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3 years ago

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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2 years ago

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 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

Complete the comparison: 5+4=?

atroni [7]

It could be
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4 0

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?

SVEN [57.7K]

Answer:

7 1/2

Step-by-step explanation:

A water tank contains 12 1/2 liters

= 25/2

Two fifth of it was consumedd

= 25/2 ×2/5

= 50/10

= 5

12.5-5

= 7.5

= 7 1/2

4 0

3 years ago