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Reika [66]
4 years ago
11

I need help on this problem please help me.

Mathematics
1 answer:
a_sh-v [17]4 years ago
3 0

Answer:

150 mL

Step-by-step explanation:

Let x represent the quantity of 75% solution needed. The total amount of alcohol in the 60% mixture will be ...

75%·x + 10%·45 = 60%·(x+45) . . . . . there will be x+45 mL of solution in the end

0.15x = 22.5 . . . . . . . . . . . . . . . . . . . . simplify, subtract .60x+4.5

22.5/0.15 = x = 150 . . . . . . . . . . . . . . mL of 75% solution needed

You will need 150 mL of the 75% solution.

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25+5 Times blank equals 35
mezya [45]
Let x = blank

25 + 5x = 35

5x = 35 - 25

5x = 10

x = 10/5

x = 2

Done!
6 0
3 years ago
At Bridgeview School,7/10 of the total students are driven to school by their parents. Another 1/6 of the total students at Brid
jarptica [38.1K]

Answer: 2/15

Step-by-step explanation:

We are given the information that at Bridgeview School, 7/10 of the total students are driven to school by their parents and that 1/6 of the total students at Bridgeview School ride the bus.

The fraction of the students that walk to school would be:

= 1 - (7/10 + 1/6)

= 1 - (21/30 + 5/30)

= 1 - 26/30

= 4/30

= 2/15

5 0
3 years ago
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
I don’t understand geometry
Talja [164]

Hope this helps u...!!!

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3 years ago
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Prime numbers are only divisible by 1 or itself; the first few are are 2, 3, 5, 7, and 11.
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