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A photographer decides to decrease a picture she took in sequence to fit it within a certain frame. She requires the picture to be one-third of the area of the original. If the original picture was 4 inches through 6 inches, how many inches is the smaller dimension of the decreased picture if each dimension changes the same amount?
Let x = the amount of reduction. Then 4 - x = the width of the decreased picture and 6 - x = the length of the decreased picture. Since area is length times width, and one-third of the old area of 24 is 8, the equation for the area of the decreased picture would be (4 - x)(6 - x) = 8. Multiply the binomials by using the distributive property: 24 - 4x - 6x + x2 = 8; combine like terms: 24 - 10x + x2 = 8. Subtract 8 from both sides: 24 - 8 - 10x + x2 = 8 - 8. Simplify and place in standard form: x2 - 10x + 16 = 0. Factor the trinomial into 2 binomials: (x - 2)(x - 8) = 0. Set each factor equal to zero and solve: x - 2 = 0 or x - 8 = 0; x = 2 or x = 8. The solution of 8 is not reasonable because it is greater than the original dimensions of the picture. Accept the solution of x = 2 and the smaller dimension of the reduced picture would be 4 - 2 = 2 inches.
Example of 3-D Coordinate System Example: Graph x = 3 in R, R 2 and R 3 . Solution In R we consist of a single coordinate system and thus x=3 is a point in a 1-D co
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if 1/x+2, 1/x+3, 1/x+5 are in AP find x Ans 1/x+2,1/x+3, 1/x+5 are in AP find x. 1/x+3 - 1/x+2 = 1/x+5-1/x+3 => 1/x 2 +5x+6 = 2/ x 2 +8x +15 => On solving we get x
Expand (1- 1/2x -x^2)^9
how to learn integration?easier
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Let a 0 , a 1 ::: be the series recursively defined by a 0 = 1, and an = 3 + a n-1 for n ≥ 1. (a) Compute a 1 , a 2 , a 3 and a 4 . (b) Compute a formula for an, n ≥ 0.
Combined mean Assume m be the combined mean Assume x 1 be the mean of first sample Assume x 2 be the mean of the second sample Assume n 1 be the size of the 1 st
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