Ask question

Ask question

All categories

3 years ago

10

J. Michael filled the watering can 6 times to water the small tomato plants. The watering can held the quarts of water. How many

gallons of water did he use?

1 answer:

Temka [501]3 years ago

3 0

Answer:

1.5 gallons

Step-by-step explanation:

You might be interested in

WILL MARK BRAINLIEST what is the standard deviation for the normal distribution shown at the right?

MrRissso [65]

Answer: 916. Because if you do 856+60=916.

Step-by-step explanation: The standard normal distribution is a normal distribution with a mean of zero and standard deviation of 1. Formula: and natural statistics x and x2. The dual expectation parameters for normal distribution are η1 = μ and η2 = μ2 + σ2.

8 0

3 years ago

The circumference of a circle is 16\pi16π inches. What is the radius of the circle?

laiz [17]

Answer: 8 inches

Step-by-step explanation: So you know that to get the circumference of a circle is 2πr, right? So you divide 16 by 2 to get your radius. So that's how you 8 inches.

4 0

3 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)

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

(c) Concentration of chlorine in the pool at time t

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

What could be the length of the third side of the triangle if it is known that the first two sides are 15 and 27?

kodGreya [7K]

It will be between the sum and difference of the two given sides.

12 < third side < 42

3 0

3 years ago

Fine the area of a circle with a radius of 7 in. Write the formula, substitute, solve.

bija089 [108]

Formula: A=(pi)r^2

Substitution: A=3.14(7^2)

A=3.14(49)

Solve: A=153.86 in^2

*I personally like to use the pi sign on my calculator to multiply the radius squared by, but my keyboard did not have a pi sign. Usually, the pi sign is more exact. If you used the pi sign, your answer would be 153.94 inches^2. Hope this helps!

8 0

3 years ago