Answer:
- Exact distance =
feet - Approximate distance = 127.2792 feet.
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Explanation:
The area of the square is 8100 square feet, abbreviated ft^2.
Apply the square root to find the distance from any base to its adjacent counterpart (eg: from 1st to 2nd base). So we get sqrt(8100) = 90 ft as that side distance. Notice that 90*90 = 8100.
If you were to draw a line from 1st base to 3rd base, then you would split the square into two congruent right triangles. Each right triangle is isosceles (the two legs being 90 ft each).
Use the pythagorean theorem to find the hypotenuse.
The distance from 1st to 3rd base is roughly 127.2792 feet.
Answer:
the answer would be choice A
Answer:12
Step-by-step explanation:
Answer:
R (t) = 60 - 60 cos (6t)
Step-by-step explanation:
Given that:
R(t) = acos (bt) + d
at t= 0
R(0) = 0
0 = acos (0) + d
a + d = 0 ----- (1)
After 
seconds it reaches a height of 60 cm from the ground.
i.e
Recall from the question that:
At t = 0, R(0) = 0 which is the minimum
as such it is only when a is negative can acos (bt ) + d can get to minimum at t= 0
Similarly; 60 × 2 = maximum
R'(t) = -ab sin (bt) =0
bt = k π
here;
k is the integer
making t the subject of the formula, we have:
replacing the derived equation of k into R(t) = acos (bt) + d
Since we known a < 0 (negative)
then d-a will be maximum
d-a = 60 × 2
d-a = 120 ----- (3)
Relating to equation (1) and (3)
a = -60 and d = 60
∴ R(t) = 60 - 60 cos (bt)
Similarly;
For 
where ;
Then b = 6
∴
R (t) = 60 - 60 cos (6t)
