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Andre45 [30]
3 years ago
6

We can use algebraic operations to rewrite expressions, including polynomial expressions. When we rewrite polynomial expressions

, the representations look different but still represent the same expression.
Where else have you seen this type of transformation—when something appears different, but its meaning remains the same?
Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
6 0

Similar transformations to the application of algebraic operations to rewrite expressions that form representations that appear different from their parent expressions, can be seen in the following fields;

Geometry; Rigid transformations such as rotation, can transform a rectangle to appear to look like a kite

Physics; The heating of a liquid, can transform its physical state from liquid to gas, however, the gas formed by heating the liquid represent the same material chemically

Learn more about transformations here;

brainly.com/question/7161333

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-5+4=3x-9 <br> My teacher told me it was negative three.. can someone help?
Alexxx [7]

Answer:

x = 8/3

Step-by-step explanation:

(When solving for a variable, always do the opposite of PEMDAS)

-5 + 4 = 3x - 9

-1 = 3x - 9

8 = 3x

8/3 = x

4 0
3 years ago
Which of the following functions has a graph that is a line? f(x) = |x| f(x) = x f(x) = x 2
sattari [20]

Answer:

f(x) = x

Step-by-step explanation:

f(x) = |x| will give you V graph

f(x) = x will give you line graph (linear)

f(x) = x² will give you a curve graph

7 0
3 years ago
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
Can anybody give real answers please? Thanks.
Kazeer [188]
I believe it is 17.0, I hope this helps some
8 0
3 years ago
Simplify 5(1 + 3)2. 400 80 50 40
Bad White [126]

Answer:

400

Step-by-step explanation:

5(1+3)^2

(5+15)^2

20^2

400

8 0
3 years ago
Read 2 more answers
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