Answer:
D) 4x + 5y = -30 = -4/5
Step-by-step explanation:
A) 5x + 4y = 28 slope = -5/4
b) 4x-5y = -25 slope = 4/5
c) 4y - 5x = -12 slope = 5/4
d) 4x + 5y = -30 slope = -4/5
Step-by-step explanation:
<em>2</em><em>:</em><em>1</em><em>0</em><em>:</em><em>1</em><em>6</em><em>=</em><em>2</em><em>/</em><em>1</em><em>0</em><em>=</em><em>1</em><em>/</em><em>5</em><em>÷</em><em>1</em><em>6</em>
<em>
</em>
<em>1</em><em>/</em><em>5</em><em>×</em><em>1</em><em>/</em><em>1</em><em>6</em>
<em>=</em><em>1</em><em>/</em><em>8</em><em>0</em>
It looks like the given equation is
sin(2x) - sin(2x) cos(2x) = sin(4x)
Recall the double angle identity for sine:
sin(2x) = 2 sin(x) cos(x)
which lets us rewrite the equation as
sin(2x) - sin(2x) cos(2x) = 2 sin(2x) cos(2x)
Move everything over to one side and factorize:
sin(2x) - sin(2x) cos(2x) - 2 sin(2x) cos(2x) = 0
sin(2x) - 3 sin(2x) cos(2x) = 0
sin(2x) (1 - 3 cos(2x)) = 0
Then we have two families of solutions,
sin(2x) = 0 or 1 - 3 cos(2x) = 0
sin(2x) = 0 or cos(2x) = 1/3
[2x = arcsin(0) + 2nπ or 2x = π - arcsin(0) + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
(where n is any integer)
[2x = 2nπ or 2x = π + 2nπ]
… … … or [2x = arccos(1/3) + 2nπ or 2x = -arccos(1/3) + 2nπ]
[x = nπ or x = π/2 + nπ]
… … … or [x = 1/2 arccos(1/3) + nπ or x = -1/2 arccos(1/3) + nπ]
QS bisects the ^PQR^ :
So -----¢ PQS = SQR = 45°
_________________________________
PQR = PQS + SQR = 45°+ 45° = 90 °
Q^ = 90° ----¢ So the PQR triangle is a
right triangle.
