Polygon Q(16, 4), R(8, 16), S(4, 4) was mapped to polygon T(–2, 5), U(–4, 8), V(–5, 5). First by the dilation: (x, y) → (0.25x, 0.25y). And then by which translation?
A M: (x, y) → (x – 18, y + 1)
B M: (x, y) → (x + 1, y – 18)
C M: (x, y) → (x + 4, y – 6)
D M: (x, y) → (x – 6, y + 4)

D M: (x, y) → (x – 6, y + 4)

Step-by-step explanation:

Polygon Q(16, 4), R(8, 16), S(4, 4) was mapped to polygon T(–2, 5), U(–4, 8), V(–5, 5). First by the dilation: (x, y) → (0.25x, 0.25y).

This means that:

Now we want to find the translation rule that takes

(4,1) to (-2,5)

(2,4) to (-4,8)

(1,1) to (-5,5)

In each case, 6 is subtracted from the first coordinates and 4 added to the second coordinates.

Therefore, the translation rule is:

(x-6,y+4)

Option D is correct.

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation:

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.