Get unknowns on one side and know on other side.
Divide both sides by -2
X= -7
Step-by-step explanation:






Taking sin²θ common in both numerator & denominator, We get :










<u>Hence</u><u>,</u><u> option</u><u> </u><u>(</u><u>a)</u><u> </u><u>2</u><u>/</u><u>3</u><u> </u><u>is </u><u>your</u><u> </u><u>correct</u><u> </u><u>answer</u><u>.</u>
Answer:
Option A.
Step-by-step explanation:
Graph of function has been given in the figure.
Coordinates of the points lying on this graph will be
x -2 -1 0 1 2
y -10 -3 -2 -1 6
Therefore, coordinates of the points which will lie on the inverse of this function will be
x -10 -3 -2 -1 6
y -2 -1 0 1 2
Therefore, Option A. will be the answer.
Answer:
20%
Step-by-step explanation:
1/5 as a percent is 20%
The 1 in 1/5 is equal to 10.
The 5 in 1/5 is equal to 50.
Therefore you would multiply 50 x 2 to get 100.
Then, you would multiply 10 x 2 to get 20/100 or 20%
It depends on how fast it was thrown/kicked but the average is around 9/10