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3 years ago

A milk container's capacity is one gallon. How many cups would be equal?

Mathematics

2 answers:

kati45 [8]3 years ago

4 0

1 gallon of milk would be 16 cups :)

aliina [53]3 years ago

3 0

16 cups equals 1 gallon.

So our answer is that a milk containers capacity is 16 cups.

You might be interested in

Each year Taylor donates 1/10 of her annual income to local charities. One-fifth of that amount goes to a charity that supports

Soloha48 [4]

Answer:

ITS CORRECT

Step-by-step explanation:

THE ANSWER IS 0.02 OR 1/50

4 0

3 years ago

Renee is playing a game online. If she gets a total of 25 points, she will have a new

Mashutka [201]

Answer:

20 points

Step-by-step explanation:

key words: difference

Take the amount of points she has and subtract what she has

25-5

4 0

3 years ago

Find the slope of the line passing through the points (-6, -3) and (4, -3).

miskamm [114]

Answer:

m = 0

Step-by-step explanation:

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Simply plug in the coordinates into the formula:

m = (-3 + 3)/(4 - 6)

m = 0/-2

m = 0

6 0

3 years ago

A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are

skelet666 [1.2K]

A) The maximum revenue is 450000$

B) The maximum profit is 216000$ when 2400 sets are manufactured and sold for 180$ each

C) When each set is taxed at $55, the maximum profit is 99125$ when 1850 sets are manufactured and sold for 207.5$ each.

A)p(x)=300−(x/20),

revenue R(x)=p*x

revenue R(x)=300x -(x2/20)

for maximum revenue dR/dx =0 ,

=>300-(2x/20)=0

=>x/10=300

=>x=3000

maximum revenue = R(3000)=300*3000 -(30002/20)

maximum revenue = R(3000)=450000$

B) profit =revenue -cost

profit P(x)=300x -(x2/20)-72000-60x

profit P(x)=240x -(x2/20)-72000

for maximum cost dP/dx =0

240 -(2x/20)=0

x=240*10

x=2400

p(2400)=300−(2400/20)=180

profit P(2400)=240*2400 -(24002/20)-72000 =216000

The maximum profit is 216000$ when 2400 sets are manufactured and sold for 180$ each

profit =revenue -cost -tax

profit P(x)=300x -(x2/20)-72000-60x-55x

profit P(x)=185x -(x2/20)-72000

for maximum cost dP/dx =0

185-(2x/20)=0

x=185*10

x=1850

p(1850)=300−(1850/20)=207.5

profit P(1850)=185*1850 -(18502/20)-72000

profit P(1850)=99125$

When each set is taxed at $55, the maximum profit is 99125$ when 1850 sets are manufactured and sold for 207.5$ each.

To know more about maximum profit check the below link:

brainly.com/question/4166660

#SPJ4

5 0

1 year ago

Uranus moves in an elliptical orbit with the sun at one of the foci. The length of the half of the major axis is 2,876,769,540 k

Alina [70]

Answer:

The minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.

Step-by-step explanation:

Consider the provided information.

The length of the half of the major axis is 2,876,769,540 kilometers, and the eccentricity is 0.0444.

The eccentricity (e) of an ellipse is the ratio of the distance from the center to the foci (c) and the distance from the center to the vertices (a).

$e=\frac{c}{a}$

Substitute a = 2,876,769,540 and e = 0.0444 in above formula and solve for c.

$0.0444=\frac{c}{2,876,769,540 }$

$c=127728567.576$

Minimum distance of Uranus from the sun is:

$a-c=2,876,769,540-127728567.576\\a-c=2749040972.424\approx2749040972$

Hence, the minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.

6 0

3 years ago