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den301095 [7]
3 years ago
11

Solve for the variable: -15r – 8 = 37 A. -5 B. -3 C. -2 D. 3

Mathematics
2 answers:
aksik [14]3 years ago
8 0
Hello!  I would be happy to help you with your math problem.  For this problem, we plug in the values until we find what fits.  In this case, we have to find what is true, or, in other words, equals 37.

The answer to your question would be B -- -3.  Here is why.

-15*-3-8=37.

Hope this helps!
4vir4ik [10]3 years ago
4 0
Answer:  [B]:  -3 .
______________________________________________________
Explanation:
______________________________________________________
Given:
__________________________________________
  "  <span>-15r – 8 = 37 " ;   Solve for " r "  ;
__________________________________________
</span>     →  Add  "8" to each side of the equation ; 
<span>__________________________________________
     </span>→  -15r – 8 + 8 = 37 + 8 ; 
__________________________________________
to get:
__________________________________________
     →  -15r  =  45 ;
__________________________________________
     → Divide EACH side of the equation by "-15" ; to isolate the variable, "r", on one side of the equation; and to solve for, "r" ; 
___________________________________________
     →  -15r / -15  =  45 / -15 ;
___________________________________________
          →   r  =  - 3  ; which is our answer; 
________________________________________
          →   which is:  Answer choice:  [B]:  "-3" .
______________________________________________________
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What is the distance between the points (−5, −9)(−5, −9) and (−5, 13)(−5, 13) ?
krok68 [10]
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6 0
3 years ago
(I've been trying to figure this out for 3 days and I really need help)
liq [111]

Check the picture below.

since the diameter of the cone is 6", then its radius is half that or 3", so getting the volume of only the cone, not the top.

1)

\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=4 \end{cases}\implies V=\cfrac{\pi (3)^2(4)}{3}\implies V=12\pi \implies V\approx 37.7

2)

now, the top of it, as you notice in the picture, is a semicircle, whose radius is the same as the cone's, 3.

\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=3 \end{cases}\implies V=\cfrac{4\pi (3)^3}{3}\implies V=36\pi \\\\\\ \stackrel{\textit{half of that for a semisphere}}{V=18\pi }\implies V\approx 56.55

3)

well, you'll be serving the cone and top combined, 12π + 18π = 30π or about 94.25 in³.

4)

pretty much the same thing, we get the volume of the cone and its top, add them up.

\bf \stackrel{\textit{cone's volume}}{\cfrac{\pi (3)^2(8)}{3}}~~~~+~~~~\stackrel{\stackrel{\textit{half a sphere}}{\textit{top's volume}}}{\cfrac{4\pi 3^3}{3}\div 2}\implies 24\pi +18\pi \implies 42\pi ~~\approx~~131.95~in^

8 0
3 years ago
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