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2 years ago

A pitcher contains 40 fluid ounces of tea Shelby pour 3 cups of tea how many pints are left

Mathematics

2 answers:

Nastasia [14]2 years ago

8 0

Answer:

1 pint of tea left in the pitcher.

Step-by-step explanation:

A pitcher contains 40 fluid ounces

1 cup = 8 fluid ounces

3 cups = 3*8 = 24 fluid ounces

After pouring tea in 3 cups,

tea left in pitcher = 40 - 24 = 16 fluid ounces

1 pint = 16 ounces

Therefore, 1 pint left.

Hope this will helpful.

Thank you.

katrin2010 [14]2 years ago

6 0

Convert 3 cups to fluid ounces.

1 cup = 8 fluid ounces

8 * 3 = 24 fluid ounces poured:

40 - 24 = 16

There are 16 fluid ounces left, now convert to pints.

1 fluid ounce = 0.0625 pints

0.0625 * 16 = 1

So there is 1 pint left.

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Approximate area under the curve f(x) =-x^2+2x+4 from x=0 to x=3 by using summation notation with six rectangles and use the the

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Answer:

Summation notation:

$\frac{1}{2}\sum_{k=1}^6f((.5k))$

or after using your function part:

$\frac{1}{2}\sum_{k=1}^6(-(.5k)^2+2(.5k)+4)$

After evaluating you get 11.125 square units.

Step-by-step explanation:

The width of each rectangle is the same so we want to take the distance from x=0 to x=3 and divide by 6 since we want 6 equal base lengths for our rectangles.

The distance between x=0 and x=3 is (3-0)=3.

We want to divide that length of 3 units by 6 which gives a length of a half per each base length.

We are doing right endpoint value so I'm going to stat at x=3. The first rectangle will be drawn to the height of f(3).

The next right endpoint is x=3-1/2=5/2=2.5, and the second rectangle will have a height of f(2.5).

The next will be at x=2.5-.5=2, and the third rectangle will have a height of f(2).

The fourth rectangle will have a height of f(2-.5)=f(1.5).

The fifth one will have a height of f(1.5-.5)=f(1).

The last one because it is the sixth one will have a height of f(1-.5)=f(.5).

So to find the area of a rectangle you do base*time.

So we just need to evaluate:

$\frac{1}{2}f(3)+\frac{1}{2}f(2.5)+\frac{1}{2}f(2)+\frac{1}{2}f(1.5)+\frac{1}{2}f(1)+\frac{1}{2}f(.5)$

or by factoring out the 1/2 part:

$\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))$

To find f(3) replace x in -x^2+2x+4 with 3:

-3^2+2(3)+4

-9+6+4

To find f(2.5) replace x in -x^2+2x+4 with 2.5:

-2.5^2+2(2.5)+4

-6.25+5+4

2.75

To find f(2) replace x in -x^2+2x+4 with 2:

-2^2+2(2)+4

-4+4+4

To find (1.5) replace x in -x^2+2x+4 with 1.5:

-1.5^2+2(1.5)+4

-2.25+3+4

4.75

To find f(1) replace x in -x^2+2x+4 with 1:

-1^2+2(1)+4

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To find f(.5) replace x in -x^2+2x+4 with .5:

-.5^2+2(.5)+4

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4.75

Now let's add those heights. After we obtain this sum we multiply by 1/2 and we have our approximate area:

$\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))$

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$\frac{1}{2}(22.25)$

$11.125$

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$\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))$

is it

$\frac{1}{2}\sum_{k=1}^{6}(f(.5+.5(k-1)))$

or after simplifying a bit:

$\frac{1}{2}\sum_{k=1}^6 f((.5+.5k-.5))$

$\frac{1}{2}\sum_{k=1}^6f((.5k))$

If you are wondering how I obtain the .5+.5(k-1):

I realize that 3,2.5,2,1.5,1,.5 is an arithmetic sequence with first term .5 if you the sequence from right to left (instead of left to right) and it is going up by .5 (reading from right to left.)

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I found the area of the two squares on the bottom
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