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kodGreya [7K]
3 years ago
11

HELP ASAP!!!!!!!!!!!!!!!!!!! PLEASE!!

Mathematics
1 answer:
vekshin13 years ago
3 0

Answer:

yes/ In 4 years, Dinah will have gained $102.00.

Step-by-step explanation:

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Hi, I don't feel very confident about my work in this calculus problem and I would appreciate it if anyone else could note any c
Lostsunrise [7]
You're pretty close. It's all in the details.

A. Correct.

B. Correct.

C. Sign error in the boxed answer. -(-0.5) = 0.5
The graph shows c ≈ 1.855. If a is more than that and decreasing, area is increasing. The sign of dA/dt will be positive. (Compare to D)

D. Units are not shown. Numerical value is correct.

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Y=-4x+4
attashe74 [19]
Steps:
1. Substitute the y in y=4x + 4 and put
X-6=-4x+4
2. Add 4x to both sides
3. 5x-6=4
4. Add 6 to both sides
5. 5x=10
6. Divide by 5 on both sides
7. X=2
8. Now to find the y, you substitute 2 for x in the problem y=x-6
9. Y=2-6
10. Y=-4
11. Your answer is
(2, -4)
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3 years ago
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3y+6+4y−7=−8<br><br> Solve for y
Dennis_Churaev [7]
The correct answer for that is y=4
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3 years ago
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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:

<em>Point a:</em>

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.

<em>Point b:</em>

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 <em>y</em> 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)

<em>Point c:</em>

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

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3 years ago
Find the mean median mode of 12,25,23,17,23
lilavasa [31]
12,17,23,23,25
mod = 23
median = 23
 
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