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djverab [1.8K]
4 years ago
6

PLEASE HELP ME ASAP!!

Mathematics
1 answer:
solong [7]4 years ago
5 0
We have to find midpoint M of the diagonal AC (or BD, there is no difference) so:

M=\left(\dfrac{x_A+x_C}{2},\dfrac{y_A+y_C}{2}\right)=\left(\dfrac{-2+4}{2},\dfrac{4+(-2)}{2}\right)=\left(\dfrac{2}{2},\dfrac{2}{2}\right)=\\\\\\=\boxed{(1,1)}
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4 years ago
I cant figure this one out at all
eduard

Answer:

a. 2^(x-2) = g^(-1)(x)

b. A, B, D

Step-by-step explanation:

the phrasing attached in the image is flagged as inappropriate, so i will be replacing it with g(x) and its inverse with g^(-1)(x)

1. replace g(x) with y and solve for x

y = log₂(x) + 2

subtract 2 from both sides to isolate the x and its log

y - 2 = log₂(x)

this text is replaced by the second image -- it was marked as inappropriate

thus, 2^(y-2) = x

replace x with g^(-1)(x) and y with x

2^(x-2) = g^(-1)(x)

2. plug this in to points A, B, C, D, E, and F

A: (2,1)

plug 2 in for x

2^(2-2) = 2⁰ = 1 so this works

B: (4, 4)

2^(4-2) = 2²= 4 so this works

C: (9, 3)

2^(9-2) = 2⁷ = 128 ≠ 3 so this doesn't work

(5, 8)

2^(5-2) = 2³ = 8 so this works

E: (3, 5)

2^(3-2) = 2¹ = 2 ≠ 5 so this doesn't work

F: (8, 5)

2^(8-2) = 2⁶ = 64 ≠ 5 so this doesn't work

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