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3 years ago

5

Solve the simultaneous equations. 3x-4y = 18 and 9x + 2y = 12. Write your final answer on one line in the form x = .... , y= ...

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2 answers:

Nat2105 [25]3 years ago

8 0

I don’t know lol just doing this to do it

qwelly [4]3 years ago

5 0

Answer:First one the x is (6, 0) and the y value is (0, -9/2)

for the second equation the x value is (4/3, 0) and they value is (0, 6)

Step-by-step explanation:

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A 26-year-old male buys 100/300/100 liability insurance, collision insurance with a $100 deductible, and comprehensive insurance

lesantik [10]

Correct me if I’m wrong, I believe you have to add all of those numbers together. Then divide after and plug in those remaining digits

7 0

3 years ago

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

2 years ago

PLSSS I NEED HELP i will give you 50 points

DIA [1.3K]

Answer:

131.2

Step-by-step explanation:

x/55=31/14

x=31/13*55

x=131.153846154

4 0

3 years ago

What are the real numbers that are greater than -6 but less than 6

Lisa [10]

-5,-4,-3,-3,-2,-1,1,2,3,4,5
hopefully that helps

6 0

3 years ago

Read 2 more answers

3/5 divided by 2 and 1/2

REY [17]

Step-by-step explanation:

6/10

divided by 2

3/10

divided by 1/2

3/5

4 0

3 years ago