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KengaRu [80]
1 year ago
6

Which shows the phrase "the product of a number and 3" as a variable expression?

Mathematics
1 answer:
JulijaS [17]1 year ago
3 0

The algebraic expression for the word phrase "the product of a number and 3" as a variable expression is 3z

<h3>How to write an algebraic expression for the word phrase "the product of a number and 3" as a variable expression?</h3>

The word phrase is given as:

"the product of a number and 3"

Represent the number with z

So, the word phrase can be rewritten as:

"the product of a number z and 3"

The product of a number z and 3 is represented as:

z * 3

When evaluated,, the expression becomes

z * 3 = 3z

Hence, the algebraic expression for the word phrase "the product of a number and 3" as a variable expression is 3z

Read more about algebraic expression at

brainly.com/question/4344214

#SPJ1

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kolezko [41]
I believe it is A. Hope this helps.
6 0
3 years ago
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Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.
Aleks [24]

Answer:

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

Step-by-step explanation:

The actual Series is::

\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}

The method we are going to use is comparison method:

According to comparison method, we have:

\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}

Now:

\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty}  \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

\sum_{n=1}^{inf}\frac{1}{n^p}

if p>1 then series converges. In oue case we have:

\sum_{n=1}^{inf}b_n=\frac{1}{n^4}

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means \sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6} also converges.

5 0
3 years ago
Easy question.<br><br> Solve the inequality -5x&lt;40
laiz [17]

\huge\text{Hey there!}

\mathsf{-5x < 40}

\large\text{DIVIDE -5 to BOTH SIDES}

\mathsf{\dfrac{-5x}{-5}

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\rm{KEEP: \dfrac{40}{-5}\ because \ that \ gives \  you \ what \ you \ are\ comparing}\\\\\rm{to \ the \ x-value}

\mathsf{\dfrac{40}{-5}=\boxed{\bf -8}}

\mathsf{x > -8}

\boxed{\boxed{\large\textsf{Answer: \large \bf x }\bf  > \large\textsf{\bf -8}}}\huge\checkmark

\boxed{\large\textsf{IT is an O(P)(E)(N)(E)(D)   circle shaded to the right}}

\large\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

3 0
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Solve each system by substitution <br> Y=2x+7<br> 3x-4y=-13
natali 33 [55]
The correct answer is (-3,1)

4 0
3 years ago
Can you help me with this? thanks made it 20 points!
liraira [26]
I agree with Roaltyjess<span />
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