1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Rina8888 [55]
3 years ago
15

What is the equation of the line that passes through (-1, 4) and ( 2, -8) written in slope-intercept form?

Mathematics
1 answer:
AfilCa [17]3 years ago
4 0
The answers tell you the y-intercept is zero, so all you need to do is choose the appropriate slope. Since y/x = -4, ...
.. selection A is appropriate.

You might be interested in
The total points scored in the games of an NBA playoff series: 195 177 215 200 189 232 201 199 192 201
Sonja [21]

Step-by-step explanation:

something is missing. maybe mean, median

3 0
3 years ago
1 On a trip, a family drove 270 kilometers in 3 hours. a Find how many kilometers were traveled in one hour. Express this as a r
wolverine [178]
90 kilometers per hour
270/3=90
7 0
2 years ago
Read 2 more answers
Find the exact value of cos(sin^-1(-5/13))
son4ous [18]

bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.

\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}

\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

8 0
3 years ago
How many solutions does the equation have?
Andreyy89

Answer:

Since it is a linear equation, it has one solution.

step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Can someone help me ?
lozanna [386]

Answer:

10 > v (if you don't trust me, check it yourself. Subsitute v for 10 in the first equation.

Step-by-step explanation:

1) 7 > v - 3

2) add 3 to both sides

3) so 7 plus 3 equals 10

So you basically do this

7 > v - 3 (ad 3 to both sides)

You get 10 > v (which is as simplified as possible)

Hope this helps! :)

8 0
3 years ago
Other questions:
  • There are 4 boxes each box has 6 granola bars your soccer team eats 18 granola bars how many are left
    6·1 answer
  • Identify the slope and Y intercept of the graph of the equation then graph The equation y=4/5x+1
    9·1 answer
  • Which of the following equations has the same solution as m - (-62) = 45?
    7·1 answer
  • A student completed 1/4 of a workbook in 3/5 hour.He plans to work for 1 more hour at the same rate.What Fraction of the workboo
    6·1 answer
  • Find the value of x and the value of y.
    11·1 answer
  • What is the LCM of 144 and 60?
    12·2 answers
  • How Many Solutions?<br> – 3 (v + 4) = 2v - 37<br> Infmute Solution<br> No Solution<br> One Solution
    13·1 answer
  • Calculate X you must include the steps in your solution and the definitions and/or conjuncture that you use
    7·2 answers
  • Helllpppp pls answer both questions !!!! 30pts
    11·1 answer
  • 64 is 4 times the difference between Sarah's age, a, and 44. Assume Sarah is older than 44. (pls show an equation to determine S
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!