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andreev551 [17]
3 years ago
14

I need help please!!!!!!!!!!!!!!!!!!!!!!!!hjdjdjdjdjsjdjjdjdjdjdjjd

Mathematics
1 answer:
BabaBlast [244]3 years ago
4 0
I think it would be A but I'm not sure ._.
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I need to find surface area of a cylinder. Formula being used: S.A= L x W +
Arturiano [62]

Answer:

90π yd²

Step-by-step explanation:

the surface area of a cylinder is the sum of the lateral area and twice the aera of one end of the cylinder:  π·d·l, where l represents the length of the cylinder.  Here, the lateral surface area is π·6 yd·12 yd, or 72π yd².

The two ends add the following to the total surface area:

2·π·(d/2)², or 2π·d²/4.

Thus, the total surface area of the cyl. is

A = 2π·(6 yd)²/4 + 72π yd², or

A = 18π yd² + 72π yd² = 90π yd²

Note:  Please check your source.  L x W +  2pi ·r ^2 is incorrect.

5 0
3 years ago
Find an equation of the line that passes through the points (-5, -3) and (3, 1)
Mkey [24]

\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{-3})\quad (\stackrel{x_2}{3}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{(-5)}}}\implies \cfrac{1+3}{3+5}\implies \cfrac{4}{8}\implies \cfrac{1}{2}

\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{1}{2}}[x-\stackrel{x_1}{(-5)}]\implies y+3=\cfrac{1}{2}(x+5) \\\\\\ y+3=\cfrac{1}{2}x+\cfrac{5}{2}\implies y=\cfrac{1}{2}x+\cfrac{5}{2}-3\implies y = \cfrac{1}{2}x-\cfrac{1}{2}

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3 years ago
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Hes friends with my dad
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What is it in standard form
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Your answer is correct. A
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Rush! please please help
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