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3 years ago

Can someone explain to me in words how to do this problem 3x+122=22x-11

Mathematics

1 answer:

inn [45]3 years ago

8 0

When you have this type of problem, you need to combine the like-terms and isolate the variable.

3x + 122 = 22x - 11

Add 11 to both sides to get rid of it

3x + 122 + 11 = 22x - 11 + 11 (-11 + 11=0)

3x + 133 = 22x

Then you would bring the 3x to the other side, so subtract 3x from both sides

3x + 133 = 22x

-3x -3x

133 = 22x - 3x

133 = 19x

Then divide both sides by 19 to isolate x

133/19 = 19x/19

133/19 = 7, so x = 7

Hope this helps!!

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2 years ago

83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi

arlik [135]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

Step-by-step explanation:

Information given

$\bar X=32.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom given by:

$df=n-1=8-1=7$

The confidence level is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

$\chi^2_{\alpha/2}=104.139$

$\chi^2_{1- \alpha/2}=62.132$

The confidence interval is given by:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

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3 years ago

71.841 to the nearest tenth

jarptica [38.1K]

Answer:
71.8
Hope this helps!

7 0

2 years ago

Read 2 more answers

Find the vertical and horizontal asymptote if they exist

tresset_1 [31]

Hello!

Vertical asymptotes are determined by setting the denominator of a rational function to zero and then by solving for x.

Horizontal asymptotes are determined by:

1. If the degree of the numerator < degree of denominator, then the line, y = 0 is the horizontal asymptote.

2. If the degree of the numerator = degree of denominator, then y = leading coefficient of numerator / leading coefficient of denominator is the horizontal asymptote.

3. If degree of numerator > degree of denominator, then there is an oblique asymptote, but no horizontal asymptote.

To find the vertical asymptote:

2x² - 10 = 0

2(x² - 5) = 0

(x - √5)(x + √5) = 0

x = √5 and x = -√5

Graphing the equation, we realize that x = -√5 is not a vertical asymptote, so therefore, the only vertical asymptote is x = √5.

To find the horizontal asymptote:

If the degree of the numerator < degree of denominator, then the line, y = 0 is the horizontal asymptote.

Therefore, the horizontal asymptote of this function is y = 0.

Short answer: Vertical asymptote: x = √5 and horizontal asymptote: y = 0

7 0

3 years ago