All categories

2 years ago

One day students drank 60 quarts of milk at lunch.how many pints of milk did the student drink

Mathematics

2 answers:

Anna11 [10]2 years ago

6 0

120 pints is your answer! :)

Lera25 [3.4K]2 years ago

5 0

It would be 160 pints of milk that they've drank!
I hope all is well, and you end up passing!
Good luck, rockstar!

You might be interested in

How does one do this? I’ve tried and still can’t get it please help

professor190 [17]

Answer:

Step-by-step explanation:

An angle bisector cuts an angle into two equal parts. So m∠ACE = m∠ECB.

m∠ACE = m∠ECB

2x − 15 = x + 5

x = 20

6 0

3 years ago

Explain how to use the sales tax rate to determine an item's total cost.

Sedbober [7]

Answer:

A. Multiply the purchase price by the tax rate. Then add the result to the purchase price.

Step-by-step explanation:

The amount of tax is found by multiplying the taxed amount by the tax rate. The customer generally pays the sales tax, so the amount the customer will pay will be the sum of the purchase price and the tax.

7 0

2 years ago

Does anybody know about this I'm stuckes it's about angles

Lera25 [3.4K]

Answer:

the answer is 25 degrees

7 0

2 years ago

Read 2 more answers

The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubi

storchak [24]

Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

From the formula

$V = \frac{1}{3}\pi r^{2}h$

Differentiate the equation with respect to time t, such that

$\frac{d}{dt} (V) = \frac{d}{dt} (\frac{1}{3}\pi r^{2}h)$

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (r^{2}h)$

To differentiate the product,

Let r² = u, so that

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)$

Then, using product rule

$\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]$

Since $u = r^{2}$

Then, $\frac{du}{dr} = 2r$

Using the Chain's rule

$\frac{du}{dt} = \frac{du}{dr} \times \frac{dr}{dt}$

∴ $\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})]$

Then,

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

Now,

From the question

$\frac{dr}{dt} = 7 m/min$

$\frac{dV}{dt} = 236 m^{3}/min$

At the instant when $r = 99 m$

and $V = 180 m^{3}$

We will determine the value of h, using

$V = \frac{1}{3}\pi r^{2}h$

$180 = \frac{1}{3}\pi (99)^{2}h$

$180 \times 3 = 9801\pi h$

$h =\frac{540}{9801\pi }$

$h =\frac{20}{363\pi }$

Now, Putting the parameters into the equation

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

$236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]$

$236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]$

$708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}$

$708 = 30790.75 \frac{dh}{dt} + 76.36$

$708 - 76.36 = 30790.75\frac{dh}{dt}$

$631.64 = 30790.75\frac{dh}{dt}$

$\frac{dh}{dt}= \frac{631.64}{30790.75}$

$\frac{dh}{dt} = 0.021 m/min$

Hence, the rate of change of the height is 0.021 meters per minute.

3 0

3 years ago

HELP ME PLZ

GarryVolchara [31]

Answer:

(-1,3)

Step-by-step explanation:

Thus;

When (-1,3) is placed in the relation 3x + y =0,the answer will be zero.

<h3>Calculations</h3>

Let;

x= -1

y= 3

3(-1) + 3 =0

-3 + 3 = 0

5 0

2 years ago