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denis-greek [22]
3 years ago
13

What is -2 + 7x = 12

Mathematics
1 answer:
8090 [49]3 years ago
5 0

Answer:

easy the answer is

Step-by-step explanation:

-2 + 7x = 12

+2 +2

7x + 14

14 / 7x = 2

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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}.

Then

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

Answer 2: correct choice is D.


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Answer:

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Step-by-step explanation:

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42 inches squared

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HELP DUE IN 15 MINS!<br><br> x =??
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Answer:

x = 32

Step-by-step explanation:

Using the Secant and Tangent Intersection Theorem we can say;

(x + 18) (18) = 30²

18x + 324 = 900

18x = 576

x = 32

Hope this helps!

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Formulating and solving inverse variation functions
andreyandreev [35.5K]

Answer:

See below.

Step-by-step explanation:

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B. The equation is y = 180/x.

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