All categories

2 years ago

What is the equation of the line through (-2, 5) and (0, 3)? A. y = x - 3 B. y=x+3 C. y = -x + 3 D. y=-x-3

Mathematics

1 answer:

Artist 52 [7]2 years ago

5 0

Answer:

Step-by-step explanation:

a typical line equation is the slope-intercept form

y = ax + b

"a" is the slope (always the factor of x) expressed as ratio of "y coordinate difference" / "x coordinate difference".

b is the y-intercept (the y value when x = 0).

going from (-2, 5) to (0, 3) we see

x changes by +2 (from -2 to 0).

y changes by -2 (from 5 to 3).

so the slope is -2/+2 = -1

using one of the points (0, 3) to solve for the missing b :

3 = -1×0 + b

b = 3

so, the equation is

y = -x + 3

You might be interested in

On a coordinate plane, a dashed straight line has a positive slope and goes through (negative 3, 1) and (0, 3). Everything to th

miss Akunina [59]

Answer:

$y> \frac{2}{3}x+3$

Step-by-step explanation:

we have the points

(-3,1) and (0,3)

step 1

Find the slope m of the line

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{3-1}{0+3}$

$m=\frac{2}{3}$

step 2

Find the equation of the line in slope intercept form

$y=mx+b$

we have

$m=\frac{2}{3}$

$b=3$ ----> the y-intercept is the point (0,3)

substitute the values

$y=\frac{2}{3}x+3$

step 3

Find the equation of the inequality

we know that

The slope is positive

Everything to the left of the line is shaded ( The inequality is of the form y > ax+b or y ≥ ax+b)

Is a dashed line (The inequality is of the form y > ax+ b or y < ax+b)

therefore

The equation of the inequality is of the form y > ax+b

The inequality is

$y> \frac{2}{3}x+3$

see the attached figure to better understand the problem

4 0

3 years ago

Read 2 more answers

Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin

ozzi

$I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt$