Answer: m = 20.
Step-by-step explanation: A triangle has to be 180 degrees. 180-135-25=20.
Use the ordered pairs to first find the slope then substitution into y=mx+b to find b
Answer:
-16/65
Step-by-step explanation:
Given sinα = 3/5 in quadrant 1;
Since sinα = opp/hyp
opp = 3
hyp = 5
adj^2 = hyp^2 - opp^2
adj^2 = 5^2 = 3^2
adj^2 = 25-9
adj^2 = 16
adj = 4
Since all the trig identity are positive in Quadrant 1, hence;
cosα = adj/hyp = 4/5
Similarly, if tanβ = 5/12 in Quadrant III,
According to trig identity
tan theta = opp/adj
opp = 5
adj = 12
hyp^2 = opp^2+adj^2
hyp^2 = 5^2+12^2
hyp^2 = 25+144
hyp^2 = 169
hyp = 13
Since only tan is positive in Quadrant III, then;
sinβ = -5/13
cosβ = -12/13
Get the required expression;
sin(α - β) = sinαcosβ - cosαsinβ
Substitute the given values
sin(α - β) = 3/5(-12/13) - 4/5(-5/13)
sin(α - β)= -36/65 + 20/65
sin(α - β) = -16/65
Hence the value of sin(α - β) is -16/65
You have to show a picture so we can solve this problem you’re having.
Answer:
To construct the Square using a compass and straight edge, you have written the steps not in proper order. I am rearranging the steps for you
1. Construct horizontal H J¯¯¯¯¯
2. Construct a circle with point H as the center and a circle with point J as a center with each circle having radius HJ .
3.Label the point of intersection of the two circles above HJ¯¯¯¯¯ , point K, and the point of intersection of the two circles below HJ¯¯¯¯¯ , point L.
4.Construct KL¯¯¯¯¯ , the perpendicular bisector of HJ¯¯¯¯¯ , intersecting HJ¯¯¯¯¯ at point M.
5.Construct a circle with point M as the center with radius MJ .
6.Label the point of intersection of circle M and KL¯¯¯¯¯ closest to point K, point N, and the point of intersection of circle M and KL¯¯¯¯¯ closest to point L, point O.
7.Construct HN¯¯¯¯¯¯ , NJ¯¯¯¯¯ , JO¯¯¯¯¯ , and OH¯¯¯¯¯¯ to complete square HNJO .
