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2 years ago

Find an equation in slope-intercept form of the line that has slope –8 and passes through point (1,5)

1 answer:

5 0

Using the equation of the line
y -yo = m (x-xo)

We substitute xo,yo and the slope

y -5 = -8(x-1) .....> y = -8x +8 +5 = -8x +13

y = -8x +13

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Someone help with question number 3 asap

mylen [45]

Answer:

Please check the attached file.

Hope this helps!

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3 years ago

Read 2 more answers

Help pretty please!

Elodia [21]

Answer:

Describe in detail how you would create a number line with the following points: 3, 2.25, the opposite of 4, and – (–1fraction of one-half). Please be sure to describe on which tick marks each point is plotted and how many tick marks are between each integer. It may help for you to draw this number line by hand on a sheet of paper first.

Step-by-step explanation:

3 0

2 years ago

Write the vector u as a sum of two orthogonal vectors, one of which is the vector projection of u onto v, projvu u=(-8,8) , v= (

Jobisdone [24]

Answer:

none of the given options is true

Step-by-step explanation:

Given: u=(-8,8) , v= (-1,2)

To find: vectors $w_1,w_2$ such that $u=w_1+w_2$

Solution:

A vector is a quantity that has both magnitude and direction.

$proj_vu=\frac{u\cdot v}{\left | v \right |^2}(v)\\=\frac{(-8,8)\cdot (-1,2)}{(\sqrt{ (-1)^2+2^2})^2}(-1,2)\\=\frac{8+16}{5}(-1,2)\\=\frac{1}{5}(-24,48)\\=(-4.8,9.6)$

Let $w_1=(-4.8,9.6)$

So,

$w_2=u-w_1\\=(-8,8)-(-4.8,9.6)\\=(-8+4.8,8-9.6)\\=(-3.2,-1.6)$

So, u = $(-4.8,9.6)+(-3.2,-1.6)$

So, none of the given options is true

8 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:

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