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3 years ago

10

Are the equations 3x - -9 and 4x - -12 equivalent? Explain. Please Help!!

1 answer:

konstantin123 [22]3 years ago

7 0

No because when you do 3*--9 = 27 and 4*--12 =48

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I will give the BRAINIEST <br><br> Find the values of x and y in the diagram below

Brrunno [24]

:=4 Y=3

-Lanaya- AT your help- Happy to help!

8 0

3 years ago

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✓ 1 N 3 4 5 6 Conv A rectangular field is 300 meters long and 250 meters wide. What is the area of the field in square kilometer

garri49 [273]

Answer:

75 km

Step-by-step explanation:

Area = Length × Width

= 300m and 250m

= 75000m

M to Km => ÷1000

= 75000m ÷ 1000

= 75 km #

8 0

3 years ago

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

4 0

3 years ago

supervisor records the repair cost for 25 randomly selected dryers. A sample mean of $93.36 and standard deviation of $19.95 are

Simora [160]

<h2><u>Answer with explanation:</u></h2>

The confidence interval for population mean (when population standard deviation is unknown) is given by :-

$\overline{x}-t^*\dfrac{s}{\sqrt{n}}< \mu$

, where n= sample size

$\overline{x}$ = Sample mean

s= sample size

t* = Critical value.

Given : n= 25

Degree of freedom : $df=n-1=24$

$\overline{x}= \$93.36$

$s=\ $19.95$

Significance level for 98% confidence interval : $\alpha=1-0.98=0.02$

Using t-distribution table ,

Two-tailed critical value for 98% confidence interval :

$t^*=t_{\alpha/2,\ df}=t_{0.01,\ 24}=2.4922$

⇒ The critical value that should be used in constructing the confidence interval = 2.4922

Then, the 95% confidence interval would be :-

$93.36-(2.4922)\dfrac{19.95}{\sqrt{25}}< \mu$

$=93.36-9.943878< \mu$

$=93.36-9.943878< \mu$

$=83.416122< \mu$

Hence, the 98% confidence interval for the mean repair cost for the dryers. = $83.4161$

4 0

3 years ago

a science fair poster is a rectangle 48 inches long and 24 inches wide what is the area of the poster in square feet

anzhelika [568]

Answer:

Area=B×H=48×24=1,152 inches^2

8 0

4 years ago

Read 2 more answers