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Mathematics

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SIZIF [17.4K]3 years ago

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One way to solve the system is to _____________ a variable.

Ivahew [28]

One way to solve the system is to <u>substitute</u> a variable.

<u>Explanation:</u>

One approach to solve an equation is by substitution of one variable. Right now, a condition for one factor, at that point substitute that arrangement in the other condition, and explain. All value(s) of the variable(s) that fulfills a condition, disparity, arrangement of conditions, or arrangement of imbalances.

The technique for tackling "by substitution" works by settling one of the conditions (you pick which one) for one of the factors (you pick which one), and afterward stopping this go into the other condition, "subbing" for the picked variable and fathoming for the other. At that point you back-explain for the principal variable.

8 0

3 years ago

Read 2 more answers

Irene purchases a refrigerator for 500.00 in lowa where the tax is 6 percent what is the sale tax

irina1246 [14]

$30
To find 6% of 500, convert 6% in to a decimal (0.06) and multiply the decimal times the amount of the purchase. Your final price will be $530 because after finding out the sales tax amount, you have to add it to the amount of purchase.

3 0

3 years ago

A.If a new series is found by differentiating a known series with included endpoints, those endpoints might not be included in t

Aleonysh [2.5K]

Consider the power series

$\displaystyle\sum_{n\ge1}\dfrac{x^n}n$

By the ratio test, this series converges for

$\displaystyle\lim_{n\to\infty}\left|\frac{x^{n+1}}{n+1}\cdot\frac n{x^n}\right|=|x|\lim_{n\to\infty}\frac n{n+1}=|x|$

though we know by the alternating series test that the series converges for $x=-1$ .

So this series converges for $-1\le x$ .

Differentiating the series yields

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n\ge1}\frac{x^n}n=\sum_{n\ge1}x^{n-1}=\sum_{n\ge0}x^n$

which is the geometric series. We know this series converges for $|x|$ , and this time the endpoints are not included.

This example shows that (A) is certainly possible; that is, $x=-1$ is valid in the first series, but not in the differentiated one.

- - -

Now consider the series

$\displaystyle\sum_{n\ge0}\dfrac{x^n}{n!}$

which we know to converge to $e^x$ .

Differentiating, we get

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n\ge1}\frac{x^{n-1}}{(n-1)!}=\sum_{n\ge0}\frac{x^n}{n!}=e^x$

as expected. But both series converge everywhere, so this serves as a counter-example to the claim of B. So B is false.

8 0

4 years ago

0.25x-8=16 what value of x makes the equation true?

prohojiy [21]

0.25x-8=16 add 8 to both sides. you get 0.25x=8. divide by .25 on each side x=32

8 0

4 years ago

Kevin divides 3 pieces of paper into fourths. How many fourths does he have? Draw a picture to support your response.

denis-greek [22]

Answer:

12 fourths

Step-by-step explanation: