Answer:
<h3>3 secs</h3>
Step-by-step explanation:
Given the height of the object as it drops from the observation deck expressed as;
h= -16t^2+152
To determine the the time it will take the object to be 8 feet above the valley floor, we will substitute h = 8 into the equation and calculate t as shown;
8 = -16t^2+152
subtract 8 from both sides
8-8 = -16t^2+152-8
0 = -16t^2+144
0-144 = -16t^2
-144 = -16t^2
16t^2 = 144
Divide both sides by 16;
16t^2/16 = 144/16
t^2 = 9
t = √9
t = 3seconds
Hence it will take 3 seconds for the object to be 8 feet above the valley floor
Answer:
Please require a picture pls
Step-by-step explanation:
Answer:
See below.
Step-by-step explanation:
So, we have:

Recall that secant is simply the reciprocal of cosine. So we can:

Now, recall the unit circle. Since cosine is negative, it must be in Quadrants II and/or III. The numerator is the square root of 3. The denominator is 2. The whole thing is negative. Therefore, this means that 150 or 5π/6 is a candidate. Therefore, due to reference angles, 180+30=210 or 7π/6 is also a candidate.
Therefore, the possible values for theta is
5π/6 +2nπ
and
7π/6 + 2nπ
Edited in the comment ............
<span>otations and translations. two shapes are congruent to each other if they have the same size and shape. dilations and stretches affect the size and shape of shapes, and so cannot be used</span>