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2 years ago

2.01 as a mixed number

Mathematics

2 answers:

Mazyrski [523]2 years ago

6 0

2.01 = 2 1/10 = 21/10

ASHA 777 [7]2 years ago

3 0

Answer:

2 1/100

Step-by-step explanation:

First you put the 2 in Then you get the 1/100 and put it next to the 2

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3 years ago

Find the measure of the missing angle.<br> 34°

Digiron [165]

The answer is 56.

As you can see, the triangle given is a right angle, meaning it is 90 degrees. To find the missing angle, we can solve 90 - 34 to get 56.

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3 years ago

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ryzh [129]

Answer: 36

Step-by-step explanation:

8x+16+6x-60=180

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6(16)-60=36

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3 years ago

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Find the length of AB. help asap !!!

Stella [2.4K]

Answer:

AB = sqrt 74 = 8.602

Step-by-step explanation:

simple:

a^2+b^2=c^2

5^2+7^2=74

ab=sqrt 74

Formula :

Distance between two points = √(xB−xA)2+(yB−yA)2(xB-xA)2+(yB-yA)2

Solution :

Distance between two points = √(8−3)2+(2−9)2(8-3)2+(2-9)2

= √52+(−7)252+(-7)2

= √25+4925+49

= √7474 = 8.6023

Distance between points (3, 9) and (8, 2) is 8.6023

3 0

2 years ago

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$

Recall that when t=0, A(t)=3000 (our initial condition)

$3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}$

3 0

3 years ago