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DENIUS [597]
3 years ago
10

A.f(x)=-3/4x+10 B.f(x)=3/4x-10 C.f(x)=4/3x+10 D.f(x)=-4/3x-10

Mathematics
1 answer:
zvonat [6]3 years ago
7 0

Answer:

Please make you photo clear

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In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank
fredd [130]

Answer:

1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) Amount \ of \, water \ remaining \ in \, the \ tank \ is \  \frac{x^3(6-\pi) }{6}

Step-by-step explanation:

1) Here we have;

First tank A

Volume of tank = x³

The  volume of the sphere = \frac{4}{3} \pi r^3

However, the diameter of the sphere = x therefore;

r = x/2 and the volume of the sphere is thus;

volume of the sphere = \frac{4}{3} \pi \frac{x^3}{8}= \frac{1}{6} \pi x^3

For tank B

Volume of tank = x³

The  volume of the spheres = 8 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 2·D = x therefore;

r = x/4 and the volume of the sphere is thus;

volume of the spheres = 8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}

For tank C

Volume of tank = x³

The  volume of the spheres = 64 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 4·D = x therefore;

r = x/8 and the volume of the sphere is thus;

volume of the spheres = 64 \times \frac{4}{3} \pi (\frac{x}{8})^3= \frac{x^3 \times \pi }{6}

Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) For the 4th tank, we have;

number of spheres on side of the tank, n is given thus;

n³ = 512

∴ n = ∛512 = 8

Hence we have;

Volume of tank = x³

The  volume of the spheres = 512 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 8·D = x therefore;

r = x/16 and the volume of the sphere is thus;

volume of the spheres = 512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}

Amount of water remaining in the tank is given by the following expression;

Amount of water remaining in the tank = Volume of tank - volume of spheres

Amount of water remaining in the tank = x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}

Amount \ of \ water \, remaining \, in \, the \ tank =  \frac{x^3(6-\pi) }{6}.

5 0
3 years ago
4x-5y=20 find the slope
Irina18 [472]
4x-5y=20 \\
-5y=-4x+20\\
y=\frac{4}{5}x-4

It's \frac{4}{5}
5 0
3 years ago
Read 2 more answers
30 POINTS!
Goshia [24]

Answer:

The correct answare would be 3.54 x 10^22 N  

7 0
2 years ago
Select the domain and range<br><br> Will give 100 points
8090 [49]
The domain is (-3,0) U (2,4)
The range is (-2,-1) U (1,2)
7 0
3 years ago
I don’t understand what’s it asking me to do!! I have done this problem 12 times and I still don’t understand it. :(
pickupchik [31]
Solve for x. so first u would subtract 25 from 25 and 298. 25 will be crossed out so now u are left with 13x<273. to solve that, divide 13 on both sides. 13 would be crossed or so now u are left with x<13. that means x can equal anything less than 13. idk about the rest, i think u can do it. :)
4 0
3 years ago
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