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emmainna [20.7K]
3 years ago
9

The Rogers family drove a total of 482 miles, starting on

Mathematics
1 answer:
Trava [24]3 years ago
7 0

you add up 138+225 and you get 363 then you subtract 482-363 this is how you set it up:

   4  8  2

- 3  6  3

------------

you have to borrow because 3 does not go into 2 so you change the 8 into a 7 then you 10 to 2 and you get 12 so this is what it looks like:


     4   7   12

-    3   6   3

-------------------

 so you then subtract 12-3=9 then 7-6=1 then 4-3=1

so this is what it looks like

    4     7    12

--- 3    6     3

--------------------

    1     1     9

so on Sunday they rode 119 miles

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A car purchased for $15,000 depreciates under a straight-line method in the amount of $950 each year. Which equation below best
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3 years ago
Read 2 more answers
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 <em>y</em> is a function of <em>x</em> :

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 ≤ <em>x</em> ≤ 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}}

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3 years ago
Complete the comparison: 5+4=?
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3 years ago
A water tank contains 12 1/2 ltres of water. Two-fifth of it was (1 mark) consumed. How much of it was left?
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Answer:

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4 0
3 years ago
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