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3 years ago

8

How many 1/3 cup servings of raisins are in 1/2 a cup of raisins?

2 answers:

sp2606 [1]3 years ago

6 0

1 1/2 or 3/2 because 1/2 divided by 1/3 is equal to that

polet [3.4K]3 years ago

6 0

<span>Just divide 1/2 by 1/3 and the improper fraction will be 3/2 and the mixed fraction will be 1 and 1/2.

Hope this helped!!!!!</span>

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a grocery store paid 298.65 for 15 racks of bread.this can be expressed by the equation y=kx.how much would they have paid for 1

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Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving

faust18 [17]

Answer:

$Volume = \frac{384}{7}\pi$

Step-by-step explanation:

Given (Missing Information):

$y = x^\frac{3}{2}$ ; $y = 8$ ; $x=0$

Required

Determine the volume

Using Shell Method:

$V = 2\pi \int\limits^a_b {p(y)h(y)} \, dy$

First solve for a and b.

$y = x^\frac{3}{2}$ and $y = 8$

Substitute 8 for y

$8 = x^\frac{3}{2}$

Take 2/3 root of both sides

$8^\frac{2}{3} = x^{\frac{3}{2}*\frac{2}{3}}$

$8^\frac{2}{3} = x$

$2^{3*\frac{2}{3}} = x$

$2^2 = x$

$4 =x$

$x = 4$

This implies that:

$a = 4$

For $x=0$

This implies that:

$b=0$

So, we have:

$V = 2\pi \int\limits^a_b {p(y)h(y)} \, dy$

$V = 2\pi \int\limits^4_0 {p(y)h(y)} \, dy$

The volume of the solid becomes:

$V = 2\pi \int\limits^4_0 {x(8 - x^{\frac{3}{2}}}) \, dx$

Open bracket

$V = 2\pi \int\limits^4_0 {8x - x.x^{\frac{3}{2}}} \, dx$

$V = 2\pi \int\limits^4_0 {8x - x^{\frac{2+3}{2}}} \, dx$

$V = 2\pi \int\limits^4_0 {8x - x^{\frac{5}{2}}} \, dx$

Integrate

$V = 2\pi * [{\frac{8x^2}{2} - \frac{x^{1+\frac{5}{2}}}{1+\frac{5}{2}}]\vert^4_0$

$V = 2\pi * [{4x^2 - \frac{x^{\frac{2+5}{2}}}{\frac{2+5}{2}}]\vert^4_0$

$V = 2\pi * [{4x^2 - \frac{x^{\frac{7}{2}}}{\frac{7}{2}}]\vert^4_0$

$V = 2\pi * [{4x^2 - \frac{2}{7}x^{\frac{7}{2}}]\vert^4_0$

Substitute 4 and 0 for x

$V = 2\pi * ([{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}] - [{4*0^2 - \frac{2}{7}*0^{\frac{7}{2}}])$

$V = 2\pi * ([{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}] - [0])$

$V = 2\pi * [{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}]$

$V = 2\pi * [{64 - \frac{2}{7}*2^2^{*\frac{7}{2}}]$

$V = 2\pi * [{64 - \frac{2}{7}*2^7]$

$V = 2\pi * [{64 - \frac{2}{7}*128]$

$V = 2\pi * [{64 - \frac{2*128}{7}]$

$V = 2\pi * [{64 - \frac{256}{7}]$

Take LCM

$V = 2\pi * [\frac{64*7-256}{7}]$

$V = 2\pi * [\frac{448-256}{7}]$

$V = 2\pi * [\frac{192}{7}]$

$V = [\frac{2\pi * 192}{7}]$

$V = \frac{\pi * 384}{7}$

$V = \frac{384}{7}\pi$

Hence, the required volume is:

$Volume = \frac{384}{7}\pi$

3 0

3 years ago

I need help on this question plz

Elis [28]

The answer for this question is chord

4 0

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