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3 years ago

Find the surface area of the cylinder with a height of 11cm and 4cm as a base

Mathematics

2 answers:

natali 33 [55]3 years ago

8 0

Answer:

Step-by-step explanation:

I believe you meant "radius of the base is 4 cm."

The vertical sides of this cylinder have the area A = 2πr*h, where r is the radius of the base and h is the height of the cylinder.

Here, that comes out to A = 2π(4 cm)(11 cm) = 88π cm^2.

If the "surface area" is to include the top and bottom as well, add the following to 88π cm²:

2[πr²] = 2(π[4 cm]² = 32π cm²

Area of side alone: 88π cm²

Area of top and bottom: 32π cm²

Total surface area: 88π + 32π cm²

Andru [333]3 years ago

5 0

I hope this helps you

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Mr. Howard is a student teacher at the local junior high. His first‐period class of 40 students averaged 96% on a recent test. H

kodGreya [7K]

Answer:

The combined average is 94%

Step-by-step explanation:

for first period, number of students =40

average =96%

for second period, no of students =20

average =90%

but $average = \frac{sum of scores}{no of students}$

let $S_1$ =sum of students for the first period

$S_2$ =sum of students for the second period

for first period;

$\frac{S_{1} }{40} =96%$

$S_{1}= 96*40=3840$

for second period,

$\frac{S_{2} }{20} =90%$

$S_{1}= 90*20=1800$

therefore the combined average = $\frac{S_1+S_2}{40+20}$

that is $\frac{3840+1800}{40+20}$ $= 94%$

so combined average is 94%

3 0

3 years ago

Read 2 more answers

Pleaseeeee helpppp!!! <br> what is BC ? how many units

balandron [24]

Answer:The answer is 25 units.

Step-by-step explanation:

here, ∡ABC=∡ACB

so, the given triangle is an isosceles triangle.

this means side AB and side AC are equal

so, you can write 4x+4=6x-14

solving this equation, we get x=9

finally, BC=2x+7=2*9+7=25

6 0

3 years ago

Marco has a 7 foot long sidewalk. He wants to divide it into 8 equal-length sections for a game. How long will each section of s

shtirl [24]

40 feet long but i not positive i will returned in one second

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3 years ago

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Suppose customers in a hardware store are willing to buy N(p) boxes of nails at p dollars per box, as given by the following fun

Viktor [21]

Answer:

(a) - 25 boxes per dollar

(b) - 20 boxes per dollar

Step-by-step explanation:

Given that,

Consumer's willing to buy boxes of nails at p dollars per box:

N(p) = 80 - 5p^{2}

(a) Change in price from $2 to $3.

N(2) = 80 - 5(2)^{2}

= 80 - 20

= 60

N(3) = 80 - 5(3)^{2}

= 80 - 45

= 35

Therefore, the average rate of change of demand is

= [N(3) - N(2)] ÷ (3 - 2)

= 35 - 60

= - 25 boxes per dollar.

(b) N(p) = 80 - 5p^{2}

Now, differentiating the above function with respect to p,

N'(p) = -10p

Therefore, the instantaneous rate of change of demand when the price is $2 is calculated as follows:

N'(p) = -10p

N'(2) = -10 × 2

= -20 boxes per dollar

7 0

3 years ago

Please help on this question?

den301095 [7]

Hi there,

θ = 180º + the angle of the right-angled triangle.

For finding the angle we know that the opposite side measures 6 units and the adjacent side measures 8 units. So, the hypotenuse is 10 units.

If we want to find the angle of the right-angled triangle we have to use the following equation.

sin(the angle of the right-angled triangle) = $\frac{6}{10}$

⇒ the angle of the right-angled triangle = $sin^{-1}(\frac{6}{10})$ ≈ 36,87º

So,

θ = 180º + the angle of the right-angled triangle

θ ≈ 180º + 36,87º

θ ≈ 216,87º

sin(θ) = sin(216,87º)

sin(θ) = $\frac{-6}{10}$

sin(θ) = $\frac{-3}{5}$

If you want to do it using properties:

θ = 180º + |the angle of the right-angled triangle|

⇒ sin(θ) = sin(180º + |the angle of the right-angled triangle|)

Using properties:

⇒ sin(θ) = sin(180º)*cos( |the angle of the right-angled triangle|) + cos(180º)*sin(|the angle of the right-angled triangle|)

Sin (180) = 0

⇒ sin(θ) = cos(180º)*sin(|the angle of the right-angled triangle|)

sin(the angle of the right-angled triangle) = - $\frac{6}{10}$

And cos(180º) = -1

⇒ sin(θ) = -1* $\frac{6}{10}$

⇒ sin(θ) = $\frac{-6}{10}$

⇒ sin(θ) = $\frac{-3}{5}$

8 0

3 years ago