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2 years ago

Write the equation of the line that passes through the points (−7,0) and (6,−4).

Mathematics

1 answer:

Over [174]2 years ago

5 0

Answer:

y=-4/13x-28/13

Step-by-step explanation:

m=y2-y1/x2-x1

m=-4-0/6-(-7)

m=-4-0/6+7

m=-4/13

y-y1=m(x-x1)

y-0=-4/13(x-(-7))

y-0=-4/13x-28/13

y=-4/13x-28/13

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Which of the following is the equation of a line perpendicular to the line

Taya2010 [7]

Answer:

A. - 2x + 3y = 21

Step-by-step explanation:

The perpendicular lines have opposite reciprocal slopes.

The given line has a slope of:

m = -3/2

The perpendicular slope is:

m = 2/3

And its y-intercept is:

9 = 2/3*3 + b
b = 7

The line is:

y = 2/3x + 7

Convert this into standard form and compare with answer choices:

3y = 2x + 21
- 2x + 3y = 21

Correct choice is A

3 0

2 years ago

What is f(x)=14−0.5x? solve for f(30)

UkoKoshka [18]

The function f(x) is a function of the variable x.

To solve the function for x = 30, just plug in 30 for x.

14 - 0.5(30) = -1

f(30) = -1

3 0

2 years ago

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term f here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

3 years ago

Home heating oil is sold by the gallon. Last winter the romano family used 370 gallons of oil at a price of $3.91 per gallon. If

shusha [124]

($3.91)(0.09) = $0.3519

$3.91 + 0.3519 = $4.2619 per gallon

if you're looking for the total expense next year:

($4.2619)(370 gallons) = $1,576.903

6 0

2 years ago

Complete the sentence. 7 is 1/10 of

Schach [20]

7 is 1/10 of.. 70, because 10 times 7 is 70 :)

5 0

2 years ago

Read 2 more answers