Answer:
•cos(s+t) = cos(s)cos(t) - sin(s)sin(t) = (-⅖).(-⅗) - (√21 /5).(⅘) = +6/25 - 4√21 /25 = (6-4√21)/25
•cos(s-t) = cos(s)cos(t) + sin(s)sin(t) = (-⅖).(-⅗) + (√21 /5).(⅘) = +6/25 + 4√21 /25 = (6+4√21)/25
cos(t) = ±√(1 - sin²(t)) → -√(1 - sin²(t)) = -√(1 - (⅘)²) = -⅗
sin(s) = ±√(1 - cos²(s)) → +√(1- cos²(s)) = +√(1 - (-⅖)²) = √21 /5
<u>ANSWER TO PART A</u>
The given triangle has vertices 
The mapping for rotation through
counterclockwise has the mapping

Therefore



We plot all this point and connect them with straight lines.
ANSWER TO PART B
For a reflection across the y-axis we negate the x coordinates.
The mapping is

Therefore



We plot all this point and connect them with straight lines.
See graph in attachment
Answer:
The height of the ball after 3 secs of dropping is 16 feet.
Step-by-step explanation:
Given:
height from which the ball is dropped = 160 foot
Time = t seconds
Function h(t)=160-16t^2.
To Find:
High will the ball be after 3 seconds = ?
Solution:
Here the time ‘t’ is already given to us as 3 secs.
We also have the relationship between the height and time given to us in the question.
So, to find the height at which the ball will be 3 secs after dropping we have to insert 3 secs in palce of ‘t’ as follows:


h(3)=160-144
h(3)=16
Therefore, the height of the ball after 3 secs of dropping is 16 feet.
Answer:
The answer would most likely to eight. If not use the formula Base x Height
Step-by-step explanation: