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Shtirlitz [24]
2 years ago
14

A 400-milliliter solution of a fabric dye contains 8 ml of hydrogen peroxide. What is the percent concentration of the hydrogen

peroxide?
Mathematics
1 answer:
just olya [345]2 years ago
7 0

Answer:

Okay so all you need to know is the percentage of the hydrogen peroxide.
We know that the total solution is 400 ml and the hydrogen peroxide solution is 8ml.
8/400

Percent means out of a hundred, this can go multiple ways.

<u>Way #1, benchmarking.</u>

So 8/400 simplified is 1/50 or 2/100.
1/50 is basically 2/100.

And percent means out of a hundred so 2/100 is basically 2percent.

<u>Way #2, proportions</u>

We want to know the percentage of hydrogen/ fabric dye

If you put it in numbers int he same fraction and put “x” as the unknown 8ml hydrogen peroxide solution you’d get.

8/400 = x/100
X/100 because percent means out of a hundred.
Now solve3

Cross multiply,

800 = 400x
divide 400 on both sides

2=x

Now plug it back in the proportion.

8/400 = 2/100

If you simplify to check if they are equal then it’s correct

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If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by
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a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by y = 70t-16t^2.

To find : The average velocity for the time period beginning when t = 2 and lasting.  a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :    

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}

(\text{Average velocity})_{0.1\ s}=696\ ft/s

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}

(\text{Average velocity})_{0.01\ s}=7536\ ft/s

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}  

(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}

(\text{Average velocity})_{0.001\ s}=75936\ ft/s

5 0
3 years ago
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