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3 years ago

The value of a car decreases by 5% each year.

1 answer:

6 0

10,000*0.95^2
=9025

0.95 is a multiplyer since 100%-5%=95% then divided by 100 giving you 0.95

You always power by the amount of years

So the format is always
Amount*multiplyer^years

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Miguel and Kat are reading the same book. Miguel reads 5 times as many

n200080 [17]

Answer:

$k=\dfrac{m}{5}$

Step-by-step explanation:

Let Kat reads k pages of the book.

Miguel reads 5 times as many pages of the book as Kat. So,

m = 5k ...(1)

Together they read 54 pages. i.e.

m+k = 54 ...(2)

From equation (1),

$k=\dfrac{m}5{}$

Hence, the equation that can be used to find how many pages Kat read is equal to $\dfrac{m}{5}$ .

6 0

2 years ago

Plz help me this one is a lil tough

Aneli [31]

Answer:

100 degrees

Step-by-step explanation:

It's going through 140 and 40, 180 - 40 = 140

180 cause a flat line is 180 degrees

8 0

3 years ago

Read 2 more answers

Sasha hikes into a canyon. She takes a break at the rest stations every 2/3 of a mile. If she started at sea level and has hiked

butalik [34]

the only right one is a, hope this helps

6 0

3 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

(50 points) NEED ANSWER ASAPPPP:)

lbvjy [14]

5-10(|0.8-3/6|)

5-10(|1/5|)

5-(10)(0.2)

5-2

=3, so C would be your best answer.

5 0

3 years ago

Read 2 more answers