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4 years ago

Please help! :) I Really need help!

Mathematics

1 answer:

ivann1987 [24]4 years ago

3 0

$3x^2(2x+3)(x-7)\\\\3x^2(2x^2 +3x-14x-21)\\\\3x^2(2x^2-11x-21)\\\\6x^4-33x^3-63x^2$

The answer is A.

$5(2x+3)^2\\\\5(4x^2+12x+9)\\\\20x^2+60x+45$

The answer is C.

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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,

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

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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

$V_{sphere}=\dfrac{4}{3}\pi R^3=\dfrac{4}{3}\pi \left(\dfrac{4.5}{2}\right)^3=\dfrac{4}{3}\pi \left(\dfrac{9}{4}\right)^3=\dfrac{243\pi}{16}\approx 47.71;$
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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:

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