What you want to do here is take this information and plug it into point-slope form. any time you're given a point and a slope, you generally want to plug it into this equation: y - y1 = m(x - x1).
in this equation, m is your slope and (x1, y1) is a given point. plug in your info--slope of -3 and (-5, 2).
y - 2 = -3(x + 5)
that is the equation of your line. however, if you want to graph it, this doesn't really make much sense to you. convert it to slope-intercept form, y = mx + b, by solving for y.
y - 2 = -3(x + 5) ... distribute -3
y - 2 = -3x - 15 ... add 2
y = -3x - 13 is your equation.
to graph this, and any other y = mx + b equation, you want to start with your y-intercept if it's present. your y intercept here is -13, which means the line you wasn't to graph crosses the y-axis at y = -13, or (0, -13). put a point there.
after you've plotted that point, you use your slope to graph more. remember that your slope is "rise over run"--you rise up/go down however many units, you run left/right however many units. if your slope is -3, you want to go down 3 units, then go to the right 1 unit. remember that whole numbers have a 1 beneath them as a fraction. -3/1 is your rise over 1.
Answer:
y = 4x - 1
Step-by-step explanation:
y = mx + b
Here m is slope and b is y-intercept
m = 4; b = -1
y = 4x - 1
Answer:
See below.
Step-by-step explanation:
A. k = 50 * 3.6 = 180.
B. The equation is y = 180/x.
C. When x = 75, y = 180/75
= 2.4.
D. Checking C: 2.4 = 180/75 = 2.4. Correct
Answer:
Use ≈ (approximately equal) sign as the scientific notation.
Step-by-step explanation:
(a) 0.001872 ≈ 0.0019
(b) 0.3411 ≈ 0.34
(c) 0.000845 ≈ 0.00085
*Zeros before non-zero numbers are not significant.
*Zeros appearing between two non-zero digits are significant.
Hope this helps!!
Answer:
2x - y - 3z = 0
Step-by-step explanation:
Since the set
{i, j} = {(1,0), (0,1)}
is a base in 
and F is linear, then
<em>{F(1,0), F(0,1)} </em>
would be a base of the plane generated by F.
F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k
F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k
Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k
We need a normal vector which is the cross product of 3i+2k and 4i-j+3k
(3i+2k)X(4i-j+3k) = 2i-j-3k
The equation of the plane whose normal vector is 2i-j-3k and contains the point (3,0,2) (the end of the vector F(1,0)) is given by
2(x-3) -1(y-0) -3(z-2) = 0
or what is the same
2x - y - 3z = 0
