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Margarita [4]
3 years ago
7

A percent is a way of a expressing a part out of 100. Remember learning about ratios and how they compare quantities? A part-to-

whole ratio is a ratio that compares a part of a whole to another part of a whole. A percent is a part-to-whole ratio where the whole is 100. 40%, or 40 out of 100, is a ratio that compares 40, the part, to 100, the whole.
Which of these could you express as a percent?

A
how many of the 100 yogurts in the refrigerator are strawberry

B
the average size of the 100 yogurts in the refrigerator

C
the maximum number of yogurts the refrigerator can hold

D
how many grams of sugar the 100 yogurts in the refrigerator have altogether
Mathematics
1 answer:
neonofarm [45]3 years ago
8 0

Answer:A

Step-by-step explanation:

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Which is a factor of the polynomial f(x) = 6x4 – 21x3 – 4x2 24x – 35? 2x – 7 2x 7 3x – 7 3x 7
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A factor of the polynomial f(x) = 6x⁴ - 21x³ - 4x² + 24x - 35 is 2x - 7

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Given the polynomial:

f(x) = 6x⁴ - 21x³ - 4x² + 24x - 35

The polynomial can be further simplified to:

f(x) = (2x - 7)(3x³ - 2x + 5)

A factor of the polynomial f(x) = 6x⁴ - 21x³ - 4x² + 24x - 35 is 2x - 7

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Serial numbers for a product are to be made using 2 letters followed by 2 digits. The letters are to be taken from the first 6 l
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Point P partitions the directed line segment from A to B into the ratio 3:4. Will P be closer to A or B? Why?
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2 years ago
Calculate the limit values:
Nataliya [291]
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
5 0
3 years ago
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