Given that the line y=ax+b passes through the points A(0; 3) and B(-2;7), how do I find the values of a and b?

1 answer:

4 0

First you have find the slope (a) in order to find the value of the y intercept (b).

To compute the slope, you simply have to find the difference of y coordinates and divide them by the difference of x coordinates:

a = (y2-y1)/(x2-x1)
a = (7-3)/(-2–0)
a = 4/-2
a = -2

To compute the y intercept denoted by b, we substitute the value of a with -2 in the formula and choose any point (x,y) and substitute their values in the formula as well. Like so:

I choose point A:

3 = (-2)(0) + b
3 = b

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What is the factors of x^2-x-6

aniked [119]

Think of two numbers that multiply to give 6 and add or take-away to give 1. Hmmm... Well 3 and 2 will work because their difference is 1 and 3 x 2 = 6 so now we need to think about the signs. We have

x²-x-6 = (x _ 3)(x _ 2).

When we times them together we get -6. To get a minus, one needs to be a + and the other needs to be a -. We need to get them to add to -1 so we can choose
I) 3-2 = 1 (nope)
II) 2-3 = -1 (yeah)
so it must be -3 and +2. This gives x²-x-6 = (x - 3)(x + 2).

3 0

3 years ago

A circle is centered at the point O (0,0) and has a radius of the square root of 38?

nekit [7.7K]

Answer:

The point (6,1) lies inside the circle.

Step-by-step explanation:

Equation of a Circle

A circle centered at the point (h,k) and with radius r, can be expressed with the equation:

$(x-h)^2+(y-k)^2=r^2$

The center of the given circle is (0,0) and the radius is $r=\sqrt{38}$ , thus the equation of the circle is:

$(x-0)^2+(y-0)^2=(\sqrt{38})^2$

Simplifying:

$x^2+y^2=38$

Each point (x,y) meeting the condition of the equation lies exactly on its circumference. For example, the point $(\sqrt{2},6)$ belongs to the circumference, but the origin (0,0) is inside the circle, and the point (5,5) is outside the circle.