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Nimfa-mama [501]
3 years ago
11

What is -7m-5(m+5)=-85

Mathematics
2 answers:
PSYCHO15rus [73]3 years ago
8 0

Answer:

m= 5

Step-by-step explanation:

-7m-5(m+5)= -85

-7m-5m-25= -85

-12m-25= -85

-12m= -60

/-12     /-12

m= 5

Pie3 years ago
8 0

Answer:

m=5

Step-by-step explanation:

−7m−5(m+5)=−85

−7m+(−5)(m)+(−5)(5)=−85

−7m+−5m+−25=−85

combine the like terms: (−7m+−5m)+(−25)=−85

−12m+−25=−85

add 25 to both sides : −12m−25+25=−85+25

divide both sides by -12

−12m /−12  =  

−60 /−12

m=5

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<h2>The perimeter of the cross section is 30 centimeters.</h2>

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Read 2 more answers
Evaluate the following limit:
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If we evaluate the function at infinity, we can immediately see that:

        \large\displaystyle\text{$\begin{gathered}\sf \bf{\displaystyle L = \lim_{x \to \infty}{\frac{(x^2 + 1)^2 - 3x^2 + 3}{x^3 - 5}} = \frac{\infty}{\infty}} \end{gathered}$}

Therefore, we must perform an algebraic manipulation in order to get rid of the indeterminacy.

We can solve this limit in two ways.

<h3>Way 1:</h3>

By comparison of infinities:

We first expand the binomial squared, so we get

                         \large\displaystyle\text{$\begin{gathered}\sf \displaystyle L = \lim_{x \to \infty}{\frac{x^4 - x^2 + 4}{x^3 - 5}} = \infty \end{gathered}$}

Note that in the numerator we get x⁴ while in the denominator we get x³ as the highest degree terms. Therefore, the degree of the numerator is greater and the limit will be \infty. Recall that when the degree of the numerator is greater, then the limit is \infty if the terms of greater degree have the same sign.

<h3>Way 2</h3>

Dividing numerator and denominator by the term of highest degree:

                            \large\displaystyle\text{$\begin{gathered}\sf L  = \lim_{x \to \infty}\frac{x^{4}-x^{2} +4  }{x^{3}-5  }  \end{gathered}$}\\

                                \ \  = \lim_{x \to \infty\frac{\frac{x^{4}  }{x^{4} }-\frac{x^{2} }{x^{4}}+\frac{4}{x^{4} }    }{\frac{x^{3} }{x^{4}}-\frac{5}{x^{4}}   }  }

                                \large\displaystyle\text{$\begin{gathered}\sf \bf{=\lim_{x \to \infty}\frac{1-\frac{1}{x^{2} } +\frac{4}{x^{4} }  }{\frac{1}{x}-\frac{5}{x^{4} }  }  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{0}=\infty } \end{gathered}$}

Note that, in general, 1/0 is an indeterminate form. However, we are computing a limit when x →∞, and both the numerator and denominator are positive as x grows, so we can conclude that the limit will be ∞.

5 0
2 years ago
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