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kumpel [21]
3 years ago
9

(Need Quick Help!)

Mathematics
2 answers:
love history [14]3 years ago
6 0

Answer:

Its A

Step-by-step explanation:

You must distribute exponents outside of parentheses to each

A = -5^3 × 4^3

B = multiplies each by 3 then raises 1 to the third, very off

C = just multiplied each by 3

D = very off, added by 3

just olya [345]3 years ago
5 0
(-5 • 4)^3 = -8000

A. (-5)^3 = -125
4^3 = 64
-125 • 64 = -8000
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7 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=x%20%2B%207%20%20%3C%20%20-%202" id="TexFormula1" title="x + 7 &lt; - 2" alt="x + 7 &lt; -
vladimir1956 [14]
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5 0
3 years ago
Solve the inequality. Check your solutions.
Komok [63]
The inequality is 

7- \frac{2}{b}\ \textless \ \frac{5}{b}

write 7 as 7b/b to have all the expressions in common denominator:

\frac{7b}{b} - \frac{2}{b}\ \textless \ \frac{5}{b}

\frac{7b-2}{b} \ \textless \ \frac{5}{b}

\frac{7b-2}{b}- \frac{5}{b}\ \textless \ 0

\frac{7b-2-5}{b}\ \textless \ 0

\frac{7b-7}{b}\ \textless \ 0

\frac{7(b-1)}{b}\ \textless \ 0

here b=1 is a root and b=0 is not in the domain of the expression, but it still has an effect in the sign of the expression.

the sign table of \frac{7(b-1)}{b} is :

+++++++[0] --------[1] +++++

this means that for values of b to the left of 0 and to the right of 1, the expression is positive, and for values of b in (0, 1), the expression is negative.

that is \frac{7(b-1)}{b}\ \textless \ 0 for b∈(0, 1)

Answer: (0, 1)
8 0
3 years ago
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