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3 years ago

Simplify each of the following expressions.

Mathematics

1 answer:

Leokris [45]3 years ago

4 0

Answer:

1. 8x

2. 5m

3. 9y

4. 0

5. 2s

6. 5x + 9y

7. w + 4t

8. 7m - n + 6

9. x + 6y

10. 12x + 6

11. 5p + 3

12. 5x - 7

13. - 4c + 82

14. 4f + 2d + 2

Step-by-step explanation:

hope this helps

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Find the domain and range of the relation.

Kisachek [45]

the answer is in the picture above

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3 years ago

A tank contains 100 gal of water and 50 oz of salt. Water containing a salt concentration of ¼ (1 + ½ sin t) oz/gal flows ito th

Likurg_2 [28]

Answer:

Part a)

$y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}$

Part b)

Check the attached figure to see the ultimate behavior of the graph.

Part c)

The level = 25, Amplitude = 0.2499

Step-by-step Solution:

Part a)

Given:

$Q(0)=50$

Rate in:

$\frac{1}{4}\left(1+\frac{1}{2}sint\right)\cdot 2\:=\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

Rate out:

$\frac{Q}{100}\cdot 2=\frac{Q}{50}$

So, the differential equation would become:

$\frac{dQ}{dt}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)-\frac{Q}{50}$

Rewriting the equation:

$\frac{dQ}{dt}+\frac{Q}{50}=\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

As $p(x)$ is the coefficient of $y$ , while $q(x)$ is the constant term in the right side of the equation:

$p\left(x\right)=\frac{1}{50}$

$q\left(x\right)=\frac{1}{2}\left(1+\frac{1}{2}sint\right)$

First it is important to determine the function $\mu$ :

$\mu \left(t\right)=e^{\int \:p\left(t\right)dt}$

$=e^{\int \:\left(\frac{1}{50}\right)dt}$

$=e^{\frac{t}{50}}$

The general solution then would become:

$y\left(t\right)=\frac{1}{\mu \left(t\right)}\left(\int \mu \left(t\right)q\left(t\right)dt+c\:\right)$

$=\frac{1}{e^{\frac{t}{50}}}\int e^{\frac{t}{50}}\:\frac{1}{2}\left(1+\frac{1}{2}sint\right)dt+\frac{1}{e^{\frac{t}{50}}}c$

$=\frac{1}{e^{\frac{t}{50}}}\left(\frac{-25e^{\frac{t}{50}}\left(50cost-sint\right)}{5002}+25e^{\frac{t}{50}}\right)+\frac{1}{e^{\frac{t}{50}}}c$

$=\frac{\left-1250cost+25sint\right}{5002}+25+\frac{1}{e^{\frac{t}{50}}}c$

Evaluate at $t=0$

$50=y\left(0\right)=\frac{\left(-1250cos0+25sin0\right)}{5002}+25+\frac{1}{e^{\frac{0}{50}}}c$

Solve to c:

$c=25+\frac{1250}{5002}$

$\mathrm{Cancel\:}\frac{1250}{5002}:\quad \frac{625}{2501}$

$c=25+\frac{625}{2501}$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:25=\frac{25\cdot \:2501}{2501}$

$c=\frac{25\cdot \:2501}{2501}+\frac{625}{2501}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$c=\frac{25\cdot \:2501+625}{2501}$

$c=\frac{63150}{2501}$

$c\approx 25.25$

Therefore, the general solution then would become:

$y\left(t\right)=\frac{-1250cost+25sint}{5002}+25+\frac{63150}{2501}\frac{1}{e^{\frac{t}{50}}}$

Part b) <em>Plot the Solution to see the ultimate behavior of the graph</em>

The graph appears to level off at about the value of $Q=25$ .

The graph is attached below.

Part c)

In the graph we note that the level is $Q=25$ .

Therefore, the level = 25

The amplitude is the (absolute value of the) coefficient of $cost\:t$ in the general solution (as the coefficient of the sine part is a lot smaller):

Therefore,

$A=\frac{1250}{5002}\:\approx 2.499$

Keywords: differential equation, word problem

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3 years ago

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Yuliya22 [10]

Answer:

#19: x=3

#20: would be n=4 i think

Step-by-step explanation:

#19- there are 4 x's and 12 +1's so 4 into 12 is 3.

#20- well the answer that i got was 5 but 4 was closer to 5 so yea.

i sure hope this helps!

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3 years ago

A scientist counted 3,921 total eggs in

uysha [10]

Answer:

The Answer is 80

Step-by-step explanation:

So I turned 3,921 to 4,000 then 49 to 50.

After that I divided the too 4,000÷50= 80

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2 years ago

Equation of a line that has a slope of -2 and passes through the point (-1,4)

EleoNora [17]

Answer:

y=mx+b

Step-by-step explanation:

thats the formula i think

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3 years ago