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sasho [114]
3 years ago
14

Please help algebra 2!!

Mathematics
1 answer:
RUDIKE [14]3 years ago
7 0
Let u = x^2

u^2 + 12u - 64

(u - 4)(u + 16)

Back-substitute for u.

(x^2 - 4)(x^2 + 16)

(x + 2)(x - 2)(x + 4i)(x - 4i)

Answer: Choice a.
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Solve for a and b<br> (6a-3)=(7b-10)
Free_Kalibri [48]

Answer:

for A=\frac{7b-7}{6} and for B=\frac{6a+7}{7}

Step-by-step explanation:

For A

\left(6a-3\right)=\left(7b-10\right)

6a-3+3=7b-10+3

6a=7b-7

\frac{6a}{6}=\frac{7b}{6}-\frac{7}{6}

a=\frac{7b-7}{6}

For B

\left(6a-3\right)=\left(7b-10\right)

7b-10=\left(6a-3\right)

7b-10=6a-3

7b-10+10=6a-3+10

7b=6a+7

\frac{7b}{7}=\frac{6a}{7}+\frac{7}{7}

b=\frac{6a+7}{7}

3 0
2 years ago
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