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Murljashka [212]
4 years ago
10

Rewrite without parentheses and simplify. (5x+6v)^ 2?

Mathematics
2 answers:
Tresset [83]4 years ago
4 0

answer: 36v^2 +60vx +25x^2

Hoochie [10]4 years ago
3 0

Expand:

(5x + 6v)² = (5x + 6v)(5x + 6v)

Remember to follow FOIL:

FOIL is the order in which to multiply when multiplying terms with differing variables:

(5x + 6v)(5x + 6v)

First:  (5x)(5x) = 25x^{2}

Outside: (5x)(6v) = 30xv

Inside: (6v)(5x) = 30xv

Last: (6v)(6v) = 36v^{2}

Combine like terms:

25x^{2} + 30xv + 30xv + 36v^{2} \\= 25x^{2} + (30xv + 30xv) + 36v^{2}\\= 25x^{2} + 60xv + 36v^{2}

Remember to order in descending power (if your teacher asks for it):

25x^{2} + 36v^{2} + 60xv

~

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Hi there!  

»»————- ★ ————-««

I believe your answer is:  

Option A

»»————- ★ ————-««  

Here’s why:  

⸻⸻⸻⸻

\boxed{\text{Option A:}}\\\\3^{-5}\\\\\rightarrow \frac{1}{3^5}\\\\\rightarrow  \frac{1}{243}\\\\\ \text{This is \textbf{NOT} an integer.}}

⸻⸻⸻⸻

\boxed{\text{Option B:}}\\\\-3^5\\\\\rightarrow -3 * -3 * -3 * -3 * -3 \\\\\rightarrow \boxed{-243}\\\\\\\text{This \textbf{IS} an integer.}

⸻⸻⸻⸻

\boxed{\text{Option C:}}\\\\\frac{1}{3}^{-5} \\\\\rightarrow \frac{1}{(\frac{1}{3})^5}\\\\\rightarrow \frac{1}{\frac{1}{243} }\\\\\rightarrow \boxed{243}\\\\\\\text{This \textbf{IS} an integer.}

⸻⸻⸻⸻

\text{Option \textbf{A} does not simplify into an integer.}

⸻⸻⸻⸻

»»————- ★ ————-««  

Hope this helps you. I apologize if it’s incorrect.  

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