Which equation satisfies all three pairs of a and b values listed in the table? a b 0 -10 1 -7 2 -4

Step-by-step explanation:

The function seems to increase at a constant rate of 3 units. This means that the equation that satisfies these points could be that of a straight line.

We use the first two points to find the slope of the line.

(0, -10)

(1, -7)

(2, -4)

The equation of a line is:

Where m is the slope of the line and b is the cutoff point with the y axis.

To find the slope of a line we use the following equation:

So:

The cut point (b) is found by replacing in the previous equation, any of the three points provided and clearing b.

Now we can write the equation of the line sought.

You can verify that the three points provided belong to this equation.

Step-by-step explanation:

An equation that satisfies all the values of a and b as listed is equation of line that passes through points A (0,-10), B (1,-7) and C (2,-4).

To understand this, first of all find points on a graph and connect the points. The result is a straight line passing through A, B and C.

The equation of straight line is given by:

where m is slope of line and b is the y-intercept.

The slope of a line is given as :

where dy is change in y and dx is change in x.

To find slope consider any two points from line. Let us consider A and C for this example. A(0,-10) is starting point of line and C (2,-4) is ending point of line ( we can also consider C as start point. It simply depends on choice).

Therefore, m= -4 -(-10) / 2-0 = -4+10/2= 6/2=3

Slope m= 3

On substituting value of m into equation.

To find b, take any point A, B or C and simply put the value of y and x into the equation. We do this as A , B and C are simply solutions of the equation and thus can be used.

Taking C and substituting values:

-4= 3*2 + b

b=-10

The resultant equation is as follows:

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

3 qrs + 3/2 xy hope this : /

step-by-step explanation: