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2 years ago

A: 37.68 B: 6.28 C: 18.84 D: 12.56

Mathematics

1 answer:

lesya [120]2 years ago

3 0

Answer:

Step-by-step explanation:

(1/3) * π * 2^2 * 3 = 4π = 12.56

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In a sale, normal prices are reduced by 20%. Andrew bought a saddle for his horse in the sale. The sale price of the saddle was

pshichka [43]

The formula for this is: (sale price) / (1 - percentage)
Percentage have to be in decimal form, so divide percentage by 100 to get the decimal form.

220/(1-.20) = $275

The original price was $275.

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3 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:

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

Point c:

To compute the error in the approximations y(0.2), y(0.6), and y(1) you need to:

Find the values y(0.2), y(0.6), and y(1) using $y=\ln \left(\frac{3t^2}{2}+e^3\right)$

$y(0.2)=\ln \left(\frac{3(0.2)^2}{2}+e^3\right)=3.002982771$

$y(0.6)=\ln \left(\frac{3(0.6)^2}{2}+e^3\right)=3.026529965$

$y(1)=\ln \left(\frac{3(1)^2}{2}+e^3\right)=3.072023504$

Next, where $y_{1}, y_{3}, \:and \:y_{5}$ are from the table.

$|y(0.2)-y_{1}|=|3.002982771-3|=0.002982771$

$|y(0.6)-y_{3}|=|3.026529965-3.017852169|=0.008677796$

$|y(1)-y_{5}|=|3.072023504-3.058523645|=0.013499859$

3 0

3 years ago

Find an equation of the circle that has center (-2, 6) and passes through (-6, 1).

Darya [45]

Answer:

(x+2)^2 + (y-6)^2 = 41

Step-by-step explanation:

The equation for a circle is (x-h)^2+(y-k)^2=r^2

So first, it is known that the circle's center is at (-2,6), Therefore, this can be filled in:

(x+2)^2+(y-6)^2=r^2

Next, we need to find the radius, and one of the points is already known, being (-6, 1)

With this, find the distance between these two points by doing the Pythagorean Theorem, a^2+b^2=c^2. The a^2 would be the x value changed and the b^2 would be the y value changed between the two numbers. Note that this is interchangeable.

To find a:

-2 to -6 = change of 4

To find b:

6 to 1 = change of 5

Next, write out the equation for this:

4^2+5^2=c^2

16+25=c^2

41=c^2

c = √41

The radius would be √41, so the equation can now be completed. Since c will be brought to the second power, this will cancel out the square root.

(x+2)^2 + (y-6)^2 = 41

Hope that helps.

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2 years ago

Simplify (7x+6)+(2x-1) final in standard form

Elodia [21]

Answer:

9x+5 this is the answer

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3 years ago

Over time the value at property on 397 west lake street increased by 275. If the initial value of the property was p which expre

kari74 [83]

Answer:

Step-by-step explanation:

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2 years ago