Please help!!!

Mathematics

1 answer:

marusya05 [52]2 years ago

5 0

Answer:

the answer is B.FALSE

Step-by-step explanation:

hope it helps

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In general. A clock that is 3 seconds slow but keeps consistent time will be more accurate than a clock that loses 1 second ever

Pani-rosa [81]

True.

If the clock loses 1 second every hour that passes then after 4 hours it will be 4 seconds off, 5 hours will be 5 seconds off, etc. It would continue and turn into minutes and hours off.

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3 years ago

Can someone help me with this problem? Thank you!

svlad2 [7]

Step-by-step explanation:

r=d/2=12/2=6

A= pi*r*r=3.14*6*6

A=113.04

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2 years ago

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Which is an example of the associative property?

Mrac [35]

I believe that the answer is C.

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3 years ago

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Assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. Based on this assumption,

kompoz [17]

If the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder, then its volume is

$V_{flask}=V_{sphere}+V_{cylinder}.$

Use following formulas to determine volumes of sphere and cylinder:

$V_{sphere}=\dfrac{4}{3}\pi R^3,\\ \\V_{cylinder}=\pi r^2h,$

wher R is sphere's radius, r - radius of cylinder's base and h - height of cylinder.

Then

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
$V_{cylinder}=\pi r^2h=\pi \cdot \left(\dfrac{1}{2}\right)^2\cdot 3=\dfrac{3\pi}{4}\approx 2.36;$
$V_{flask}=V_{sphere}+V_{cylinder}\approx 47.71+2.36=50.07.$

Answer 1: correct choice is C.

If both the sphere and the cylinder are dilated by a scale factor of 2, then all dimensions of the sphere and the cylinder are dilated by a scale factor of 2. So

R'=2R, r'=2r, h'=2h.

Write the new fask volume:

$V_{\text{new flask}}=V_{\text{new sphere}}+V_{\text{new cylinder}}=\dfrac{4}{3}\pi R'^3+\pi r'^2h'=\dfrac{4}{3}\pi (2R)^3+\pi (2r)^2\cdot 2h=\dfrac{4}{3}\pi 8R^3+\pi \cdot 4r^2\cdot 2h=8\left(\dfrac{4}{3}\pi R^3+\pi r^2h\right)=8V_{flask}.$