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SAT MATH PREPERATION

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P roblems Related to SAT Geometry Algebra Precalculus Additional problems in the blog

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Problems 1. Question: In an isosceles triangle ABC, AM & CM are angle bisectors of angles BAC and BCA respectively. What is the measure of angle AMC? (A)110’ (B)115’ (C)120’ (D)125’ (E)130’

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Solution: Isosceles triangle means that any 2 sides are equal Here we consider AB and CB to be equal Therefore angles BAC and BCA are equal and the value is (180-40)/2 = 70’ We are given that AM and CM are angle bisectors. Angles MAC and MCA = 35’ Finally angle AMC is 180-(2*35) = 110’

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2. Question: In a circle a square is inscribed. What is the degree measure of arc ST? (A)45’ (B)60’ (C)90’ (D)120’ (E)180’

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Solution: OS and OT are angle bisectors Angles of sides of a square are 90’ Therefore the angle bisectors cut the square at 45’ each. The angle of arc ST is 180-(2*45)=90’

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3.5/x = (5 + a)/(x + a) ; If a not equal to 0, find the value of x? (A)-5 (B)-1 (C)1 (D)2 (E)5 Solution: Multiply x on both sides to get 5 = (5 + a)*x/(x + a) Multiply (x + a) on both sides to get 5(x + a) = x*(5 + a) Simplify to get 5x + 5a = 5x + x*a 5a – x*a = 0

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a(5 - x) = 0 Given that a is not zero. Therefore (5 - x)=0 This implies x = 5

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4.If y = 2x + 3 and x < 2 which of the following represents all possible values of y? (A)y 7 (C)y 5 (E)5

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5.Let the function f(x) = 5x – 2a, where a is a constant. If f(10) + f(5) = 55, what is the value of a? (A)-5 (B)0 (C)5 (D)10 (E)20 Solution: f(10) = 5*10 – 2a = 50 – 2a f(5) = 5*5 – 2a = 25 – 2a Substitute in the above results in the Given function f(10) + f(5) = 55 50 – 2a + 25 – 2a = 55 -4a = -20 a = 5

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6.If f(x) = 2x and g(x) = x + 2. Find the answer for fog(x)? (A)2x + 4 (B)x + 2 (C)2x + 2 (D)3x + 2 (E)x + 2 Solution: The above problem is a function within function fog(x) means f(g(x)). This means we substitute the value of g(x) in place of the x term in the f(x) equation We get f(x + 2) = 2(x + 2) = 2x + 4

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Extra Problems in the blog The following problems have the solutions also being provided

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1.Given that x + k = 6 and p(x + k) = 36. Find the value of p? (A)8 (B)6 (C)-8 (D)-6 (E)-2 Solution: Given that (x + k) = 6 Substitute this value in the second equation p * 6 = 36 Therefore p = 6

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2. y = h/x ; h is a constant. Initially the value of y is 3 and x is 4. Find the value of y when x is 6? (A)-2 (B)5 (C)-1 (D)1 (E) 2 Solution: We find the value of h by substituting values of y = 3 and x = 4 in the main equation We get h = 4 * 3 = 12 Now we find y = h/x = 12/6 = 2

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3. A rectangle is 4 times as long as it is wide. If the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches. What were the dimensions of the original rectangle? Solution: Draw a rectangle with the required dimensions

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Let x = original width of rectangle Area of rectangle = Length * Width Plug in the values from the question and from the sketch 60 = (4x + 4)(x –1) Solving we get 4x 2 – 4 – 60 = 0 (2x – 8)(2x + 8) = 0 x= 4 or x = -4 We consider only the positive value. So the width of the original rectangle is 4 and the length is 16

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4. Find the domain and range for function f(x) = x 2 + 2. Solution: The function f(x) = x 2 + 2 is defined for all real values of x. Therefore the domain is all real values of x Since x 2 is never negative, x 2 + 2 is never less than 2. Hence, the range of f(x) is "all real numbers f(x) ≥ 2".

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5. Find the domain and range for the function Solution: g(s) is not defined for real numbers greater than 3 which would result in imaginary no. Hence, the domain for g(s) is "all real numbers, s ≤ 3". Also >= 0. The range of g(s) is "all real numbers g(s) ≥ 0"

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6. Graph the function y = x − x 2 Solution: Determine the y-values for a typical set of x- values and write them in a table. Plot the values on a x-y plane. For a better graph have more number of points

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7. Graph the function Solution: Note: y is not defined for values of x < -1 We determine x and its corresponding y- values and write them in a table Draw the graph

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8. Consider the function Solution: Factoring the denominator gives We observe that the function is not defined for x = 0 and x = 1. Here is the graph of the function.

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We say the function is discontinuous when x = 0 and x = 1.

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