY":"2","CONT_ORDER":"0","X_ROTATION":null,"Y_ROTATION":null,"Z_ROTATION":null,"BG_COLOR":"0x000000","X_POSITION":null,"Y_POSITION":null,"Z_POSITION":null,"TEMP_DESC":"\u0026lt;div\u0026gt;Overview:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;div\u0026gt;A circle is defined as a round closed figure whose boundary consists of points equidistant from a fixed point.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;Its radius is the line connecting the center to the outer boundary of circle and diameter is twice of the radius. The circumference of a circle is defined as the outer boundary of circle. 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Circles","CONT_DESC":"\u003Ch3\u003EOverview:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EA sector is the part of a circle enclosed by two radii and an intercepted arc of that circle. The segment of a circle is the region bounded by a chord. An annulus is a flat ring shaped object. To find the area of an annulus, a sector, and a segment we actually need to find the fractional part of the area of the entire circle.\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to:\u003C\/div\u003E \r\n\u003Cdiv\u003E- Find the area of a sector.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Find the area of a segment.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Find the area of an annulus.\u003C\/div\u003E","CONT_DESC_AR":"When finding the area of an annulus, sector, and segment you are actually finding a fractional part of the area of the entire circle.\u003C\/br\u003E\u003C\/br\u003E\r\n\u003Cstrong\u003ELearning Objectives\u003C\/strong\u003E\u003C\/br\u003E\u003C\/br\u003E\r\nAt the end of this simulation, you will be able to apply the formula of:\u003C\/br\u003E\r\n\u2022 the area of a sector\u003C\/br\u003E\r\n\u2022 the area of a segment\u003C\/br\u003E\r\n\u2022 the area of an annulus","BACKING_FILE":"hs300020.apk","FILE_UID":null,"SCORM_COURSE_ID":null,"CONT_SRC":"","MOD_FILES":null,"FOLDER_NAME":null,"CONTTYPE_ID":"9","ANDROID_PKG":"com.umety.vr.hs300020","TOPIC_ID":"hs300020","IS_PUBLISH":"Y","IS_PUBLIC":"Y","CONT_PRICE":null,"PUBLIC_IMG":"thumb_HS300020.jpg","PUBLIC_BANNER_IMG":"HS300020.jpg","PUBLIC_VIDEO":"pvideo_hs300020.mp4","PUBLIC_VIDEO_URL":"https:\/\/youtu.be\/z4XP6Ift0Yc","DIST":null,"SHOW_ON_HOME":"N","CONTROLLER_REQUIRED":"Y","DOMAIN":"3","CONCEPT":"0","STATUS":"A","EXPIRY_DAYS":null,"CREATED_ON":"0000-00-00 00:00:00","CREATED_BY":"1","UPDATED_ON":"0000-00-00 00:00:00","UPDATED_BY":"2","CONT_ORDER":"0","X_ROTATION":null,"Y_ROTATION":null,"Z_ROTATION":null,"BG_COLOR":"0x000000","X_POSITION":null,"Y_POSITION":null,"Z_POSITION":null,"TEMP_DESC":"\u0026lt;div\u0026gt;Overview:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;A sector is the part of a circle enclosed by two radii and an intercepted arc of that circle. The segment of a circle is the region bounded by a chord. An annulus is a flat ring shaped object. To find the area of an annulus, a sector, and a segment we actually need to find the fractional part of the area of the entire circle.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;Learning Objectives:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;div\u0026gt;After completing this module, you will be able to:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Find the area of a sector.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Find the area of a segment.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Find the area of an annulus.\u0026lt;\/div\u0026gt;\u0026lt;\/div\u0026gt;","IS_ANALYTICS":"Y","VR_ENABLE":"Y","VR_SESSION_ENABLE":"Y","YOUTUBE_URL":null,"CONT_TYPE":"VR Module","CAT_NAME":"\u0627\u0644\u0645\u0633\u0627\u062d\u0629 \u0627\u0644\u0645\u0631\u062a\u0628\u0637\u0629 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A part of a line that is bounded by two distinct end points is defined as line segment. A ray is defined as a line with one endpoint.\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to:\u003C\/div\u003E \r\n\u003Cdiv\u003E- Define intersecting lines.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Define parallel lines.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Define a point.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Define rays and line segments.\u003C\/div\u003E","CONT_DESC_AR":"A line is defined as a line of points that extends infinitely in two directions. It has one dimension, length. Points that are on the same line are called collinear points\u0026lt;br \/\u0026gt;\n\u0026lt;br \/\u0026gt;\n\u0026lt;strong\u0026gt;Learning Objectives\u0026lt;\/strong\u0026gt;\u0026lt;br \/\u0026gt;\n\u0026amp;nbsp;\u0026lt;br \/\u0026gt;\nIn this topic, you will be able to:\u0026lt;br \/\u0026gt;\n\u0026amp;bull; explore intersecting lines\u0026lt;br \/\u0026gt;\n\u0026amp;bull; explore parallel lines\u0026lt;br \/\u0026gt;\n\u0026amp;bull; explore what a point is\u0026lt;br \/\u0026gt;\n\u0026amp;bull; explore rays and line segments","BACKING_FILE":null,"FILE_UID":null,"SCORM_COURSE_ID":null,"CONT_SRC":"","MOD_FILES":null,"FOLDER_NAME":null,"CONTTYPE_ID":"9","ANDROID_PKG":"com.umety.vr.ms300045","TOPIC_ID":"ms300045","IS_PUBLISH":"Y","IS_PUBLIC":"Y","CONT_PRICE":null,"PUBLIC_IMG":"thumb_MS300045.jpg","PUBLIC_BANNER_IMG":"MS300045.jpg","PUBLIC_VIDEO":"pvideo_ms300045.mp4","PUBLIC_VIDEO_URL":"https:\/\/youtu.be\/qsCqLjwf7P8","DIST":null,"SHOW_ON_HOME":"N","CONTROLLER_REQUIRED":"Y","DOMAIN":"3","CONCEPT":"0","STATUS":"A","EXPIRY_DAYS":null,"CREATED_ON":"0000-00-00 00:00:00","CREATED_BY":"1","UPDATED_ON":"0000-00-00 00:00:00","UPDATED_BY":"2","CONT_ORDER":"0","X_ROTATION":null,"Y_ROTATION":null,"Z_ROTATION":null,"BG_COLOR":"0x000000","X_POSITION":null,"Y_POSITION":null,"Z_POSITION":null,"TEMP_DESC":"\u0026lt;div\u0026gt;Overview:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;A line is defined as a line of points that extend infinitely in two directions. A part of a line that is bounded by two distinct end points is defined as line segment. A ray is defined as a line with one endpoint.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;Learning Objectives:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;div\u0026gt;After completing this module, you will be able to:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Define intersecting lines.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Define parallel lines.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Define a point.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Define rays and line segments.\u0026lt;\/div\u0026gt;\u0026lt;\/div\u0026gt;","IS_ANALYTICS":"Y","VR_ENABLE":"Y","VR_SESSION_ENABLE":"Y","YOUTUBE_URL":null,"CONT_TYPE":"VR Module","CAT_NAME":"\u0627\u0644\u062e\u0637\u0648\u0637","ADMSUBJECT_ID":"733","ADMCOURSE_ID":"132","DISPLAY_NAME":"UAE - Grade 10 - Mathematics","DISPLAY_NAME_AR":"UAE - Grade 10 - 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\r\n\u003Cdiv\u003EDirect variation is the relationship between two variables in which one is a constant multiple of the other. In other words, we can say when one variable changes the other changes in proportion to the first. If b is directly proportional to a, the equation is of the form b = ka (where k is a constant).