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3 years ago

Urgent im not sure look at the picture

Mathematics

2 answers:

andreev551 [17]3 years ago

8 0

It’s would be letter B 43.4 miles

Dmitrij [34]3 years ago

3 0

Answer:

43.4 mph

Step-by-step explanation:

super simple, you can disregard all the speed stuff, since they are both hours, it is useless information. all you have to do is convert distances, so multiply 70 by .62 to get 43.4milers per hour.

if you have any questions, leave them in the comments, i will try to answer them, pls give brainliest if this helps.

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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

Given directed line segment PR below, find the coordinates of Q on PR

g100num [7]

Answer:

$(\frac{4}{3},-3)$

Step-by-step explanation:

The location of a point O(x, y) which divides line segment AB in the ratio a:b with point A at ( $x_1,y_1$ ) and B( $x_2,y_2$ ) is given by the formula:

$x=\frac{a}{a+b}(x_2-x_1)+x_1\\ \\y=\frac{a}{a+b}(y_2-y_1)+y_1$

From the graph, the location of the point is P(2, 5), R(1, -7). Let point Q be (x, y) which divides PR in ratio 2:1. Coordinates of point Q is at:

$x=\frac{a}{a+b}(x_2-x_1)+x_1=\frac{2}{2+1}(1-2)+2=\frac{2}{3}(-1)+2=\frac{4}{3} \\ \\y=\frac{a}{a+b}(y_2-y_1)+y_1=\frac{2}{2+1}(-7-5)+5= \frac{2}{3}(-12)+5=-3$