Answer:
C=y=sin1/2x
Step-by-step explanation:
As given in the graph:
Amplitude= 1
period=2π
Finding function of sin that have period of 4π and amplitude 1
A: y=1/2sinx
Using the formula asin(bx-c)+d to find the amplitude and period
a=1/2
b=1
c=0
d=0
Amplitude=|a|
=1/2
Period= 2π/b
=2π
B: y=sin2x
Using the formula asin(bx-c)+d to find the amplitude and period
a=1
b=2
c=0
d=0
Amplitude=|a|
=1
Period= 2π/2
=π
C: y=sin1/2x
Using the formula asin(bx-c)+d to find the amplitude and period
a=1
b=1/2
c=0
d=0
Amplitude=|a|
=1
Period= 2π/1/2
=4π
D: y=sin1/4x
Using the formula asin(bx-c)+d to find the amplitude and period
a=1
b=1/4
c=0
d=0
Amplitude=|a|
=1
Period= 2π/1/4
=8π
Hence only c: y=sin1/2x has period of 2π and amplitude 1
By the angle difference formula for sines, we have:
By the angle sum formula for tangents, we have:
.
Rationalizing the denominator gives
as the final answer.
Don’t believe them it’s a scam.
Answer:
x = 13
Step-by-step explanation:
Take the square root of both sides
Its fairly straightforward. Since the bottom equation only has one unknown,x, because y=1.3, you can plug y in and solve for x. Once you find the value of x, you then have the value for two variables, x and y, and again have one unknown coefficient a. To solve for the coefficient you just plug in your y value (1.3) and your x value (which can be rounded to 0.42). Using a little bit of algebra, you can then solve for a which should be a=2.108. I am not sure if your teacher wants you to solve it this way but you could also use the elimination method or substitution method that you would of learned when discussing system of equations. But no matter which way you do it, the math follows the rules. Hope this helps. I’d suggest you solve it yourself to double check my work.
To verify my credibility,
I am a Mechanical Engineering major w/ minor in mathematics
