Answer:
maximum height is 4.058 metres
Time in air = 0.033 second
Step-by-step explanation:
Given that the equation height h
h = -212t^2 + 7t + 4
What is the toy's maximum height?
Let us assume that the equation is a perfect parabola
Time t at Maximum height will be
t = -b/2a
Where b = 7 and a = - 212
t = -7/ - 212 ×2
t = 7/ 424 = 0.0165s
Substitute t in the main equation
h = - 212(7/424)^2 + 7(7/424) + 4
h = - 0.05778 + 0.115567 + 4
h = 4.058 metres
Therefore the maximum height is 4.058 metres
How long is the toy in the air?
The object will go up and return to the ground.
At ground level, h = 0
-212t^2 + 7t + 4 = 0
212t^2 - 7t - 4 = 0
You can factorize the above equation and pick the positive time t since time can't be negative
Or
Since we have assumed that it's a perfect parabola,
Total time in air = (-b/2a) × 2
Time in air = 0.0165 × 2 = 0.033 s
Givens
y = 2
x = 1
z(the hypotenuse) = √(2^2 + 1^2) = √5
Cos(u) = x value / hypotenuse = 1/√5
Sin(u) = y value / hypotenuse = 2/√5
Solve for sin2u
Sin(2u) = 2*sin(u)*cos(u)
Sin(2u) = 2(
) = 4/5
Solve for cos(2u)
cos(2u) = - sqrt(1 - sin^2(2u))
Cos(2u) = - sqrt(1 - (4/5)^2 )
Cos(2u) = -sqrt(1 - 16/25)
cos(2u) = -sqrt(9/25)
cos(2u) = -3/5
Solve for Tan(2u)
tan(2u) = sin(2u) / cos(2u) = 4/5// - 3/5 = - 0.8/0.6 = - 1.3333 = - 4/3
Notes
One: Notice that you would normally rationalize the denominator, but you don't have to in this case. The formulas are such that they perform the rationalizations themselves.
Two: Notice the sign on the cos(2u). The sin is plus even though the angle (2u) is in the second quadrant. The cos is different. It is about 126 degrees which would make it a negative root (9/25)
Three: If you are uncomfortable with the tan, you could do fractions.

10*3 tens= 30 tens= 300 units
Answer:
17
Step-by-step explanation:
12+5