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3 years ago

Plz help me with this

Mathematics

1 answer:

ivann1987 [24]3 years ago

6 0

Answer:

Step-by-step explanation:

You can find the cube root of 2 and multiply it by the 6th root of 5.

cube root(2) = 2^(1/3) = 1.2599

6th root of (5) = 0.7418

Product = 1.2599 * 0.7418 = .9346

You need a key that will do fractional powers. Look around on your calculator for ^ that key y^x or x^y.

Using one of those keys, you would put in 2 ^ 0.33333333 =

If your calculator does not have this key (I recommend you buy one that does. They are less than 10 dollars), then you can use the calculator on windows.

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What is 9911 divided by 62

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The answer is 159.85

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Morgan uses 2 oz of dog shampoo to bathe her dog each week. After 4 wk, 34 oz of shampoo remains.

Flauer [41]

The amount of shampoo required by Morgan each week to bathe her dog = 2 oz
So
The amount of shampoo required by Morgan in 7 days to bathe her dog = 2 oz
The amount of shampoo remaining after 4 weeks = 34 oz
So the amount of shampoo remaining after (4 * 7) days = 34 oz
The amount of shampoo remaining after 28 days = 34 oz
The amount of shampoo that Morgan uses in 28 days = (2/7) * 28 oz
= 2 * 4 oz
= 8 oz
Then
8 oz of shampoo is required by Morgan in = 28 days
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6 0

3 years ago

If -48 equals 6(a) then a is what<br>

Oksi-84 [34.3K]

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>

<em>Expression:</em>

$\frac{-48}{6}=\frac{6}{6}a$

$-8 = a$

Your answer is -8.

I hope this helps!

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3 years ago

What’s the degree 2xy5

frutty [35]

Answer:

Step-by-step explanation:

The degree of the term

$2xy^5$

is the sum of the exponents of the variables, so is 1+5 = 6.

The degree of the term is 6.

6 0

3 years ago

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

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