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m_a_m_a [10]
2 years ago
15

X/5 -8=4 find the value of x

Mathematics
1 answer:
elena-s [515]2 years ago
6 0

Answer:

x=60

Step-by-step explanation:

I'm assuming you meant this: (x/5)-8=4

In which case you would add 8 to both sides to get rid of the 8 on the left (your goal is to get x by itself so you want to move the numbers on the x side to the other side of the equal sign)

(x/5)=12

Then you would multiply 5 on both sides to get rid of the fraction with the 5 on the bottom on the left side.

x=60

There's your answer.

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I need help...............
V125BC [204]

Answer:

2 (a) d=3

(b) e= 21

(c) f= 4

3.(A) 3cm

This is your answer

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:

<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

3 0
3 years ago
5. Give two equations that will cross at (3, 4)
Anton [14]

Answer: 4x=3y, 8x=6y

Step-by-step explanation: We can use a simplified form of an equation for this specific problem, since a solution is (3,4), if you plugin 3 for x and 4 for y you get the same thing. Since 4 times 3 equals 3 times 4 you can call x as 4 and y as 3. You can simply multiply this by 2 for another equation since it doesn't affect the ratio. I hope this helps!

6 0
2 years ago
Consider the functions f(x) = 2x + 1 and g(x) = x^2 - 10. What is the value of f[g(3)]?
ziro4ka [17]

Answer:

f[g(x)]= 2(x^2-10)+1

2x^2-20+1

f[g(3)]=2(3)^2-20+1

18-20+1

18-19=-1

7 0
2 years ago
Evaluate (8+t)³-6 when t = 2.
Sophie [7]

Answer:994

Step-by-step explanation: (8+t) -6, t=2

(8+2)^3 -6

10^3-6

1000-6

994

6 0
1 year ago
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