\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to :\u003C\/div\u003E \r\n\u003Cdiv\u003E- Explain the concept of direct variation.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Identify direct variation.\u003C\/div\u003E","CONT_DESC_AR":"A relationship between two variables in which one is a constant multiple of the other. In particular, when one variable changes the other changes in proportion to the first. If b is directly proportional to a, the equation is of the form b = ka (where k is a constant)\u0026lt;br \/\u0026gt;\n\u0026amp;nbsp;\u0026lt;br \/\u0026gt;\n\u0026lt;strong\u0026gt;Learning Objectives\u0026lt;\/strong\u0026gt;\u0026lt;br \/\u0026gt;\n\u0026amp;nbsp;\u0026lt;br \/\u0026gt;\nAt the end of this simulation, you will:\u0026lt;br \/\u0026gt;\n\u0026amp;bull; know the concept of direct variation\u0026lt;br \/\u0026gt;\n\u0026amp;bull; identify direct variation","BACKING_FILE":null,"FILE_UID":null,"SCORM_COURSE_ID":null,"CONT_SRC":"","MOD_FILES":null,"FOLDER_NAME":null,"CONTTYPE_ID":"9","ANDROID_PKG":"com.umety.vr.ms300077","TOPIC_ID":"ms300077","IS_PUBLISH":"Y","IS_PUBLIC":"Y","CONT_PRICE":null,"PUBLIC_IMG":"thumb_MS300077.jpg","PUBLIC_BANNER_IMG":"ms300077.jpg","PUBLIC_VIDEO":"pvideo_ms300077.mp4","PUBLIC_VIDEO_URL":"https:\/\/youtu.be\/ccOglkJsAqI","DIST":null,"SHOW_ON_HOME":"N","CONTROLLER_REQUIRED":"Y","DOMAIN":"3","CONCEPT":"0","STATUS":"A","EXPIRY_DAYS":null,"CREATED_ON":"0000-00-00 00:00:00","CREATED_BY":"1","UPDATED_ON":"0000-00-00 00:00:00","UPDATED_BY":"2","CONT_ORDER":"0","X_ROTATION":null,"Y_ROTATION":null,"Z_ROTATION":null,"BG_COLOR":"0x000000","X_POSITION":null,"Y_POSITION":null,"Z_POSITION":null,"TEMP_DESC":"\u0026lt;div\u0026gt;Overview:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;Direct variation is the relationship between two variables in which one is a constant multiple of the other. 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Graphically","CONT_DESC":"\u003Ch3\u003EOverview:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EA linear inequality is an inequality which involves a linear function. A linear inequality contains one of the symbols of inequality: \u003C is less than, \u003E is greater than, \u2264 is less than or equal to, \u2265 is greater than or equal to. Linear inequality in two variables can be solved in a similar manner as we solve linear inequality.\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to:\u003C\/div\u003E \r\n\u003Cdiv\u003E- Explain the concept of inequality.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Distinguish between graphs of inequalities.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Solve a system of linear inequalities graphically.\u003C\/div\u003E","CONT_DESC_AR":"A linear inequality is an inequality which involves a linear function.\u003C\/br\u003E\r\nA linear inequality contains one of the symbols of inequality: \u003C is less than, \u003E is greater than, \u2264 is less than or equal to.\u003C\/br\u003E\r\nLinear inequality in two variables can be solved in a similar manner as we solve system of linear equations.