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Arturiano [62]
3 years ago
5

What is the value of the expression?? PLEASE HELPPPPP

Mathematics
2 answers:
ycow [4]3 years ago
5 0

Answer:

18

Step-by-step explanation:

= -4 x (-5) - 2

= 20 - 2

= 18

Anton [14]3 years ago
3 0

Answer:

18

Step-by-step explanation:

-4x - 2

-4 (-5) - 2

(20)  - 2

   18

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riadik2000 [5.3K]
300 is the answer. Hope I helped :)
5 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
Can someone help with this
xenn [34]

Answer:

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Step-by-step explanation:

3 0
3 years ago
A plant foreman wanted to know how many hours had been spent on a particular project. He asked his supervisors to report the hou
anygoal [31]

Answer:

828.7 hours.

Step-by-step explanation:

To answer a question in which the total time has to be calculated of all the activities, first of all, all the times have to be converted to a single form. There are two types: decimals and fractions. Convert the fractions into decimals.

Department A = 202.50 hours.

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Therefore, the total number of hours spent on this project by all four departments. is the sum of all the above numbers. 202.50 hours + 212.75 hours + 198.25 hours + 215.20 hours = 828.7 hours.

Therefore, the total time is 828.7 hours!!!

3 0
3 years ago
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She will pay $1.50 for the hairbrush and her mum paid $1.50 as well
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