What is the slope and y- intercept of the line passing through (11,-3) and (7,9)

Mathematics

1 answer:

Misha Larkins [42]3 years ago

7 0

Answer:

Slope is -3, y-intercept is 30

Step-by-step explanation:

You take the two points (11,-3) and (7,9), plug them into the formula y2-y1/x2-x2=m. Now, you have 9-(-3)/7-11 which gives you 12/-4, 12/-4 can be simplified into -3 which is your slope. To find y-intercept you must use the formula y=mx+b. Plug -3, your slope, into m, and any of the coordinates on your line into the y and x; I used (7,9). You solve for b from 9=-3*7+b. -3*7 gives you -21, now you have 9=-21+b, to get b by itself, you add 21 to both sides, 30=b. B is your y-intercept.

You might be interested in

if 40% of the students in the class are boys , and if there are 18 girls in the class , then how many students are in the class?

chubhunter [2.5K]

30

60% were girls so you put this as a ratio

60/100= 18/x

You calculate x to determine the total number which would equal 30.

5 0

3 years ago

Write the equation of the line in slope-intercept form using y=mx+b

artcher [175]

Answer:
y=2x+0

Explanation:
To find the “b” of this equation you simply see where the line hits the y- intercept , in this case it will be 0. And to find the “m” of this equation you find the rise over run . Hope that helps.

3 0

3 years ago

What is the distance of KF. round to the nearest tenths

snow_tiger [21]

Answer:

its 9.2

Step-by-step explanation:

Well i counted 7 units (x-axis) and 6 units (y axis). 6 squared is 36 and 7 squared is 49. Add them together and get 85. Square root 85 and youll get 9.21954445729. Round it to the nearest tenths and get 9.2

5 0

3 years ago

Solve this identifying holes, vertical asymptotes, and horizontal asymptotes for

USPshnik [31]

$x^2+7x+12=x^2+4x+3x+12=x(x+4)+3(x+4)=(x+4)(x+3)\\\\-x^2-3x+4=-(x^2+3x-4)=-(x^2+4x-x-4)\\\\=-[x(x+4)-1(x+4)]=-(x+4)(x-1)\\\\f(x)=\dfrac{x^2+7x+12}{-x^2-3x+4}=\dfrac{(x+4)(x+3)}{-(x+4)(x-1)}\\\\\text{Vertical asymptotes:}\\\\(x+4)(x-1)=0\iff x+4=0\ \vee\ x-1=0\\\\\boxed{x=-4\ and\ x=1}\\\\\text{Horizontal asymptotes:}$

$\lim\limits_{x\to\pm\infty}\dfrac{x^2+7x+12}{-x^2-3x+4}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\dfrac{7}{x}+\dfrac{12}{x^2}\right)}{x^2\left(-1-\dfrac{3}{x}+\dfrac{4}{x^2}\right)}=\dfrac{1}{-1}=-1\\\\\boxed{y=-1}$

6 0

4 years ago

Unit Test Review

LiRa [457]

Answer:

x - 1 < n < 3x + 5

Step-by-step explanation:

If you have two sides of a triangle, the length of the third side is at least the difference between the other two sides, and at most the sum of the other two sides.

3 0

3 years ago