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3 years ago

How do you round 18.386 to the nearest hundredth

1 answer:

4 0

Your answer would be 18.39

And here's why:

Behind the decimal your numbers are in the following places.

The (3) is in the tenths place.
The (8) is in the hundredths place.
And the (6) is in the thousandths place.

Whatever number is AFTER the place your rounding, is going to help you out.

If the number is 1-4, you round down.
If the number is 5-9, you round up.

Hope this helps! (:

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A business consultant earns a flat fee for his work as well as an hourly fee. He charges his clients at a rate of $75 per hour.

gregori [183]

The answer would be D. y=75x+300 because if you multiply 75 and 20 you would get 1500 and subtract 1800 for how much was paid and 1500 for the result of multiplying 75 and 20
Answer is D. y=75x+300

5 0

3 years ago

Two balls are drawn in turn without replacement from a box containing three green and two white balls.Find the probability of dr

AURORKA [14]

Answer:

A. 2/5

B.2/5

C. 3/5

D. 5/5=1

7 0

2 years ago

The perimeters of the triangles shown below are equal

Naya [18.7K]

Answer:

x=19

Step-by-step explanation:

The perimeter of the right triangle is

P=9+12+15 =36

The perimeter of the left triangle is

P =x-7+ x-7+ x-7 = 3x-21

Set them equal

36 = 3x-21

Add 21 to each side

36+21 = 3x-21+21

57 = 3x

Divide each side by 3

57/3 = 3x/3

19 =x

6 0

2 years ago

Read 2 more answers

Write 305,501 in expanded form

WINSTONCH [101]

That would be

300,000 + 5,000 + 500 + 1

4 0

2 years ago

Find the surface area of the surface given by the portion of the paraboloid z=3x2+3y2 that lies inside the cylinder x2+y2=4. (hi

natta225 [31]

Parameterize the part of the paraboloid within the cylinder - I'll call it $S$ - by

$\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\left\langle u\cos v,u\sin v,3u^2\right\rangle$

with $0\le u\le2$ and $0\le v\le2\pi$ . The region's area is given by the surface integral

$\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=2}\int_{v=0}^{v=2\pi}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv$
$=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du\,\mathrm dv$
$=\displaystyle2\pi\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du$

Take $w=1+36u^2$ so that $\mathrm dw=72u\,\mathrm du$ , and the integral becomes

$=\displaystyle\frac{2\pi}{72}\int_{w=1}^{w=145}\sqrt w\,\mathrm dw$
$=\displaystyle\frac\pi{36}\frac23w^{3/2}\bigg|_{w=1}^{w=145}$
$=\dfrac\pi{54}(145^{3/2}-1)\approx101.522$

7 0

2 years ago