Answer:
x=5.6
y=-5.6
Step-by-step explanation:
3x - 5y = 14 Equation 1
– 2x + 2y = 0 Equation 2
Simultaneous equation can be solved either through elimination method or substitution method. But we use elimination method for this question
Multiply equation 1 by -2 (the coefficient of x in equation 2) and multiply equation 2 with 3 (the coefficient of x in equation 1), so that x will have the same coefficient in the new equations, and easy to eliminate
-2(3x - 5y = 14)
-6x+10y=-28 Equation 3
3(– 2x + 2y = 0)
-6x+6y=0 Equation 4
Subtract equation 4 from 3 to eliminate x
-6x+10y=-28
-6x+6y=0
-6x-(-6x)=-6x+6x=0
10y-6y=5y
-28-0=-28
5y=-28
y=-28/5
y=-5.6
Substitute for y in equation 2
– 2x + 2y = 0
– 2x + 2(-5.6) = 0
-2x-11.2=0
-2x=11.2
x=-11.2/2
x=5.6
Answer:
Right skewed Histogram.
Explanation:
As the peak of the graph lies to the left side of the centre.
Answer:
Step-by-step explanation:
A1. C = 104°, b = 16, c = 25
Law of Sines: B = arcsin[b·sinC/c} ≅ 38.4°
A = 180-C-B = 37.6°
Law of Sines: a = c·sinA/sinC ≅ 15.7
A2. B = 56°, b = 17, c = 14
Law of Sines: C = arcsin[c·sinB/b] ≅43.1°
A = 180-B-C = 80.9°
Law of Sines: a = b·sinA/sinB ≅ 20.2
B1. B = 116°, a = 11, c = 15
Law of Cosines: b = √(a² + c² - 2ac·cosB) = 22.2
A = arccos{(b²+c²-a²)/(2bc) ≅26.5°
C = 180-A-B = 37.5°
B2. a=18, b=29, c=30
Law of Cosines: A = arccos{(b²+c²-a²)/(2bc) ≅ 35.5°
Law of Cosines: B = arccos[(a²+c²-b²)/(2ac) = 69.2°
C = 180-A-B = 75.3°
