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Tanya [424]
3 years ago
13

Cody wants to watch a new movie that is three hours long.

Mathematics
2 answers:
FrozenT [24]3 years ago
8 0

Answer:

1/2 of an hour

Step-by-step explanation:

the movie is three hours long, I would divide it into 3 parts, each being an hour long. because it says he watched 1/6 of the movie so far you would want to split those 3 parts in half to make 6 parts, each being 30 minutes. so if we took one of those hours, and the time he watched 1/6 of the movie which would be 30 minutes, an hour is 60 minutes, so 30 mintues is half an hour


Ne4ueva [31]3 years ago
3 0

Answer:

1/2 of an hour

Step-by-step explanation:

Well firstly we must find the amount of minutes spent watching. Which is 3 * 60  = 180mins.Then we find 1/6 of 180 = 30We now know Cody has watch 30 minute of the movie.To find what fraction this is of an hour we divide it by minutes in an hour:30/60 which is 1/2Therefore the answer is 1/2 of an hour


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Vince is trying to fit two boxes into the trunk of his car. One box is 1/2 of a foot tall and the other is 2 4/5 feet tall. Stac
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Answer:

3 3/10

Step-by-step explanation:

In order to do this, you need to find the least common multiple. When you do that, you get 5/10 and 2 8/10. Next, what you what to do is turn 2 and 8/10 to an improper fraction,  and when you do that you get 28/10. Add 28/10 and 5/10 to get 33/10. Simplify 33/10, and the correct answer is 3 3/10.

4 0
3 years ago
Read 2 more answers
What is the value of the expression - x [ 6 + ( 4 x 2 ) ] - 5
jonny [76]
1.-x[6+(4*2)]-5, first do the things in the parenthesis: 4*2=8, then +6=14 then *-x=-14x and keep the -5
-14x-5
6 0
3 years ago
A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter
faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.

So the amount of chlorine in the tank changes according to

\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):

\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0

\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0

\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0

\dfrac{c(t)}{(200-3t)^{5/3}}=C

c(t)=C(200-3t)^{5/3}

There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}

\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}

7 0
3 years ago
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garri49 [273]
The answer is (2,889 mi) hope it helps and have a great day!
7 0
3 years ago
Read 2 more answers
What is the approximate perimeter?
vladimir1956 [14]
300^2 + 500^2 = c^2
9000 + 250000=c^2
340000=c^2
sqrt (340000) = c
c = 583
300 + 500 + 583 = 1383
1380 is the perimeter
8 0
3 years ago
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