A restaurant offers 6 choices of appetizer, 8 choices of main meal, and 5 choices of dessert. A customer can choose to eat just
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The coordinates for the pre-image are P(1,3), Q(4,4), R(4,1), and S(1, -1).
The transformation is 4 units left, and 4 units down.
That means we must subtract 4 units to x-coordinates, and subtract 4 units from y-coordinates. So, the image has coordinates P'(-3,-1), Q'(0,0), R'(0, -3), and S'(-3, -5).
The image below shows the image and pre-image.
Answer:
According to tangents secant segments theorem,
11) x(16+x) = (x + 6)^2
x (16+x) = (x + 6)^2
16x + x^2 = x^2 + 2(6)(x) + 6^2
16x + x^2 = x^2 + 12x + 36
16x - 12x + x^2 - x^2 = 36
4x = 36
x = 36/4
x = 4
13) x(x + 5) = (x + 2)^2
x^2 + 5x = x^2 + 2x 2(2)(x) + 2^2
x^2 + 5x = x^2 + 4x + 4
x^2 -x^2 + 5x - 4x = 4
x = 4
14) x + 8( x + 8 + 32) = (3x) ^2
x + 8(x + 40) = 9x^2
x(x + 8) + 40(x + 8) = 9x^2
x^2 + 8x + 40x + 320 = 9x^2
x^2 + 48x + 320 =9x^2
48x + 320 = 9x^2 - x^2
48x + 320 = 8x^2
Dividing the whole eq. by 8,
6x = 40 = x^2
0 = x^2 - 6x - 40
0 = x^2 - 10x + 4x - 40
0 = x(x - 10) + 4(x - 10)
x - 10 = 0 OR x + 4 = 0
x = 10 OR x = - 4
length cannot be negative, so,
x = 10
15) (x + 3) ( x + 3 + 15) = (2x) ^2
(x + 3) (x + 18) = 4x^2
x(x + 18) + 3(x + 18) = 4x^2
x^2 + 18x + 3x + 54 = 4x^2
x^2 + 21x + 54 = 4x^2
0 = 4x^2 - x^2 - 21x - 54
0 = 3x^2 - 21x - 54
Dividing the whole eq. by 3,
x^2 - 7x - 18 = 0
x^2 - 7x - 18 = 0
x^2 - 9x + 2x - 18 = 0
x(x - 9) + 2 (x - 9)
x + 2 = 0 Or x - 9 = 0
x = -2 or x = 9
length cannot be negative,so,
x = 9
