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3 years ago

Helpppp Plsssss Asap!! Show your work

Mathematics

2 answers:

satela [25.4K]3 years ago

8 0

Answer:

Step-by-step explanation:

The anwser is C

the inverse of 5x is x/5 and the inverse of +4 is -4 so the answer would be x-4/5

solmaris [256]3 years ago

7 0

Answer:

B. f-1(x) = (x - 4)/5.

Step-by-step explanation:

Let y = 5x + 4

5x = y - 4

x = (y - 4)/5

f-1(x) = (x - 4)/5.

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Write the following number in standard decimal form.<br> seven and sixty-one thousandths

amid [387]

Answer:

0.761

Step-by-step explanation:the 7 is in the tenths the 6 is in the hundredths and the 1 is in the thousandths

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3 years ago

Help! Match each linear equation with its graph

Bond [772]

Answer:

y = -1/4 x+4 is blue

y = 1/2 x+4 is green

y = 3x is red

y = -1/2x is purple

y = 3 is black

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3 years ago

Ashley went to HEB to buy the apples to make her famous apple pie. if she bought 22 pounds of apples for $1.43 per pound. How mu

PilotLPTM [1.2K]

You have to multiply the price by the quantity and you will find the total cost.

1.43 * 22 = 31.46

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3 years ago

Watch help video

ollegr [7]

Answer: 15309

hope this helps, the nth term calulation is (7/3) times 3^n

6 0

2 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

Helpppp Plsssss Asap!! Show your work​

Helpppp Plsssss Asap!! Show your work