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Sever21 [200]
3 years ago
9

I need help with this problem quick.

Mathematics
1 answer:
KatRina [158]3 years ago
8 0

The absolute value - or distance from zero, is fairly simple to find. Make your final answer positive to find it.


|\frac{10-7}{4}| = |\frac{3}{4}| = |.75| = .75

|\frac{-15+8}{6}| = |-\frac{7}{6}| = |-1.2| = 1.2

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Harrizon [31]
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natka813 [3]

Answer:

12 ( B )

Step-by-step explanation:

mode means the number that appears the most often

in that case the mode is 12 ( B )

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A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so
Debora [2.8K]

Let A(t) = amount of salt (in pounds) in the tank at time t (in minutes). Then A(0) = 11.

Salt flows in at a rate

\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}

and flows out at a rate

\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}

which I'll solve with the integrating factor method.

\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95

-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}

\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}

Integrate both sides. By the fundamental theorem of calculus,

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)

\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}

\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}

\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}

After 1 hour = 60 minutes, the tank will contain

A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131

pounds of salt.

7 0
1 year ago
Use the Polynomial Identity below to help you create a list of 10 Pythagorean Triples:
Stolb23 [73]

Answer:

(3,4,5)

(6,8,10)

(5,12,13)

(8,15,17)

(12,16,20)

(7,24,25)

(10,24,26)

(20,21,29)

(16,30,34)

(9,40,41)

Just choose 2 numbers from {1,2,3,4,5,6,7,8,...} and make sure the one you input for x is larger.

Post the three in the comments and I will check them for you.

Step-by-step explanation:

We need to choose 2 positive integers for x and y where x>y.

Positive integers are {1,2,3,4,5,6,7,.....}.

I'm going to start with (x,y)=(2,1).

x=2 and y=1.

(2^2+1^2)^2=(2^2-1^2)^2+(2\cdot2\cdot1)^2

(4+1)^2=(4-1)^2+(4)^2

(5)^2=(3)^2+(4)^2

So one Pythagorean Triple is (3,4,5).

I'm going to choose (x,y)=(3,1).

x=3 and y=1.

(3^2+1^2)^2=(3^2-1^2)^2+(2\cdot3\cdot1)^2

(9+1)^2=(9-1)^2+(6)^2

(10)^2=(8)^2+(6)^2

So another Pythagorean Triple is (6,8,10).

I'm going to choose (x,y)=(3,2).

x=3 and y=2.

(3^2+2^2)^2=(3^2-2^2)^2+(2\cdot3\cdot2)^2

(9+4)^2=(9-4)^2+(12)^2

(13)^2=(5)^2+(12)^2

So another is (5,12,13).

I'm going to choose (x,y)=(4,1).

(4^2+1^2)^2=(4^2-1^2)^2+(2\cdot4\cdot1)^2

(16+1)^2=(16-1)^2+(8)^2

(17)^2=(15)^2+(8)^2

Another is (8,15,17).

I'm going to choose (x,y)=(4,2).

(4^2+2^2)^2=(4^2-2^2)^2+(2\cdot4\cdot2)^2

(16+4)^2=(16-4)^2+(16)^2

(20)^2=(12)^2+(16)^2

We have another which is (12,16,20).

I'm going to choose (x,y)=(4,3).

(4^2+3^2)^2=(4^2-3^2)^2+(2\cdot4\cdot3)^2

(16+9)^2=(16-9)^2+(24)^2

(25)^2=(7)^2+(24)^2

We have another is (7,24,25).

You are just choosing numbers from the positive integer set {1,2,3,4,... } and making sure the number you plug in for x is higher than the number for y.

I will do one more.

Let's choose (x,y)=(5,1).

(5^2+1^2)^2=(5^2-1^2)^2+(2\cdot5\cdot1)^2

(25+1)^2=(25-1)^2+(10)^2

(26)^2=(24)^2+(10)^2

So (10,24,26) is another.

Let (x,y)=(5,2).

(5^2+2^2)^2=(5^2-2^2)^2+(2\cdot5\cdot2)^2

(25+4)^2=(25-4)^2+(20)^2

(29)^2=(21)^2+(20)^2

So another Pythagorean Triple is (20,21,29).

Choose (x,y)=(5,3).

(5^2+3^2)^2=(5^2-3^2)^2+(2\cdot5\cdot3)^2

(25+9)^2=(25-9)^2+(30)^2

(34)^2=(16)^2+(30)^2

Another Pythagorean Triple is (16,30,34).

Let (x,y)=(5,4)

(5^2+4^2)^2=(5^2-4^2)^2+(2\cdot5\cdot4)^2

(25+16)^2=(25-16)^2+(40)^2

(41)^2=(9)^2+(40)^2

Another is (9,40,41).

5 0
3 years ago
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