Answer:
Angle y and Angle x are complementary angles
Step-by-step explanation:
we know that
If
sin(y°)=cos(x°)
then
Angle y and Angle x are complementary angles
so
∠y+∠x=90°
∠x=90°-∠y
sin(y°)=cos(90°-y°)
therefore
The sine of an angle is equal to the cosine of its complementary angle
2√20=2((√4)(√5)=2((2)(√5)=4√5
now we have
4√5-3√5=1√5=√5
PART A
The given equation is

In order to find the maximum height, we write the function in the vertex form.
We factor -16 out of the first two terms to get,

We add and subtract

to get,

We again factor -16 out of the first two terms to get,

This implies that,

The quadratic trinomial above is a perfect square.

This finally simplifies to,

The vertex of this function is

The y-value of the vertex is the maximum value.
Therefore the maximum value is,

PART B
When the ball hits the ground,

This implies that,

We add -29 to both sides to get,

This implies that,




or

Since time cannot be negative, we discard the negative value and pick,
F(x)=5x
normal domain: all real numbers
practical domain: <span>all positive integers
</span>becasue we can substituent with any positive integer in the place of x
In the data shown, rearranging the data [<span>9.4, 9.2, 9.7, 9.8, 9.4, 9.7, 9.6, 9.3, 9.2, 9.1, 9.4] from the least to the greatest would give us the following data set:
9.1, 9.2, 9.2, 9.3, 9.4, 9.4, 9.4, 9.6, 9.7, 9.7, 9.8
The box-plots uses a 5-number summary. The minimum value, then Q1 which is the media of the lower half of the set, Q2 which is the median of the total set, Q3 which is the median of the upper half of the set, and Q4 which is the highest number. Among the choices, the correct answer is B.</span>