Jeff has a cube train that is 22 inches long. he removes 8 inches of cubes from the train. How long is Jeff's cube train now?

Write the equation in slope-intercept form through the point (-4, -7) and is perpendicular to the line y = -(7/4)x + 4 and graph

Scilla [17]

You have to write the equation for a line that crosses the point (-4, -7) and is perpendicular to the line

$y=-\frac{7}{4}x+4$

When you have to determine a line that is perpendicular to a known line, you have to keep in mind that the slope of the perpendicular line will be the negative inverse of the first one.

If for exampla you have two lines, the first one being:

$y_1=mx+b$

And the second one, that is perpedicular to the one above:

$y_2=nx+c$

The slope of the second one is the negative inverse of the first one:

$n=-\frac{1}{m}$

The slope of the given line y=-7/4+4 is m=-7/4

So the slope of the perpendicular line has to ve the inverse negative of -7/4

$\begin{gathered} n=-(-\frac{4}{7}) \\ n=\frac{4}{7} \end{gathered}$

Considering it has to pass through the point (-4,-7) and that we already determined its slope, you can unse the point slope formula to determine the equation of the perpendicular line:

$y-y_1=m(x-x_1)$

replace with the coordinates of the point and the slope and calculate:

$\begin{gathered} y-(-7)=\frac{4}{7}(x-(-4)) \\ y+7=\frac{4}{7}(x+4) \\ y+7=\frac{4}{7}x+\frac{16}{7} \end{gathered}$

Subtract 7 to both sides of the equation to write it in slope-intercept form:

$\begin{gathered} y+7-7=\frac{4}{7}x+\frac{16}{7}-7 \\ y=\frac{4}{7}x-\frac{33}{7} \end{gathered}$

Now you can graph both lines

8 0

1 year ago

!!HELP ME PLEASE!!

gulaghasi [49]

Answer:

answers

Step-by-step explanation:

A. Vertex at (−6, 1)

6 0

3 years ago

The graph of the function f(x)=|3x| is translated 4 units up.

siniylev [52]

Translating a graph upwards just means adding or subtracting a value from the initial function.

So to push a graph up, you just add (+) the number of units onto the original function.

7 0

2 years ago

Read 2 more answers

There is 3 4 of an apple pie left from dinner. Tomorrow, Victor plans to eat 1 4 of the pie that was left. Enter how much of the

lora16 [44]

Answer:

3/16

Step-by-step explanation:

Let the size of the whole apple be T

Given that 3/4 of the apple is left, it means the size left

= 3T/4

Since Victor plans to eat 1/4 of the pie that was left, this amounts to

1/4 * 3T/4

= 3T/16

This means that Victor will eat 3/16 of the whole pie tomorrow

5 0

2 years ago

Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).

I am Lyosha [343]

Take the homogeneous part and find the roots to the characteristic equation:

$y''+4y=0\implies r^2+4=0\implies r=\pm2i$

This means the characteristic solution is $y_c=C_1\cos2x+C_2\sin2x$ .

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form $y_p=ax\cos2x+bx\sin2x$ . Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With $y_1=\cos2x$ and $y_2=\sin2x$ , you're looking for a particular solution of the form $y_p=u_1y_1+u_2y_2$ . The functions $u_i$ satisfy

$u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx$

where $W(y_1,y_2)$ is the Wronskian determinant of the two characteristic solutions.

$W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2$

So you have

$u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx$
$u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x$

$u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx$
$u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x$

So you end up with a solution

$u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x$

but since $\cos2x$ is already accounted for in the characteristic solution, the particular solution is then

$y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x$

so that the general solution is

$y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x$

7 0

3 years ago