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Arada [10]
2 years ago
15

Who can help me on this ?

Mathematics
2 answers:
Anna [14]2 years ago
5 0

Answer: 3

Step-by-step explanation:

Marat540 [252]2 years ago
3 0

Answer:

b= -4

Step-by-step explanation:

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7x + 10x + 20 = what​
RSB [31]

Answer:

the answer is 17x+20

Step-by-step explanation:

7x+10x+20

17x+20

hope it helps

mark me brainliest pls

6 0
2 years ago
Read 2 more answers
5.Gabby is fishing from a small boat. Her fishing hook is 36 feet below her, and a fish is swimming at the same depth as the hoo
Darina [25.2K]

Answer:

113 feet away

Step-by-step explanation:

This one is just simple maths! 36 + 77.

Hava a great day!

5 0
2 years ago
Is this equations linear?
Svetlanka [38]

Answer:

No

Step-by-step explanation:

There can not be x and y both attached to one coefficient.

5 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
The sum of 33 consecutive even numbers is 78. What is the second number in this sequence?
hjlf

Answer: The second number is 26


Step-by-step explanation:

To solve this problem you must apply the proccedure shown below:

1. Let's call:

x: The first even number.

x+2: the second consecutive even number.

x+4: the third consecutive even number.

2. So, you know that the sum ot the 3 consecutive numbers is 78, therefore, you must add them, as you can see below:

x+(x+2)+(x+4)=78\\3x+6=78\\3x=78-6\\3x=72\\x=\frac{72}{3}\\x=24

3. Substitute the value obtain into x+2. Therefore, the second number is:

x+2=24+2=26


3 0
3 years ago
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