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

answer: i think it's a

step-by-step explanation:

walter needs to make at least 5 more cookies but no more than 40. 5 ≤ x ≤ 40

step-by-step explanation:

20 ≤ x + 15 ≤ 55

subtract 15 from all parts

20-15 ≤ x + 15-15 ≤ 55-15

5 ≤ x ≤ 40

this means how many more cookies he has to make

walter needs to make at least 5 more cookies, but no more than 40 cookies; 5 ≤ x ≤ 40