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3 years ago

I need help and how do u get the answer to all of them

Mathematics

1 answer:

Vitek1552 [10]3 years ago

4 0

It says to simplify so number 3 would be 12 $x^{3}$ $y^{4}$

Number 4 is $u^{24}$

Number 5 is $- \frac{7}{ c^{8} }$

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Yesenia is investing money in the stock market, in which she is looking at growth over the first month. She invested $300 and in

mixer [17]

Answer:

1.25 because when in a calculator i add 1.25% i subtract 300 and that is all that is left.

7 0

3 years ago

Read 2 more answers

Does 6/4 = 1 2/4? I need awnsers by tmrw

Gala2k [10]

Answer:

Yes

Step-by-step explanation:

There is definetly a whole number so thats correct

1. subtract 4 from 6 equals 2

2. turn that 4 to a 1 whole and leave what's alone

3. Look at what you did... 1 2/4

4. Take pride

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3 years ago

After a new $28,000 car is driven off the lot, it begins to depreciate at a rate of 18.9% annually.

Rainbow [258]

Answer:

Step-by-step explanation:

We get 0.811 as our depreciating value because we take 18.9% and turn it into a decimal. Then, we subtract that from 1. If we did 0.189 instead, it would depreciate at a rate of 81.1% annually.

5 0

2 years ago

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$

We can simplify this further to

$\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}$

7 0

3 years ago

What do i do here I dont know

Mrrafil [7]

159mi/3hr = 53 mi/hr. Hope this helps

4 0

3 years ago