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dexar [7]
3 years ago
10

If the radius of a cylinder is 2/3 the height and the height is 6, what is the area of the cylinder

Mathematics
1 answer:
Zolol [24]3 years ago
3 0

Answer:

This is the answer if the question is asking for surface area. If the question is asking for lateral area or base area the formula will be different..

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Find the slope of the line passing through the pair of given points.<br> (5, 7) and (2,7)
gulaghasi [49]

Answer:

m = 0

Step-by-step explanation:

m = \frac{7 - 7}{5 -2} \\m = \frac{0}{3}\\m = 0

8 0
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Doug regularly mows his neighbors lawn
Yuki888 [10]
Okay , what about it?
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10m radius. What is the area of the circle? Round your answer to the nearest tenth.
Evgen [1.6K]

Answer:

D

Step-by-step explanation:

The area of a circle can be found using:

a=πr^2

We know that the radius is 10 meters. Therefore, we can substitute 10 in for r.

a=π*10^2

Evaluate the exponent

a=π*100

Multiply pi and 100

a=314.159265359

Round to the nearest tenth: there is a 5 in the hundreth place, which indicates we round up the 1 in the tenths place to a 2.

a=314.2

The area is 314.2 square meters, or choice D

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3 years ago
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One highlighter has a mass of 11 grams. Another highlighter
sergejj [24]
10,800 milligrams equals 10.8 grams so that the first highlighter has the greater mass
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One leg of a right triangle is 7 inches longer than the smaller leg and the hypotenuse is 8 inches longer than the smaller leg.
Makovka662 [10]
Smaller leg = x
longer leg = y
hypotenuse = z

y=x+7
z=x+8

due to
{z}^{2}  =  {x}^{2}  +  {y}^{2}  \\ (x + 8) ^{2}  =  {x}^{2}  + (x + 7) ^{2}  \\  {x}^{2}  + 16x + 64 =  {x}^{2}  +  {x}^{2}  + 14x + 49 \\  {x}^{2}  + 16x + 64 = 2 {x}^{2}  + 14x + 49 \\  {x}^{2}  - 2 {x}^{2}  + 16x - 14x + 64 - 49 = 0 \\  -  {x}^{2}  + 2x + 15 = 0 \\  {x}^{2}  - 2x - 15 = 0 \:  \: (multiplied \: by \:  - 1) \\ (x - 5)(x + 3) = 0 \\ then \:  \: x = 5 \:  \: or \:  \: x =  - 3
but x cannot be -3
so smaller leg is 5 inches

then longer leg = 5+7 = 12 inches
and hypotenuse = 5+8 = 13 inches
3 0
3 years ago
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