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riadik2000 [5.3K]
3 years ago
5

5 times the quantity x plus 6

Mathematics
1 answer:
seropon [69]3 years ago
3 0
So your question is what is "5 times the quantity x plus 6"

Ok so 5 times the quantity x plus 6, (the "quantity" means to group the "x + 6" into parenthesis) 

<span>So basically you will get: 5(x + 6) as the answer.

=====End=====

Hope you understand :D

-Sky/Crystal
7th Grade
Age 13.</span>
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1,807 is the answer:)
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3 years ago
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Calculate the limit values:
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A) This particular limit is of the indeterminate form,
\frac{ \infty }{ \infty }
if we plug in infinity directly, though it is not a number just to check.

If a limit is in this form, we apply L'Hopital's Rule.

's
Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_ {x \rightarrow \infty } \frac{( ln(x ^{2} + 1 ) ) '}{x ' }
So we take the derivatives and obtain,

Lim_ {x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ \frac{2x}{x^{2} + 1} }{1}

Still it is of the same indeterminate form, so we apply the rule again,

Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 2 }{2x}

This simplifies to,

Lim_{x \rightarrow \infty } \frac{ ln(x ^{2} + 1 ) }{x} = Lim_{x \rightarrow \infty } \frac{ 1 }{x} = 0

b) This limit is also of the indeterminate form,

\frac{0}{0}
we still apply the L'Hopital's Rule,

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ (tanx)'}{x ' }

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (x) }{1 }

When we plug in zero now we obtain,

Lim_ {x \rightarrow0 }\frac{ tanx}{x} = Lim_ {x \rightarrow0 } \frac{ \sec ^{2} (0) }{1 } = \frac{1}{1} = 1
c) This also in the same indeterminate form

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ ({e}^{2x} - 1 - 2x)'}{( {x}^{2} ) ' }

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (2{e}^{2x} - 2)}{ 2x }

It is still of that indeterminate form so we apply the rule again, to obtain;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = Lim_ {x \rightarrow0 } \frac{ (4{e}^{2x} )}{ 2 }

Now we have remove the discontinuity, we can evaluate the limit now, plugging in zero to obtain;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } = \frac{ (4{e}^{2(0)} )}{ 2 }

This gives us;

Lim_ {x \rightarrow0 }\frac{ {e}^{2x} - 1 - 2x}{ {x}^{2} } =\frac{ (4(1) )}{ 2 }=2

d) Lim_ {x \rightarrow +\infty }\sqrt{x^2+2x}-x

For this kind of question we need to rationalize the radical function, to obtain;

Lim_ {x \rightarrow +\infty }\frac{2x}{\sqrt{x^2+2x}+x}

We now divide both the numerator and denominator by x, to obtain,

Lim_ {x \rightarrow +\infty }\frac{2}{\sqrt{1+\frac{2}{x}}+1}

This simplifies to,

=\frac{2}{\sqrt{1+0}+1}=1
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Answer:

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Given a right triangle with legs a, b and hypotenuse c, solve for a if b= 20 and c= 29
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Answer:

C) 21

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a² = 441                  Take the square root of both sides

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A) 5/12×3=1 1/4 I know this for sure
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