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3 years ago

Please I need this answer very quick . I need help with #14

Mathematics

2 answers:

jeyben [28]3 years ago

5 0

I hope this helps you

DIA [1.3K]3 years ago

3 0

The answer is a not b

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The area of the continent of North America is about 9,540,000square miles. Write 9,540,000 in expanded form using exponents to s

Kobotan [32]

9,000,000+500,000+40,000

8 0

3 years ago

Does anyone know the answer to this?

Klio2033 [76]

Try photo math
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7 0

3 years ago

Help please n ty :P<br> sdhgfahedgewkgfzndb

Ugo [173]

Answer:

$y=-3.2\\\\y=-2.2\\\\y=-1.8$

Step-by-step explanation:

Insert the given x values for the given equation to find the corresponding y coordinate:

$y=0.2x-2$

$x:-6,-1,1$

Insert -6:

$y=0.2(-6)-2$

Simplify multiplication (when multiplying a negative and a positive, the result will always be negative):

$y=-1.2-2$

Simplify (when subtracting from a negative number, add the values, but keep the negative symbol):

$y=-3.2$

Insert -1:

$y=0.2(-1)-2\\\\y=-0.2-2\\\\y=-2.2$

Insert 1:

$y=0.2(1)-2\\\\y=0.2-2\\\\y=-1.8$

:Done

3 0

3 years ago

Find the following integral

ololo11 [35]

There's nothing preventing us from computing one integral at a time:

$\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2$

$\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2$

$\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx$

Expand the integrand completely:

$x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x$

Then

$\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}$

4 0

2 years ago

I need the answers please help 60points

enyata [817]

In 22, you're looking for the vertical height of the triangle. You're given the angle opposite the side you want to find (which I'll call $x$ ) and the length of the hypotenuse. This sets you up with the relation

$\sin2^\circ=\dfrac x{1600}\implies x\approx55.84\text{ m}$

In 23, you're given a similar situation, except now you're looking for the angle (I'll call it $\theta$ ) in the triangle opposite the side denoting the height of the airplane. So this time,

$\sin\theta=\dfrac{1500}{5000}=\dfrac3{10}\implies\theta=\arcsin\dfrac3{10}\approx17.46^\circ$

6 0

3 years ago

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