All categories

4 years ago

What is 3 1/8 as decimal

Mathematics

1 answer:

FromTheMoon [43]4 years ago

7 0

Answer: 25/3

Step-by-step explanation: 8/8•3=24/8 then add 24/8 + the 1/8 = 25/8.

You might be interested in

PLEASE HELP ASAP 30 POINTS

Yuki888 [10]

Answer:

I don't know how to do please let me I will try solve the question

8 0

3 years ago

PLEASE HELP. i have one hour to get this done and i have no clue what to do on this bonus question i need to pass!! look in the

Scilla [17]

Answer: the perimeter is 214. AB's half is 32.5 and BC's whole is 65

Step-by-step explanation: 10x-5 and 12x-26 are equal, as shown by the marks on their lines, so we can set them equal to each other to solve for x. when solved x=7. then plug 7 into each equation to get 65. lastly add up 84+65+65 to get 214.

5 0

3 years ago

Read 2 more answers

I need to know what the equation is for this graph..

Lostsunrise [7]

Answer:

The choice second

$y = \frac{1}{3} \times ( \frac{1}{4} ) ^{x}$

3 0

2 years ago

Show the number on a number line:<br> -12 ÷ 4 =

Gwar [14]

Answer:

well if we take -12 and divide it by four we will get -3.

Step-by-step explanation:

If we have a total of negative 12, just divide it by 4 and the difference between each division is the sum.

Your answer will be -3 on the number line.

3 0

2 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

2 years ago