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4 years ago

Here are 946 milliliters in a quart. There are 22 pints in a quart. How many milliliters are in a pint?

Mathematics

2 answers:

mash [69]4 years ago

7 0

Answer:

473 for khan

Step-by-step explanation:

Fed [463]4 years ago

6 0

Answer: The answer is 43.

Step-by-step explanation: Given that there is a quart, in which there are 946 millilitres and 22 pints. We are to find how many millilitres are there in a pint.

For that, we just need to divide number of milliliters by the number of pints in the quart.

Let us start as follows -

In 22 pints of the quart = 946 millilitres.

Therefore, in 1 pint of the quart = 946/22 = 43 millilitres.

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One batch of muffins require 1/4 cup of sugar. Kelly wants to make 3/2 batch of muffins. Will she use more than, less than, or a

Alla [95]

She will use more than 1/4 cup of sugar to make 1.5 batches of muffins. If one batch of muffins requires 1/4 cup of sugar, you can think of the amount of sugar needed being 1/4 cup per batch. That being said multiply 1.5 batches by 1/4 cup per batch to find cups of sugar.
(1.5 batches)x((1/4 cups)/(1 batch))=0.375 cups of sugar.

You can also think of it as if one batch takes 1/4 cup of sugar, half of a batch would take 1/8 cup of sugar. Since 1.5 batches is just 1 batch+1/2 batch, you can say the amount of sugar needed is 1/4 cup+1/8 cup which equals 3/8 cup sugar.

I hope this helps you.

5 0

3 years ago

A cylindrical can has a volume of 16 cm 3. What dimensions yield the minimum surface area? 1. The radius of the can with the mi

kykrilka [37]

Answer:

radius = 1.37 cm

height = 2.71 cm

Step-by-step explanation:

We are given volume = 16 m³.

Formula for volume of a cylinder is;

V = πr²h

Thus,

πr²h = 16

h = 16/πr²

Now formula for the surface area is;

S = 2πr² + 2πrh

Putting 16/πr² for h gives;

S = 2πr² + 2πr(16/πr²)

S = 2πr² + 2π(16/πr)

S = 2π(r² + 16/πr)

To minimize, we will find the derivative of S and equate to zero

S' = 2π(2r - 16/πr²) = 0

4πr - 32/r² = 0

4πr = 32/r²

r³ = 32/4π

r = ∛(32/4π)

r = 1.37 cm

From h = 16/πr²;

h = 16/(π × 1.37²)

h = 2.71 cm

5 0

4 years ago

The weight of 1000 identical samples of a substance is 0.1 pounds. What is the weight of 100 smaples?

mrs_skeptik [129]

0.010000000000000000000000000

4 0

4 years ago

I need help plz anyone

Contact [7]

Answer:

Step-by-step explanation:

7 0

3 years ago

Find the exact area of the surface obtained by rotating the curve about the x-axis. y = 1 + ex , 0 ≤ x ≤ 9

tekilochka [14]

The surface area is given by

$\displaystyle2\pi\int_0^9(1+e^x)\sqrt{1+e^{2x}}\,\mathrm dx$

since $y=1+e^x\implies y'=e^x$ . To compute the integral, first let

$u=e^x\implies x=\ln u$

so that $\mathrm dx=\frac{\mathrm du}u$ , and the integral becomes

$\displaystyle2\pi\int_1^{e^9}\frac{(1+u)\sqrt{1+u^2}}u\,\mathrm du$

$=\displaystyle2\pi\int_1^{e^9}\left(\frac{\sqrt{1+u^2}}u+\sqrt{1+u^2}\right)\,\mathrm du$

Next, let

$u=\tan t\implies t=\tan^{-1}u$

so that $\mathrm du=\sec^2t\,\mathrm dt$ . Then

$1+u^2=1+\tan^2t=\sec^2t\implies\sqrt{1+u^2}=\sec t$

so the integral becomes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec t}{\tan t}+\sec t\right)\sec^2t\,\mathrm dt$

$=\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\left(\frac{\sec^3t}{\tan t}+\sec^3 t\right)\,\mathrm dt$

Rewrite the integrand with

$\dfrac{\sec^3t}{\tan t}=\dfrac{\sec t\tan t\sec^2t}{\sec^2t-1}$

so that integrating the first term boils down to

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\frac{\sec t\tan t\sec^2t}{\sec^2t-1}\,\mathrm dt=2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\frac{s^2}{s^2-1}\,\mathrm ds$

where we substitute $s=\sec t\implies\mathrm ds=\sec t\tan t\,\mathrm dt$ . Since

$\dfrac{s^2}{s^2-1}=1+\dfrac12\left(\dfrac1{s-1}-\dfrac1{s+1}\right)$

the first term in this integral contributes

$\displaystyle2\pi\int_{\sqrt2}^{\sqrt{1+e^{18}}}\left(1+\frac12\left(\frac1{s-1}-\frac1{s+1}\right)\right)\,\mathrm ds=2\pi\left(s+\frac12\ln\left|\frac{s-1}{s+1}\right|\right)\bigg|_{\sqrt2}^{\sqrt{1+e^{18}}}$

$=2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}$

The second term of the integral contributes

$\displaystyle2\pi\int_{\pi/4}^{\tan^{-1}(e^9)}\sec^3t\,\mathrm dt$

The antiderivative of $\sec^3t$ is well-known (enough that I won't derive it here myself):

$\displaystyle\int\sec^3t\,\mathrm dt=\frac12\sec t\tan t+\frac12\ln|\sec t+\tan t|+C$

so this latter integral's contribution is

$\pi\left(\sec t\tan t+\ln|\sec t+\tan t|\right)\bigg|_{\pi/4}^{\tan^{-1}(e^9)}=\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

Then the surface area is

$2\pi\sqrt{1+e^{18}}+\pi\ln\dfrac{\sqrt{1+e^{18}}-1}{1+\sqrt{1+e^{18}}}+\pi\left(e^9\sqrt{1+e^{18}}+\ln(e^9+\sqrt{1+e^{18}})-\sqrt2-\ln(1+\sqrt2)\right)$

$=\boxed{\left((2+e^9)\sqrt{1+e^{18}}-\sqrt2+\ln\dfrac{(e^9+\sqrt{1+e^{18}})(\sqrt{1+e^{18}}-1)}{(1+\sqrt2)(1+\sqrt{1+e^{18}})}\right)\pi}$

4 0

4 years ago