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3 years ago

Can anybody help me or already know the answer to this ? :)

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

4 0

The bottom one is the answer I’m pretty sure ….hope it helps

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YOULL GET BRAINLEST JUST ANSWER THIS CORRECTLY!!!

trapecia [35]

Answer:

area of circle=π $r^{2}$

452.1=3.14 $r^{2}$

452.1/3.14= $r^{2}$

143.98= $r^{2}$

$\sqrt{143.98}$ =r

11.99=r

now ,

diameter=2r

=2*11.99

=23.98

Step-by-step explanation:

5 0

3 years ago

What is the domain of the following relation?

e-lub [12.9K]

The answer is a, because I need more

6 0

3 years ago

What is the solution to this equation? 3[x] - 1 = 8

Yanka [14]

Answer:

Step-by-step explanation:

The whole goal is to isolate the variable.

3x-1=8

1+3x-1=8+1 first add 1 to both sides of the equation to cancel out the -1

3x=9 now just divide 3 to both sides of the equation because the opposite of multiplication is division. So dividing 3 on both sides cancels out the 3 to isolate the x

3x/3=9/3

x=3

8 0

3 years ago

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:

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