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3 years ago

Round 16.748 to the ones place

Mathematics

2 answers:

IceJOKER [234]3 years ago

4 0

17 for ones place.. 16.7 for tenths place.. 16.75 for hundredths place and so on..

Aleonysh [2.5K]3 years ago

3 0

So , what helps me in math is my teachers was us say 4 or lest let it rest 5 or more let it sore so what dose the 8 tell the 4 USE A ARROW if needed. thank you

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12, 17, 22, ...<br> Find the 36th term.

pychu [463]

Answer:

187

Step-by-step explanation:

a=12,d=17-12=5

nth term =a+(n-1)d

36th term=12+(36-1)5

=12+(35×5)

12+175

36th term=187

8 0

2 years ago

Diane has $334 in her checking account. She writes a check for $112, makes a deposit of $100, and then writes another check for

Dmitrij [34]

Answer:

334 - 112 + 100 - 98

Step-by-step explanation:

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2 years ago

41 kilometers to meter

g100num [7]

Answer:

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Step-by-step explanation:

multiply the length value by 1000

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3 years ago

Read 2 more answers

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

The solution set of the quadratic inequality x squared minus 5 x plus 4 less or equal than 0 is

BigorU [14]

Answer:

it is equal to 0

7 0

3 years ago