Answer:
6x² + 72x - 170 = -2
<=> 6x² + 72x - 168 = 0
<=> x² + 12x - 28 = 0
<=> x² + 12x + 36 = 64
<=> (x+6)² = 64
Step-by-step explanation:
Answer:
Step-by-step explanation:
2) ∠A + ∠B = 146 {Exterior angle theorem}
5y + 3 + 4y + 8 = 146
5y + 4y + 3 + 8 = 146 {combine like terms}
9y + 11 = 146
9y = 146 -11
9y = 135
y = 135/9
y = 15
3) m∠A = 5y + 3
= 5*15 + 3
= 75 + 3
∠A = 78
4) m∠B = 4y + 8
= 4*15 + 8
= 60 + 8
m∠B = 68
5) m∠ACB + m∠A + m∠B = 180 {Angle sum property of triangle}
m∠ACB + 78 + 68 = 180
m∠ACB + 146 = 180
m∠ACB = 180 - 146
m∠ACB = 34
True
False
Answer:
True dat my dudes
Answer:
sin α = =
cos α = =
tan α = =
cot α = =
sec α = =
csc α = =
Step-by-step explanation:
If the point is given on the terminal side of an angle, then:
Calculate the distance between the point given and the origin:
r =
Here it is: = =
So we have:
x = 3
y = 1
r =
Now we can calculate all 6 trig, functions:
sin α = =
cos α = =
tan α = =
cot α = =
sec α = =
csc α = =
The question involves calculating the six trigonometric functions (Sin, Cos, Tan, Csc, Sec, Cot) for an angle in standard position with its terminal side passing through point (3,1). We find opposite and adjacent sides from the point's coordinates, the hypotenuse from Pythagoras' theorem, and calculate each ratio accordingly.
In the context of the problem, the point (3,1) specifies the terminal side of angle θ in standard position. This point is effectively the definition of your right triangle's opposite (y) and adjacent (x) sides in the trigonometric calculation, with the hypotenuse (h) obtained from Pythagoras' Theorem (h² = y² + x²). In this case, we get h = sqrt(3² + 1²) = sqrt(10).
The six trigonometric functions are defined as follows for this scenario:
#SPJ3