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arlik [135]
3 years ago
11

What is the surface area of a conical grain storage tank that has a height of 43 meters and a diameter of 14 meters? Round the a

nswer to the nearest square meter

Mathematics
2 answers:
Vinil7 [7]3 years ago
8 0

Answer:

1112 m^{3}

Step-by-step explanation:

The formula for the surface area of a conical grain storage is:

S= \pi *r\sqrt{r^{2}+ h^{2} } +\pi *r^{2}

Now you just have to put the values inside the formula, knowin that the radius is half the diameter the radius would be:

\frac{d}{2}=\frac{14}{2}=7

ANd the height is 43.

So the formula would be:

S= \pi *7\sqrt{7^{2}+ 43^{2} } +\pi *7^{2}

S=1112m^{3}

8_murik_8 [283]3 years ago
4 0
Check the picture below

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LMNO is a parallelogram. If NM = x + 33 and OL = 4x + 9, find the value of x and then find NM and OL.
anyanavicka [17]
Because these are on opposite sides of the parallelogram, x + 33 = 4x + 9.

This gives you x = 8; OL = 41, NM = 41. 
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If two sides of a triangle are 9cm and 15cm in length, which COULD be the measure of the third side?
MArishka [77]
The lemght could be 15 because a triangle has two long sides and one short one and the two are equivalent
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URGENT!!LAST QUESTION!!!! PLS JUST TAKE A LOOK! EASY BUT I AM DUMB!!!!!! WILL GIVE BRANLIEST!!
Luden [163]

Answer:

x = 28

Step-by-step explanation:

If the quadrilaterals are similar, there is a proportionality among their sides:

The top side in the large figure (70) is to the top side in the small figure (10) in the same ratio as the left side (x) in the large figure is to the left side in the small figure (4). This in math terms is written as:

\frac{70}{10} =\frac{x}{4}

We can then solve for the unknown "x" by multiplying both sides by 4:

\frac{70}{10} =\frac{x}{4}\\\frac{70\,*\,4}{10} =x\\x=28

4 0
3 years ago
What is the answer to -5(-3w-8)=21 for w
Sauron [17]

Answer:

w = -1.266

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8 0
3 years ago
Does anybody know how to do time with exponential decay
My name is Ann [436]
<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

and so the amount of air remaining after 50 breaths is

b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}

which is a very small number close to zero.</span>
5 0
3 years ago
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