How do you solve (7^-1)^-1 and 5^2 * 5^4 * 5^-3

Step-by-step explanation:

exponents raised to a - power  is an inverse function probably not the right word

       a^-b    =  1/(a^b)     example       4^-2   = 1/(4^2)     =  1/16

so   (7^-1)^-1  is a double inverse which is back to the original number  7

      inverted 7 twice

      (7^-1)^-1     = (1/7)^-1     =  1/(1/7)     =   7     trust me or use your calculator

since the coeffients are the same  just add the powers...  some exponent rule

     5^2 * 5^4 * 5^-3   = 5^(2+4-3)   = 5³

here is it written out

5^2 * 5^4 * 5^-3    = 5²  ×   5^4   ×  5^-3    =   (5×5) × (5×5×5×5)  × (1/5×1/5×1/5)

    5×5 × 5× (5×5×5  × 1/5×1/5×1/5)           (5×5×5  × 1/5×1/5×1/5)  = 1

    5×5×5    = 5³

   so

22

step-by-step explanation:

-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