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QveST [7]
3 years ago
5

Im bored in this quarentine and have a lot of work to catch up on. One of my math problems is "How much water should be added to

20 ounces of a 20% acid solution to reduce it to a 10% solution?" Can you help me?
Mathematics
1 answer:
Kazeer [188]3 years ago
8 0

Answer:

20/3  ounces of water is added.

Step-by-step explanation:

how many ounces of pure water should be added to 20 ounces of 40% solution of muriatic acid to obtain a 30% solution of muriatic acid?

--------------

The acid is 40% of 20 ounces = 8 ounces.

0.4*20 = 8

----------

 

That same 8 ounces is to be 30%. The water will be the other 70%.

8/30 = x/100

30x = 800

x = 80/3, the total amount of the solution.

We started with 20, so 80/3 - 20 = 20/3

20/3 ounces of water is added.

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Answer:

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Step-by-step explanation:

Start of by doing -2-1

4 0
2 years ago
P = 3t² + 7t<br> Work out the value of P when t = -4
Tanzania [10]

Answer:

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Step-by-step explanation:

Hello!

We can solve this by substituting -4 for t.

<h3>Evaluate</h3>
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<u>Remember that any negative number squared is the absolute value of that number squared.</u>

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2 years ago
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7nadin3 [17]

Answer:

C: (3,1)

Step-by-step explanation:

y = -4x² + 24x - 35

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7 0
3 years ago
What is the value of x ??
Zinaida [17]

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Step-by-step explanation:

hope this helped :)

6 0
2 years ago
Read 2 more answers
X/2 is greater than or equal to -4
irga5000 [103]
\frac{x}{2}≥-4

Pretend the greater than or equal to sign is an equal sign, we will substitute it back in later.

x/2 = -4
Multiply both sides by 2 to get x alone
x = -8
Substitute the greater than or equal to sign back in and voila!
x≥-8
3 0
3 years ago
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