\u003C\/br\u003E\u003C\/br\u003E\r\n\u003Cstrong\u003ELearning Objectives\u003C\/strong\u003E\u003C\/br\u003E\u003C\/br\u003E\r\nIn this simulation,you will be able to:\u003C\/br\u003E\r\n\u2022 explain the concept of inequality\u003C\/br\u003E\r\n\u2022 distinguish between the graphs of inequalities\u003C\/br\u003E\r\n\u2022 solve the system of linear inequalities graphically","BACKING_FILE":"ss300049.apk","FILE_UID":null,"SCORM_COURSE_ID":null,"CONT_SRC":"","MOD_FILES":null,"FOLDER_NAME":null,"CONTTYPE_ID":"9","ANDROID_PKG":"com.umety.vr.ss300049","TOPIC_ID":"ss300049","IS_PUBLISH":"Y","IS_PUBLIC":"Y","CONT_PRICE":null,"PUBLIC_IMG":"thumb_SS300049.jpg","PUBLIC_BANNER_IMG":"SS300049.jpg","PUBLIC_VIDEO":"pvideo_ss300049.mp4","PUBLIC_VIDEO_URL":"https:\/\/youtu.be\/H6wES_wtrQ4","DIST":null,"SHOW_ON_HOME":"N","CONTROLLER_REQUIRED":"Y","DOMAIN":"3","CONCEPT":"0","STATUS":"A","EXPIRY_DAYS":null,"CREATED_ON":"0000-00-00 00:00:00","CREATED_BY":"1","UPDATED_ON":"0000-00-00 00:00:00","UPDATED_BY":"2","CONT_ORDER":"0","X_ROTATION":null,"Y_ROTATION":null,"Z_ROTATION":null,"BG_COLOR":"0x000000","X_POSITION":null,"Y_POSITION":null,"Z_POSITION":null,"TEMP_DESC":"\u0026lt;div\u0026gt;Overview:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;A linear inequality is an inequality which involves a linear function. A linear inequality contains one of the symbols of inequality: \u0026amp;lt; is less than, \u0026amp;gt; is greater than, \u2264 is less than or equal to, \u2265 is greater than or equal to. Linear inequality in two variables can be solved in a similar manner as we solve linear inequality.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;Learning Objectives:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;div\u0026gt;After completing this module, you will be able to:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Explain the concept of inequality.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Distinguish between graphs of inequalities.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Solve a system of linear inequalities graphically.\u0026lt;\/div\u0026gt;\u0026lt;\/div\u0026gt;","IS_ANALYTICS":"Y","VR_ENABLE":"Y","VR_SESSION_ENABLE":"Y","YOUTUBE_URL":null,"CONT_TYPE":"VR Module","CAT_NAME":"\u062d\u0644 \u0646\u0638\u0627\u0645 \u0645\u062a\u0628\u0627\u064a\u0646\u0627\u062a \u062e\u0637\u064a\u0629 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Matrices can be added, subtracted and multiplied. There are different types of matrices: row matrix, column matrix, identity matrix, zero matrix, diagonal matrix, symmetric matrix and skew symmetric matrix.\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to:\u003C\/div\u003E \r\n\u003Cdiv\u003E- Create a matrix.\u003C\/div\u003E \r\n\u003Cdiv\u003E- List types of matrices.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Transpose a matrix.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Distinguish between a symmetric and a skew symmetric matrices.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Perform operations on a matrix.\u003C\/div\u003E","CONT_DESC_AR":"A matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.\u0026lt;br \/\u0026gt;\nMatrices can be added, subtracted and multiplied.\u0026lt;br \/\u0026gt;\nThere are different types of matrices: row matrix, column matrix, identity matrix, zero matrix, diagonal matrix, symmetric matrix and skew symmetric matrix.