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3 years ago

A square with an area of 49 in2 is rotated to form a cylinder. What is the volume of the cylinder

1 answer:

8 0

343 pi in^3 ; exact volume 1078 in^3 ; Approximate volume Since it's not specified, I will assume the axis of rotation will be one edge of the square. With that in mind, here's the solution. Since the shape specified is a square with an area of 49 in^2, the length of any edge will be sqrt(49) = 7 inches. Since we're rotating along one edge, we will create a cylinder with a radius of 7 inches and a height of 7 inches. The volume will be the area of a circle with a 7 inch radius multiplied by the height of the cylinder. So V = pi*h*r^2 V = pi*7*7^2 V = pi*7*49 V = pi*343 V = 343pi V = 1077.56628; approximately. So the exact volume is 343pi in^3 and an approximate volume is 1078 in^3

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I need help on this

Naily [24]

Answer:x=90° y=58° z=32°

Step-by-step explanation:

x=90

y=180-(90+32) sum of angles in a triangle is 180

y=58

y+90+z=180 they are supplementary

58+90+z=180

148+z=180

z=180-148

z=32

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3 years ago

What is 10 1/12 as a decimal

cricket20 [7]

Answer:

10.083

since 1/12 is .0833333333 and so on, it’s shortened

Step-by-step explanation:

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3 years ago

Read 2 more answers

Solve for x. Write your answer in the space below, x + 5 —— = 10 2

Travka [436]

Answer:

x=15

Step-by-step explanation:

15+5=20

20/2=10

So, 15 =x

8 0

2 years ago

Read 2 more answers

Please help I don't under stand how to do this

stealth61 [152]

You want to figure out what the variables equal to, all of these are parallelograms meaning opposite sides and angles are equal to each other.

In question 1 start with 3x+10=43, this means that 3x is 10 less than 43 which is 33, 33 divided by 3 is 11 meaning x=11.

Same thing can be done with the sides 124=4(4y-1), start by getting rid of the parentheses with multiplication to get 124=16y-4, this means that 16y is 4 more than 124, so how many times does 16 go into 128? 8 times, so x=11 and y=8

Question 2 can be solved because opposite angles are the same in a parallelogram, so u=66 degrees

You can find the sum of the interial angles with the formula 180(n-2) where n is the number of sides the shape has, a 4 sided shape has a sum of 360 degrees, so if we already have 2 angles that add up to a total of 132 degrees and there are only 2 angles left and both of those 2 angles have to be the same value then it’s as simple as dividing the remainder in half, 360-132=228 so the other 2 angles would each be 114, 114 divided into 3 parts is 38 so u=66 and v=38

Question 3 and 4 can be solved using the same rules used in question 1 and 2, just set the opposite sides equal to each other

6 0

2 years ago

Let y(t) be the solution to y˙=3te−y satisfying y(0)=3 . (a) Use Euler's Method with time step h=0.2 to approximate y(0.2),y(0.4

OLEGan [10]

Answer:

$y(0.2)=3$ , $y(0.4)=3.005974448$ , $y(0.6)=3.017852169$ , $y(0.8)=3.035458382$ , and $y(1.0)=3.058523645$

The general solution is $y=\ln \left(\frac{3t^2}{2}+e^3\right)$

The error in the approximations to y(0.2), y(0.6), and y(1):

$|y(0.2)-y_{1}|=0.002982771$

$|y(0.6)-y_{3}|=0.008677796$

$|y(1)-y_{5}|=0.013499859$

Step-by-step explanation:

Point a:

The Euler's method states that:

$y_{n+1}=y_n+h \cdot f \left(t_n, y_n \right)$ where $t_{n+1}=t_n + h$

We have that $h=0.2$ , $t_{0}=0$ , $y_{0} =3$ , $f(t,y)=3te^{-y}$

We need to find $y(0.2)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{1}=t_{0}+h=0+0.2=0.2$

$y\left(t_{1}\right)=y\left(0.2)=y_{1}=y_{0}+h \cdot f \left(t_{0}, y_{0} \right)=3+h \cdot f \left(0, 3 \right)=$

$=3 + 0.2 \cdot \left(0 \right)= 3$

$y(0.2)=3$

We need to find $y(0.4)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{2}=t_{1}+h=0.2+0.2=0.4$

$y\left(t_{2}\right)=y\left(0.4)=y_{2}=y_{1}+h \cdot f \left(t_{1}, y_{1} \right)=3+h \cdot f \left(0.2, 3 \right)=$

$=3 + 0.2 \cdot \left(0.02987224102)= 3.005974448$

$y(0.4)=3.005974448$

The Euler's Method is detailed in the following table.

Point b:

To find the general solution of $y'=3te^{-y}$ you need to:

Rewrite in the form of a first order separable ODE:

$e^yy'\:=3t\\e^y\cdot \frac{dy}{dt} =3t\\e^y \:dy\:=3t\:dt$

Integrate each side:

$\int \:e^ydy=e^y+C$

$\int \:3t\:dt=\frac{3t^2}{2}+C$

$e^y+C=\frac{3t^2}{2}+C\\e^y=\frac{3t^2}{2}+C_{1}$

We know the initial condition y(0) = 3, we are going to use it to find the value of $C_{1}$

$e^3=\frac{3\left(0\right)^2}{2}+C_1\\C_1=e^3$

So we have:

$e^y=\frac{3t^2}{2}+e^3$

Solving for y we get:

$\ln \left(e^y\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y\ln \left(e\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y=\ln \left(\frac{3t^2}{2}+e^3\right)$