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Gekata [30.6K]
3 years ago
15

Use the substitution method to solve the system. {m+n=5m−n=3 What is the solution? The solution is (4, 1). There is no solution.

There are an infinite number of solutions. The solution is (2, −1).
Mathematics
2 answers:
katrin2010 [14]3 years ago
8 0

The answer would be

The solution is (4, 1)

mixer [17]3 years ago
4 0
Hint:  \bf \begin{cases}
m+n=5\to &m=\boxed{5-n}
\\ \quad \\ \quad \\

m-n=3\to &\boxed{5-n}-n=3\quad \impliedby \textit{solve for "n"}
\end{cases}
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Which expression is equivalent to sqrt8x^7y^8? assume x>0
Ilia_Sergeevich [38]

Answer:

\sqrt{8x^7y^8= 2x^3y^4 * \sqrt{2x}

Step-by-step explanation:

Given

\sqrt{8x^7y^8

Required

Evaluate

\sqrt{8x^7y^8

Separate each factor with multiplication sign

\sqrt{8*x^7*y^8

Express 8 as 4 * 2 and x^7 as x^6 * x

\sqrt{4 * 2 *x^6 * x*y^8

Split each factor

\sqrt{4} * \sqrt{2} *\sqrt{x^6} * \sqrt{x}*\sqrt{y^8}

Evaluate \sqrt4

2 * \sqrt{2} *\sqrt{x^6} * \sqrt{x}*\sqrt{y^8}

Apply law of indices:

2 * \sqrt{2} * x^{6*\frac{1}{2}} * \sqrt{x} * y^{8*\frac{1}{2}}

2 * \sqrt{2} * x^3 * \sqrt{x} * y^4

Reorder

2* x^3 * y^4 * \sqrt{2}  * \sqrt{x}

2x^3y^4 * \sqrt{2}  * \sqrt{x}

2x^3y^4 * \sqrt{2*x}

2x^3y^4 * \sqrt{2x}

2x^3y^4 \sqrt {2x

Hence:

\sqrt{8x^7y^8= 2x^3y^4 * \sqrt{2x}

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Slav-nsk [51]

Answer:

x = -2

Step-by-step explanation:

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x = -2

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Since it is a terminating decimal, 4.9 is rational?
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Answer:

Yes it is because it's terminating.

Step-by-step explanation:

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