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3 years ago

10

Omar is 14 years old and 6 feet tall. If he has grown .5 feet each year for the past 3 years, how tall was he when he was 12 yea

rs old?

2 answers:

Verizon [17]3 years ago

7 0

He was 4 foot. With this I just took .5 and multiplyed it by 4, then subtracted it from 6. Hope this can help! Feel free to message me if you have anymore questions!

liq [111]3 years ago

5 0

He was 12 years old 2 years ago, which means he grew 1 ft (2*.5=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

2 years ago

A particular car is said to depreciate 15% a year. If the new car was valued at $20,000, what will it be worth after 6 years?

Margarita [4]

Answer:7542.99 dollars

Step-by-step explanation:20000 (base price) x (1 - .15) (depreciation value) to the 6th power

20000(0.85)⁶

7 0

3 years ago

Please please answer these two (16-17) questions

valkas [14]

This would be C and B

3 0

3 years ago

Which number is equivalent to (1/3)^-4<br><br> A.12<br> B.81<br> C.1/12<br> D.1/81

marishachu [46]

Answer:

B

Step-by-step explanation:

(1/3)^-4

=1/[(1/3)^4]

=1/(1/81)

=81

6 0

2 years ago

The cost of K-12 education per student was $2200 in 1978 and increased to $10,300 in 2008. (Source: Department of Education). (a

kobusy [5.1K]

Answer: $C=270x+2200$

Step-by-step explanation:

Given

Cost per student in the year 1978 is $2200

The cost increases to $10,300 in 2008

Suppose $C=mx+a$ is the linear equation defining the cost per student after x years

Substitute the value for 1978

$\Rightarrow 2200=m(0)+c\\\Rightarrow c=2200$

After 30 years it becomes

$\Rightarrow 10,300=m(30)+2200\\\Rightarrow 30m=8100\\\Rightarrow m=270$

Thus, the linear equation becomes

$\Rightarrow C=270x+2200$

6 0

2 years ago