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Paraphin [41]
2 years ago
6

Sara sells fruit juices. She made 400 cups of fruit juice. By the end of the day, 34th of the quantity was sold. How many pints

of fruit juice was sold?
Mathematics
2 answers:
vredina [299]2 years ago
7 0

To answer the question, we need to convert cups to pints, now here is a way to remember the gallon, quarts, to pints, and to cups

In the Kingdom of Gallons (Write a big G on your paper)

Inside the G write 4 Qs

There were four Queens

Inside the Qs write 2 ps

Each Queen had to princess

In each princess write 2 Cs

And each princess has two Cats

Now we know that there is 2 Cups in each Pints so

since she sold 34 cups of juice in cups then we divide 34 by 2 to get

17 PINTS

lana [24]2 years ago
4 0

Answer:

150 pints

Step-by-step explanation:

I believe you meant "By the end of the day, 3/4th of the quantity was sold."

Recall that 2 cups = 1 pint, so 400 cups = 200 pints, and

3/4 th of that would be 150 pints.

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anzhelika [568]

Answer: Choice C

\displaystyle \frac{1}{2}\left(1 - \frac{1}{e^2}\right)

============================================================

Explanation:

The graph is shown below. The base of the 3D solid is the blue region. It spans from x = 0 to x = 1. It's also above the x axis, and below the curve y = e^{-x}

Think of the blue region as the floor of this weirdly shaped 3D room.

We're told that the cross sections are perpendicular to the x axis and each cross section is a square. The side length of each square is e^{-x} where 0 < x < 1

Let's compute the area of each general cross section.

\text{area} = (\text{side})^2\\\\\text{area} = (e^{-x})^2\\\\\text{area} = e^{-2x}\\\\

We'll be integrating infinitely many of these infinitely thin square slabs to find the volume of the 3D shape. Think of it like stacking concrete blocks together, except the blocks are side by side (instead of on top of each other). Or you can think of it like a row of square books of varying sizes. The books are very very thin.

This is what we want to compute

\displaystyle \int_{0}^{1}e^{-2x}dx\\\\

Apply a u-substitution

u = -2x

du/dx = -2

du = -2dx

dx = du/(-2)

dx = -0.5du

Also, don't forget to change the limits of integration

  • If x = 0, then u = -2x = -2(0) = 0
  • If x = 1, then u = -2x = -2(1) = -2

This means,

\displaystyle \int_{0}^{1}e^{-2x}dx = \int_{0}^{-2}e^{u}(-0.5du) = 0.5\int_{-2}^{0}e^{u}du\\\\\\

I used the rule that \displaystyle \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx which says swapping the limits of integration will have us swap the sign out front.

--------

Furthermore,

\displaystyle 0.5\int_{-2}^{0}e^{u}du = \frac{1}{2}\left[e^u+C\right]_{-2}^{0}\\\\\\= \frac{1}{2}\left[(e^0+C)-(e^{-2}+C)\right]\\\\\\= \frac{1}{2}\left[1 - \frac{1}{e^2}\right]

In short,

\displaystyle \int_{0}^{1}e^{-2x}dx = \frac{1}{2}\left[1 - \frac{1}{e^2}\right]

This points us to choice C as the final answer.

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Step-by-step explanation:

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