Solve for c. a−b+c‾‾‾‾‾‾‾‾‾√=3

A. c=3−a+b

B. c=9−a‾√+b‾√

C. c=3−a‾√+b‾√

D. c=9−a+b

The solution for c in the given equation is . The correct option is the fourth option

From the question,

We are to solve for c in

To do this,

First, we will take the squares of both sides

That is,

Then, we get

Now, we will add (b - a) to both sides of the equation

That is,

Then, we can write that

∴

Hence, the solution for c in the given equation is . The correct option is the fourth option

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-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3