All categories

3 years ago

Does two eighths equal one fourths ?

Mathematics

2 answers:

Aliun [14]3 years ago

8 0

It does. 1/4 equals 2/8 2 times.

KonstantinChe [14]3 years ago

7 0

Yes all that you have the need to do is divide it by 2 and then you have your answer

You might be interested in

Find the area. 11 cm 8 cm 7 cm

HACTEHA [7]

66.5cm

Split the shape into a square and triangle and find the area separately

Square: 8*7=57
Triangle: 7*3/2 = 10.5
=66.5

7 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

1 year ago

I need this answered by today 9-3-18

Talja [164]

we;retk;tykh'tk';rekw'kr'e

6 0

3 years ago

Will mark brainly!!

Kruka [31]

Is there supposed to be an x before 4x?

5 0

2 years ago

A large container holds 8 gallons of water. It begins leaking at a constant rate. After 48 minutes, the container has 5 gallons

Juliette [100K]

Answer:

The water is leaking at the rate of 0.0625 gal/min.

Step-by-step explanation:

Rate of change

The rate of change (ROC) is a measure that compares two quantities, usually to know how fast one variable changes in time.

The container holds 8 gallons of water and is leaking at a constant rate to be calculated later.

When 48 minutes have passed, the container has 5 gallons of water left. This means it has leaked 8 gallons - 5 gallons = 3 gallons.

Calculating the rate at which it's leaking:

$\displaystyle r=\frac{3\ gallons}{48\ min}$

r = 0.0625 gal/min

The water is leaking at the rate of 0.0625 gal/min.

5 0

2 years ago