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allsm [11]
2 years ago
12

How much water should we add to 20 kg of seawater to change its concentration from 3% to 2%?

Mathematics
1 answer:
zhannawk [14.2K]2 years ago
6 0

Answer:

10 kg

Step-by-step explanation:

In 3% concentration of 20 kg of seawater, there will be \frac{20 \times 3}{100} = 0.6 kg of salt.

So, if it is a 2% solution of seawater,

Then 2 kg salt will be there in 100 kg solution.

So, 0.6 kg salt will be there in \frac{100 \times 0.6}{2}= 30 kg of solution.

So, to make a solution from 3% to 2% we have to add (30 - 20) = 10 kg of water with a 20 kg solution. (Answer)

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Mrs B spent 99.12 altogether

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If C=5/9(F-32), what is F when C=20?
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5 0
3 years ago
What are the solutions to the following system?
bija089 [108]

The solutions are (\sqrt{2},-1) \text { and }(-\sqrt{2},-1).

Solution:

Given equation:

-2 x^{2}+y=-5

Add 2 x^{2} on both sides.

-2 x^{2}+y+2 x^{2}=-5+2 x^{2}

y=-5+2 x^{2} -------- (1)

y=-3 x^{2}+5 (given)  -------- (2)

Equate (1) and (2).

-5+2 x^{2}=-3 x^{2}+5

Add 3 x^{2} on both sides.

-5+2 x^{2}+3 x^{2}=-3 x^{2}+5 +3 x^{2}

-5+5x^{2}=5

Add 5 on both sides.

-5+5x^{2}+5=5+5

5x^{2}=10

Divide by 5 on both sides, we get

x^{2}=2

Taking square root on both sides, we get

x=\sqrt{2}, x=-\sqrt{2}

Substitute x=\sqrt{2} in (1).

y=-5+2(\sqrt{2})^2

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Substitute x=-\sqrt{2} in (1).

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Therefore the solutions are (\sqrt{2},-1) \text { and }(-\sqrt{2},-1).

Option C is the correct answer.

6 0
3 years ago
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