\u0026lt;br \/\u0026gt;\n\u0026amp;nbsp;\u0026lt;br \/\u0026gt;\n\u0026lt;strong\u0026gt;Learning Objectives\u0026lt;\/strong\u0026gt;\u0026lt;br \/\u0026gt;\n\u0026lt;br \/\u0026gt;\nIn this simulation, You will be able to:\u0026lt;br \/\u0026gt;\n\u0026amp;bull; create a matrix\u0026lt;br \/\u0026gt;\n\u0026amp;bull; list types of matrices\u0026lt;br \/\u0026gt;\n\u0026amp;bull; transpose a matrix\u0026lt;br \/\u0026gt;\n\u0026amp;bull; distinguish between symmetric and skew symmetric matrices\u0026lt;br \/\u0026gt;\n\u0026amp;bull; perform operations on a matrix","BACKING_FILE":"ss300009.apk","FILE_UID":null,"SCORM_COURSE_ID":null,"CONT_SRC":"","MOD_FILES":null,"FOLDER_NAME":null,"CONTTYPE_ID":"9","ANDROID_PKG":"com.umety.vr.ss300009","TOPIC_ID":"ss300009","IS_PUBLISH":"Y","IS_PUBLIC":"Y","CONT_PRICE":null,"PUBLIC_IMG":"thumb_SS300009.jpg","PUBLIC_BANNER_IMG":"SS300009.jpg","PUBLIC_VIDEO":"pvideo_ss300009.mp4","PUBLIC_VIDEO_URL":"https:\/\/youtu.be\/aCPP3rt6pYM","DIST":null,"SHOW_ON_HOME":"N","CONTROLLER_REQUIRED":"Y","DOMAIN":"3","CONCEPT":"0","STATUS":"A","EXPIRY_DAYS":null,"CREATED_ON":"0000-00-00 00:00:00","CREATED_BY":"1","UPDATED_ON":"0000-00-00 00:00:00","UPDATED_BY":"2","CONT_ORDER":"0","X_ROTATION":null,"Y_ROTATION":null,"Z_ROTATION":null,"BG_COLOR":"0x000000","X_POSITION":null,"Y_POSITION":null,"Z_POSITION":null,"TEMP_DESC":"\u0026lt;div\u0026gt;Overview:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;A matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. 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With the first term (a), common difference (d) and (n) number of terms, we can find the last term using the formula, an = a+(n-1)d. Arithmetic mean between two numbers is a+b\/2 and if the sequence is a, A1, A2, A3, A4, b then A1, A2, A3, A4 are four arithmetic mean between a and b.\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to:\u003C\/div\u003E \r\n\u003Cdiv\u003E- Assess arithmetic sequence and series.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Apply a formula for arithmetic sequence and series.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Describe the concept of arithmetic sequence and series.\u003C\/div\u003E","CONT_DESC_AR":"An arithmetic sequence is a sequence in which the successive terms have common differences.\u0026lt;br \/\u0026gt;\nWith the first term (a), common difference (d) and (n) number of terms, we can find the last term using the formula, an = a+(n-1)d. Arithmetic mean between two numbers is a+b\/2 and if the sequence is a, A1, A2, A3, A4, b then A1, A2, A3, A4 are four arithmetic mean between a and b.\u0026lt;br \/\u0026gt;\n\u0026lt;strong\u0026gt;\u0026amp;nbsp;\u0026lt;br \/\u0026gt;\nLearning Objectives\u0026lt;\/strong\u0026gt;\u0026lt;br \/\u0026gt;\n\u0026lt;br \/\u0026gt;\nIn this simulation, you will be able to:\u0026lt;br \/\u0026gt;\n\u0026amp;bull; assess arithmetic sequence and series\u0026lt;br \/\u0026gt;\n\u0026amp;bull; apply a formula for arithmetic sequence and series\u0026lt;br \/\u0026gt;\n\u0026amp;bull; describe the concept of arithmetic sequence and series\u0026lt;br \/\u0026gt;\n\u0026amp;nbsp;","BACKING_FILE":"ss300002.apk","FILE_UID":null,"SCORM_COURSE_ID":null,"CONT_SRC":"","MOD_FILES":null,"FOLDER_NAME":null,"CONTTYPE_ID":"9","ANDROID_PKG":"com.umety.vr.ss300002","TOPIC_ID":"ss300002","IS_PUBLISH":"Y","IS_PUBLIC":"Y","CONT_PRICE":null,"PUBLIC_IMG":"thumb_SS300002.jpg","PUBLIC_BANNER_IMG":"ss300002.jpg","PUBLIC_VIDEO":"pvideo_ss300002.mp4","PUBLIC_VIDEO_URL":"https:\/\/youtu.be\/A8TSAmwj-ac","DIST":null,"SHOW_ON_HOME":"N","CONTROLLER_REQUIRED":"Y","DOMAIN":"3","CONCEPT":"0","STATUS":"A","EXPIRY_DAYS":null,"CREATED_ON":"0000-00-00 00:00:00","CREATED_BY":"1","UPDATED_ON":"0000-00-00 00:00:00","UPDATED_BY":"2","CONT_ORDER":"0","X_ROTATION":null,"Y_ROTATION":null,"Z_ROTATION":null,"BG_COLOR":"0x000000","X_POSITION":null,"Y_POSITION":null,"Z_POSITION":null,"TEMP_DESC":"\u0026lt;div\u0026gt;Overview:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;An arithmetic sequence is a sequence in which the successive