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2 years ago

Can someone plz correct me I got that wrong plz I beg u

Mathematics

1 answer:

Tatiana [17]2 years ago

4 0

Answer:

2400

Step-by-step explanation:

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What’s the abbreviation of 78 degrees below Celsius

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NISA [10]

The absolute value equation for the scores within one standard deviation of the mean is:

|x - 20.9| = 5.3.

The minimum score that is within one standard deviation of the mean is of 15.6.

<h3>What is the absolute value function?</h3>

The absolute value function is defined by:

|x| = x, x >= 0.

|x| = x, x < 0.

It measures the distance of a point x to the origin of x = 0.

One standard deviation is of 5.3, while the mean is of 20.9, hence the absolute value equation for the scores within one standard deviation of the mean is:

|x - 20.9| = 5.3.

Which means that the minimum score is:

x - 20.9 = -5.3

x = 15.6.

More can be learned about the absolute value function at brainly.com/question/24734454

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1 year ago

Identify the place value of digit 9 in each given number.

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Step-by-step explanation:

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2 years ago

Evaluate the integral using the indicated trigonometric substitution. (use c for the constant of integration.) x^3 / sqrt x^2 +

slava [35]

$\displaystyle\int\frac{x^3}{\sqrt{x^2+49}}\,\mathrm dx$

Taking $x=7\tan\theta$ gives $\mathrm dx=7\sec^2\theta\,\mathrm d\theta$ , so that the integral becomes

$\displaystyle\int\frac{(7\tan\theta)^3}{\sqrt{(7\tan\theta)^2+49}}(7\sec^2\theta)\,\mathrm d\theta$
$=\displaystyle7^4\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{49\tan^2\theta+49}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\tan^2\theta+1}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\sec^2\theta}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{|\sec\theta|}\,\mathrm d\theta$

When $\sec\theta>0$ , we have

$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sec\theta}\,\mathrm d\theta$
$=\displaystyle7^3\int\tan^3\theta\sec^2\theta\,\mathrm d\theta$

and from here we can substitute $u=\tan\theta$ to proceed from here.

Quick note: When we set $x=7\tan\theta$ , we are implicitly enforcing $-\dfrac\pi2$ just so that the substitution can be undone later via $\theta=\tan^{-1}\dfrac x7$ . But note that over this domain, we automatically guarantee that $\sec\theta>0$ , so the absolute value bars can be dropped immediately.

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2 years ago

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