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3 years ago

Use the net to find the lateral area of the cylinder 9yd 4yd

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

6 0

The formula of a lateral area of a cylinder:

$LA=2\pi rH$

We have r = 9 yd and H = 4 yd. Substitute:

$LA=2\pi(9)(4)=72\pi\ yd^2$

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sweet-ann [11.9K]

Ok, so we know that teh diameters of the 3 tennis balls added up equals the height of the canister

also, since the tennis balls touch the sides, the diameter of the tennis ball=diameter of the base

so what we do is
canister volume-3tennisballvolume=wasted space

normally alot of the equations are using radius so
2.7=diameter
diameter=2r
d/2=r
2.7/2=r
1.35=r

since we are to aprox volume to hundreth. use 3.14=pi

volume of cylinder=hpir^2
volume of sphere (tennis ball)=volume of spere= $\frac{4}{3}pir^{3}$

first find volume of cylinder

h=3 times diameter of 1 tennis ball=3 times 2.7=8.1
r=1.35
vofcylinder=8.1 times 3.14 times 1.35^2
v=25.434 times1.8225
v=46.353465 (keep as many digits a possible for accuracy)

volume of tennis balls=volume of shpere times 3
v= $\frac{4}{3}pir^{3}$ times 3
and we can see that the 4/3 and the 3 cancel to get 4
v= $4pir^{3}$
v= $4pi1.35^{3}$
v=4(3.14)(2.460375)
v=30.90231

so we have
volume of canister=46.353465 in^3
volume of tennis balls=30.90231 in^3
subtract
wasted space=46.353465-30.90231
wasted space=15.451155
round to hundreth
15.45

wasted space=15.45 in^3

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3 years ago

What would happen to the variance of a data set if we multiplied every observation by S?

kipiarov [429]

The variance of a constant is zero. Rule 2. Adding a constant value, c, to a random variable does not change the variance, because the expectation (mean) increases by the same amount. ... Multiplying a random variable by a constant increases the variance by the square of the constant.

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Scorpion4ik [409]

Answer:

Option D is correct.

Step-by-step explanation:

f(x)= 9x^2 -12

Let y = 9x^2 - 12

Switch places of x and y and solve for y

x = 9y^2 - 12

Adding 12 on both sides

x + 12 = 9y^2

Dividing both sides by 9

x+12/9 = y^2

=> Taking square root on both sides

=> $\sqrt{y^2}= \sqrt{\frac{x+12}{9}}\\y=\pm\frac{\sqrt{x+12}}{3}$

Replace y with g(x)

$g(x)=\pm\frac{\sqrt{x+12}}{3}$

Option D is correct.

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