terms have common differences. With the first term (a), common difference (d) and (n) number of terms, we can find the last term using the formula, a\u0026lt;span style=\u0026quot;font-size: 9.75px; line-height: 0; position: relative; vertical-align: baseline; bottom: -0.25em; color: rgb(0, 77, 64); font-family: \u0026amp;quot;Open Sans\u0026amp;quot;, \u0026amp;quot;Lucida Sans\u0026amp;quot;, sans-serif; text-align: justify;\u0026quot;\u0026gt;n\u0026lt;\/span\u0026gt;\u0026amp;nbsp;= a+(n-1)d. Arithmetic mean between two numbers is a+b\/2 and if the sequence is a, A1, A2, A3, A4, b then A1, A2, A3, A4 are four arithmetic mean between a and b.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;Learning Objectives:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;br\u0026gt;\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;\u0026lt;div\u0026gt;After completing this module, you will be able to:\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Assess arithmetic sequence and series.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Apply a formula for arithmetic sequence and series.\u0026lt;\/div\u0026gt;\u0026lt;div\u0026gt;- Describe the concept of arithmetic sequence and series.\u0026lt;\/div\u0026gt;\u0026lt;\/div\u0026gt;","IS_ANALYTICS":"Y","VR_ENABLE":"Y","VR_SESSION_ENABLE":"Y","YOUTUBE_URL":null,"CONT_TYPE":"VR Module","CAT_NAME":"\u0645\u0642\u062f\u0645\u0629 \u0644\u0644\u0645\u062a\u062a\u0627\u0628\u0639\u0627\u062a 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Triangle","Base","Perpendicular","Hypotenuse","Pythagoras Theorem"],"contData":{"CONT_ID":"550","CATEGORY_ID":"1","CONT_TITLE":"Use of the Pythagorean Theorem","CONT_SLUG":"use-of-pythagoras-theorem","CONT_TITLE_AR":"","CONT_DESC":"\u003Ch3\u003EOverview:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EThe Pythagorean theorem deals with the lengths of the sides of a right triangle. The theorem states that the sum of the squares of the lengths of the legs of a right triangle is equal to the square of the length of its hypotenuse .\u003C\/div\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Ch3\u003ELearning Objectives:\u003C\/h3\u003E \r\n\u003Cdiv\u003E \r\n \u003Cbr\u003E \r\n\u003C\/div\u003E \r\n\u003Cdiv\u003EAfter completing this module, you will be able to:\u003C\/div\u003E \r\n\u003Cdiv\u003E- Explain the Pythagorean theorem.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Calculate the unknown dimensions of any right triangle.\u003C\/div\u003E \r\n\u003Cdiv\u003E- Apply the Pythagorean theorem in real life situations.\u003C\/div\u003E","CONT_DESC_AR":"","BACKING_FILE":null,"FILE_UID":null,"SCORM_COURSE_ID":null,"CONT_SRC":null,"MOD_FILES":null,"FOLDER_NAME":null,"CONTTYPE_ID":"9","ANDROID_PKG":"com.umety.vr.hs300186","TOPIC_ID":"hs300186","IS_PUBLISH":"Y","IS_PUBLIC":"Y","CONT_PRICE":null,"PUBLIC_IMG":"thumb_HS300186.jpg","PUBLIC_BANNER_IMG":"HS300186.jpg","PUBLIC_VIDEO":"pvideo_hs300186.mp4","PUBLIC_VIDEO_URL":"https:\/\/youtu.be\/ftnG5We0TUc","DIST":null,"SHOW_ON_HOME":"N","CONTROLLER_REQUIRED":"Y","DOMAIN":"3","CONCEPT":"0","STATUS":"A","EXPIRY_DAYS":null,"CREATED_ON":"2017-10-06 05:33:46","CREATED_BY":"0","UPDATED_ON":"2024-10-07 12:26:50","UPDATED_BY":"2","CONT_ORDER":"0","X_ROTATION":null,"Y_ROTATION":null,"Z_ROTATION":null,"BG_COLOR":"0x000000","X_POSITION":null,"Y_POSITION":null,"Z_POSITION":null,"TEMP_DESC":"Overview:\u0026lt;br\u0026gt;\u0026lt;br\u0026gt;The Pythagorean theorem deals with the lengths of the sides of a right triangle. 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