Answer:
Width of the rectangle = 10 feet
Length of the rectangle = 25 feet
Step-by-step explanation:
Perimeter of a rectangle = 2(length + width)
Let
Width of the rectangle = x feet
Length of the rectangle = 3x - 5 feet
Perimeter of the rectangle = 70 feet
Perimeter of a rectangle = 2(length + width)
70 = 2{x + (3x - 5)}
70 = 2{x + 3x - 5}
70 = 2(4x - 5)
70 = 8x - 10
70 + 10 = 8x
80 = 8x
Divide both sides by 8
x = 80 / 8
= 10
Width of the rectangle = 10 feet
Length of the rectangle = 3(10) - 5
= 30 - 5
= 25 feet
Answer:
3.843
Step-by-step explanation:
(3x1)= 1
(8x1/10= 0.8
(4x1/100)=0.04
(3x1/1000)=0.003
Then, add together to get 3.843.
We know that
and this is the only point when sin and cos are equal lengths. Because both 
Now if the sin of 30° is a half that would mean that cos of 60° is also a half.
Hope this helps.
<u>r3t40</u>
You start by cross multiplying:
9 x 8 = 72
This is considered the fraction on the left's value.
You do the same for the other side:
10 x 7 = 70
This is considered the value of the fraction 7/8
72 has a greater value than 70.
Therefore, 9/10 is larger.
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

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 d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5B%5Cdfrac%7Bc%28t%29%7D%7B%28200-3t%29%5E%7B5%2F3%7D%7D%5Cright%5D%3D0)


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

![\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}](https://tex.z-dn.net/?f=%5Cimplies%5Cboxed%7Bc%28t%29%3D%5Cdfrac1%7B200%7D%5Csqrt%5B3%5D%7B%5Cdfrac%7B%28200-3t%29%5E5%7D5%7D%7D)