Select all ordered pairs that satisfy the function y= -1/2x + 4 a) (-1, 3&1/2) b) (1, 3&1/2) c) (2&1/2, 3) d) (2, 3)

The correct options are:

Option b) (1, 3 + 1/2)

Option d) (2,3)

Step-by-step explanation:

Suppose that we have a function:

y = f(x)

A ordered pair (a, b) satisfies this function if:

b = f(a).

In this case, the function is:

y = -(1/2)*x + 4

Let's try with each one of the options:

a) (-1, 3 + 1/2)

Replacing these values we get:

3 + 1/2 = -(1/2)*-1 + 4

3 + 1/2 = 1/2 + 4

3 + 1/2 = 4 + 1/2

This is false, then the point (-1, 3 + 1/2) does not satisfy the function.

b) (1, 3 + 1/2)

Replacing these values we get:

3 + 1/2 = -(1/2)*1 + 4

3 + 1/2 = 4 - 1/2 = (3 + 1) - 1/2 = 3 + (1 - 1/2) = 3 + 1/2

3 + 1/2 = 3 + 1/2

This ordered pair satisfies the function.

c) (2&1/2, 3)

Replacing these values we get:

3 = (-1/2)*(2 + 1/2) + 4

3 = -(1/2)*2 - (1/2)*(1/2) + 4

3 = -1 - 1/4 + 4

3 = 3 - 1/4

This is false, then the point (2 + 1/2, 3) does not satisfy the function.

d) (2, 3)

Replacing these values in the function we get:

3 = -(1/2)*2 + 4

3 = - 1 + 4

3 = 3

This is true, then (2, 3) satisfies the function.

what graph

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)