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Ilia_Sergeevich [38]

3 years ago

10

mel's dress shop received an invoice dated july 17 for $1,200 with terms of 2/10, 1/15, n/60. on july 20, mel dress shop sent a

partial payment of $650. the actual amount that should be credited is

2 answers:

pogonyaev3 years ago

7 0

$663.27 i think thats the answer plzz make me the brainliest

DENIUS [597]3 years ago

3 0

$663.27 will be the answer.

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Pumps a, b, and c operate at their respective constant rates. pumps a and b, operating simultaneously, can fill a certain tank i

MrMuchimi

Let a, b, and c be the times each pump will fill the tank when working alone.

Therefore, in 1 hour;

1/a +1/b = 1/(6/5) = 5/6 ---- (1)
1/a+1/c = 1/(3/2) = 2/3 ---- (2)
1/b+1/c = 1/(2) = 1/2 ---- (3)

From equation (1)
1/a = 5/6-1/b

Substituting for 1/a in eqn (2)
5/6-1/b+1/c = 2/3
-1/b +1/c = -1/6 => 1/c = 1/b - 1/6 --- (4)

Using eqn (4) in eqn (3)
1/b+1/b-1/6 = 1/2
2/b-1/6 = 1/2
2/b =1/2+1/6 = 2/3
1/b = 1/3
Then,
1/c = 1/3 - 1/6 = 1/6
1/a = 5/6 - 1/3 = 1/2

This means, in 1 hour and with all the pumps working together, the tank will be filled to;
1/a+1/b+1/c = 1/2+1/3+1/6 = 1 (filled fully).

Therefore, it will take 1 hour to fill the tank when all pumps are working together.

7 0

2 years ago

1.<br> What is the solution to the equation x+3(x-5)=9

solmaris [256]

Answer:

PEMDAS

Step-by-step explanation:

PEMDAS

6 0

3 years ago

Read 2 more answers

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

Please answer all of these questions I will give 50 points and brainiest. (The slash means divide) Show all your work, please!

Mnenie [13.5K]

Answer:

1) 14.6

2) 1.36

3) .384

4) 37.2

5) 3.125

6) 4.75

7) 2.5

8) 3.04230769231

5 0

3 years ago

The voltage V of an audio speaker can be represented by V= 4 sqrt P , where P is the power of the speaker. An engineer wants to

Marysya12 [62]

Answer:

$Voltage=76\ volts$

Step-by-step explanation:

$Given\ V=4\sqrt{P}\\\\where\ V\ is\ voltage\ and\ P\ is\ Power.$

$Here\ power\ is\ given\\\\P=361\ watts\\\\Substitute\ P\ in\ the\ given\ equation\\\\V=4\sqrt{P}\\\\V=4\sqrt{361}\\\\V=4\sqrt{19\times 19}\\\\V=4\times 19\\\\V=76\ volts$

8 0

3 years ago