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g100num [7]
3 years ago
11

What are the inequality solutions for 3 ≤ 3 x − 4 < 2 x + 1

Mathematics
1 answer:
krek1111 [17]3 years ago
3 0

Answer:

7/3 ≤ x < 5.

Step-by-step explanation:

3 ≤ 3 x − 4 < 2 x + 1

Split this up into 2 inequalities:

3x - 4 ≥ 3

3x ≥ 7

x ≥ 7/3.

and

3x - 4 < 2x + 1

3x - 2x < 1 + 4

x < 5.

So it's 7/3 ≤ x < 5.

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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

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;
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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}.

Then

\dfrac{V_{\text{new flask}}}{V_{\text{flask}}} =\dfrac{8}{1}=8.

Answer 2: correct choice is D.


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Step-by-step explanation:

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