Answer:
Step-by-step explanation:
This is your position equation:
There's a whole lot of information in that equation, but what we are concerned about right now is the height of the ball after t = 3 seconds. If this is the position of the ball at any time t, we will sub in 3 for t to find out where the ball is at 3 seconds.
which simplifies to
s(3) = -44.1 + 54 + 10 which is
s(3) = 19.9 meters
That's how high the ball is in the air at 3 seconds.
Answer:
The base is 4cm and the height is 5cm.
Step-by-step explanation:
This is a solve the system question. Call H the height of the triangle and B the base. The question tells us:
and
Sub the first equation into the second (as H is already isolated). You will end up with a quadratic equation - solve that any way you wish (e.g. quadratic formula). I've provided the factored form below which shows you the roots:
In this question, we take B=4. You can't have a negative side length so the other answer is eliminated. Now sub the value for B into either of the original equations. I'll use the first, again because H is already isolated:
Answer:
1/2sin(6x) + 1/2sin(2x)
Step-by-step explanation:
You can look up the formulas for the product identities for sine and cosine, or you can guess and check using a graphing calculator. I did the calculator solution first (see the first attachment), then looked up the identities so I can tell you what they are (see the second attachment).
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These identities are based on the sum and difference angle identities:
sin(α+β) +sin(α-β) = (sin(α)cos(β) +sin(β)cos(α)) + (sin(α)cos(β) -sin(β)cos(α))
= 2sin(α)cos(β)
Dividing by 2 gives the identity of interest in this problem:
sin(4x)cos(2x) = (1/2)(sin(4x +2x) +sin(4x -2x))
sin(4x)cos(2x) = (1/2)(sin(6x) +sin(2x))
