I need help with my math

the local movie theatre collected a total of $8,250 last night, of which 45% came from sales at the concession stand. how much m

Wewaii [24]

45%=0.45

8,250x0.45
=3712.50

The theater collected $3712.50 at the concession stand.

3 0

3 years ago

The temperature in Lucy's refrigerator is 2.74 degrees celious.the temperature in the freezer section of her refrigerator is -19

irinina [24]

2.74 - (-19.4) = 22.14, so the freezer is 22.14°c colder. I hope that helps!

8 0

3 years ago

Read 2 more answers

Jayden is downloading songs onto his MP3 player. One song is 5 minutes long,another is 2 minutes long,and a third is between the

Sladkaya [172]

Answer:

Jayden is downloading songs onto his MP3 player. One song is 5 minutes long,another is 2 minutes long,and a third is between the lengths of the other two songs. What is the total length of all three songs?

Step-by-step explanation:

Jayden is downloading songs onto his MP3 player. One song is 5 minutes long,another is 2 minutes long,and a third is between the lengths of the other two songs. What is the total length of all three songs?

so the answer iis 3.5 minutes long

BRAINLIEST PLS??????????

6 0

3 years ago

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Given A is between Y and Z and YA = 2x, AZ = 16x, and YZ = 4x+56, find AZ. Plsss help

cupoosta [38]

Answer:

64

Step-by-step explanation:

4 0

3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

aliya0001 [1]

Answer:

$A=1500-1450e^{-\dfrac{t}{250}}$

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

$R_{in}$ =(concentration of salt in inflow)(input rate of brine)

$=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}$

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, $C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}$

$R_{out}$ =(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}$

Now, the rate of change of the amount of salt in the tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{250}$

We solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}$

Recall that when t=0, A(t)=50 (our initial condition)

$50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}$

4 0

3 years ago