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Olegator [25]
2 years ago
14

Bob has five times as much money as John and together they have 60.00 what would this look like as an equation

Mathematics
1 answer:
Vanyuwa [196]2 years ago
6 0
The equation would look like 60 = 5x
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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
1 year ago
There are four steps for converting the equation x2 + y2 + 12x + 2y – 1 = 0 into standard form by completing the square. complet
goblinko [34]
Let's first write each step of the procedure:
 Step 1: 
 group the x terms together and the terms and together, and move the constant term to the other side of the equation:
 x² + 12x + y² + 2y = 1
 Step 2:
 
determine (b ÷ 2) 2 for the x and y terms.
 (12 ÷ 2) 2 = 36
 and
 (2 ÷ 2) 2 = 1
 Step 3:
 add the values to both sides of the equation.
 x2 + 12x + 36 + y2 + 2y + 1 = 1 + 36 + 1
 Step 4:
 write each trinomial to binomial squared, and simplify the right side.
 (x + 6) 2 + (y + 1) 2 = 38
 Answer:
 the last step is:
 (x + 6) 2 + (y + 1) 2 = 38
3 0
1 year ago
A bag of trail mix weighs 1.625 pounds round 1.625 to the nearest hundredths
Andrew [12]
If you round 1.625 pounds to nearest hundredths its 1.63 pounds.
7 0
2 years ago
The planetarium, which is not marked on the map, is halfway between the historic village and the science center. What are its co
lutik1710 [3]
As I have seen this question before:

Coordinates of historic village: (-4 1/2 , 5 1/2)
Coordinates of science center: (-4 1/2 , -1 3/4)

To find the midpoint, we use:
((x₂ + x₁)/2) , (y₂ + y₁)/2) 
= ( (-4 1/2 + -4 1/2)/2 , (-1 3/4 + 5 1/2) / 2)
= ( -4 1/2 , 1 7/8)
6 0
1 year ago
2 A pharmacist has a 13 % alcohol solution and another 18 % alcohol solution. How much of each must he use to make 50 grams of a
nalin [4]
<h2>Therefore he took 40 gram of 1^{st} type solution and 10 gram of 2^{nd} type solution.</h2>

Step-by-step explanation:

Given that , A pharmacist 13% alcohol solution another 18% alcohol solution .

Let he took x gram solution of 1^{st} type solution

and he took (50-x) gram of 2^{nd} type solution.

Total  amount of alcohol =[x\times\frac{13}{100}] +[(50 -x) \times\frac{18}{100} ] gram

Total amount of solution = 50 gram

According to problem

⇔\frac{ [x\times\frac{13}{100}] +[(50 -x) \times\frac{18}{100} ]}{50}= \frac{14}{100}

⇔\frac{13x +900-18x}{100\times50} =\frac{14}{100}

⇔- 5x= 700 - 900

⇔5x = 200

⇔x = 40 gram

Therefore he took 40 gram of 1^{st} type solution and (50 -40)gram = 10 gram of 2^{nd} type solution.

7 0
2 years ago
Read 2 more answers
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