Question

I WILL FAN AND MEDAL YOU!!!
Is anyone REALLY good at algebra? I got a super hard long equation i need help w

Answer 1

I am

Answer 2

I'm ready let'z go!

Answer 3

Okay thanks so much let me write it out :)

Answer 4

ok brb in ten mins, @danieldjpon3 when you done

Answer 5

im redy letz go! h

Answer 6

\[(3^{8} * 2^{-5} * 9^{0})^{-2} *\left( 2^{-2} \right)/3{^{3}} )^{4} * 3^{28}\]

Answer 7

ok

Answer 8

(38 ⋅ 2−5 ⋅ 90)−2 ⋅ 2 to the power of negative 2 over 3 to the power of 3, whole to the power of 4 ⋅ 328 this is it n written form

Answer 9

i have to find the value

Answer 10

the answer or value of what?

Answer 11

the expression

Answer 12

(38*2 - 5*90) - 2 * ( 2^-2 / 3^3 ) - 4* 3^28
( 190 - 450 ) - 2 * (0.25/27) - 9.15^13
-260 * -2 * (0.25/27 ) - 9.15 ^ 13
(-260 * -0.02) - 9.15 ^13
5.2 - 9.15^ 13
-3 . 15^12
I hope that helps!!!

Answer 13

well... @kaybabyluvzyhu16

Answer 14

???

Answer 15

@danieldjpon3 the first 3 is on exponent of 8 not 3 time 8 and these all time 2 on exponent of minus 5 time 9 on exponent of zero and all these inside parentheses on exponent of minus 2
- this is the first part of this exercise inside first parentheses
so these all time 2 on exponent of minus 2 divide 3 on exponent of 3
@KayBabyLuvzYhu16.and after these there is one parentheses and exponent of 4 but i think that you missed the pair of this parenthese so bc. i dont see it nowhere please clarify
and this time 3 on exponent of 28

are these above wrote correct in this way please ?

Answer 16

well the 1st thing you need to know is that

\[a^0 = 1\]

or any base to a power of zero is 1

so that should simplify the 1st set of brackets

next, still looking at the 1st set of brackets apply the index law for power of a power

\[(a^x)^y = a^{x \times y}\]

so you multiply the power of -2

\[(3^8 \times 2^{-5})^{-2} = (3^8)^{-2} \times (2^{-5]})^{-2} \]

in the next part of the equation re-write it as



just simplify the power on the right hand side of the equation.

apply the same rule to the next part of the
\[\frac{2^{-2}}{3^4} = 2^{-2} \times 3^{-4}\]

fractions can be written with negative powers s you now need to use the power of a power rule.... for

\[(2^{-2} \times 3^{-4})^4 \]

then the solution is simply a case of adding the powers of 3 and adding the powers of 2

to help get the final solution