What is the slope of a trend line that passes through the points (1.3) and (10, 25)?
-15/2
-1/2
22/9
24/7

22/9

Step-by-step explanation :slope of a trend line that passes through the points (1, 3) and (10, 25) is  22/9

Formula of slope(m) for two points is given by:

slope m: y^2-y^1/x^2-x^1

Given the two points:

(1,3) and (10,25)

Substitute these points in the given formula we have;

m= 25-3/10-1

Simplify:

m= 22/9

Therefore, the slope of a trend line that passes through the points (1, 3) and (10, 25) is  22/9

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solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw   congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.

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step-by-step explanation: