Which point is a solution to the inequality shown in this graph? 5
(3,-1)
(-3,-3)
O A. (5,-5)
O B. (1,5)
C. (-3,-3)
D. (3, -1)

We want to see which point is a solution for the graphed inequality.

We will find that the correct option is B, (1, 5)

Notice that the line that defines the inequality contains the points (-3, -3) and (3, -1)

Then the slope of that line is:

Then the line is something like:

y = (1/2)*x + b

To find the value of b, we use the fact that this line passes through the point (3, -1), then we have:

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

-1 - 3/2 + b

-5/2 = b

So the line is:

y = (1/2)*x - 5/2

And we can see that the line is slashed, and the shaded area is above the line, then we have:

y > (1/2)*x - 5/2

Now that we have the inequality, we can just input the values of the points in the inequality and see if this is true.

First, options C and D can be discarded because these points are on the line, and the points on the line are not solutions.

So we only try with A and B.

A) x = 5

   y = -5

then we have:

-5 > (1/2)*5 - 5/2

-5 > 0

Which clearly is false.

B) x = 1

   y = 5

Then we have:

5 > (1/2)*1 - 5/2 = -4/2

5 > -4/2

This is true, then the point (1, 5) is a solution.

Thus the correct option is B.

If you want to learn more, you can read:

link

Hey there!

To find the answer, we just need to see which point falls in this blue, which represents the inequality.

We see that the point (5,-5) is not on the blue.

We see that the point (1,5) is on the blue.

(-3,-3) is on the dotted line but not a solution of the inequality. The dotted line is excluded from the inequality. If it were a bold line, then it would be a solution of the inequality.

(3,-1) is also on the dotted line so it is not a solution.

Therefore, the answer is B. (1,5)

I hope that this helps!

1.677

step-by-step explanation:

i'm not good at explaining stuff but if you need an answer i'm here for you.

< 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)