Step-by-step explanation:
Let x represent theta.
![\sin( \frac{\pi}{4} - x )](https://tex.z-dn.net/?f=%20%5Csin%28%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20-%20x%20%29%20)
Using the angle addition trig formula,
![\sin(x - y) = \sin(x) \cos(y) - \cos(x) \sin(y)](https://tex.z-dn.net/?f=%20%5Csin%28x%20-%20y%29%20%20%3D%20%20%5Csin%28x%29%20%20%5Ccos%28y%29%20%20-%20%20%5Ccos%28x%29%20%20%5Csin%28y%29%20)
![\sin( \frac{\pi}{4} ) \cos(x) - \cos( \frac{\pi}{4} ) \sin(x)](https://tex.z-dn.net/?f=%20%5Csin%28%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%29%20%20%5Ccos%28x%29%20%20-%20%20%5Ccos%28%20%5Cfrac%7B%5Cpi%7D%7B4%7D%20%29%20%20%5Csin%28x%29%20)
![( \frac{ \sqrt{2} }{2}) \cos(x) - (\frac{ \sqrt{2} }{2} )\sin(x)](https://tex.z-dn.net/?f=%28%20%5Cfrac%7B%20%5Csqrt%7B2%7D%20%7D%7B2%7D%29%20%20%5Ccos%28x%29%20%20-%20%20%28%5Cfrac%7B%20%5Csqrt%7B2%7D%20%7D%7B2%7D%20%20%29%5Csin%28x%29%20)
Multiply one side at a time
Replace theta with x , the answer is
![\frac{ \sqrt{2} \cos(x) }{2} - \frac{ \sin(x) \sqrt{2} }{2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Csqrt%7B2%7D%20%5Ccos%28x%29%20%20%7D%7B2%7D%20%20-%20%20%5Cfrac%7B%20%5Csin%28x%29%20%5Csqrt%7B2%7D%20%20%7D%7B2%7D%20)
2. Convert 30 degrees into radian
![\frac{30}{1} \times \frac{\pi}{180} = \frac{\pi}{6}](https://tex.z-dn.net/?f=%20%5Cfrac%7B30%7D%7B1%7D%20%20%5Ctimes%20%20%5Cfrac%7B%5Cpi%7D%7B180%7D%20%20%3D%20%20%5Cfrac%7B%5Cpi%7D%7B6%7D%20)
Using tangent formula,
![\tan(x + y) = \frac{ \tan(x) + \tan(y) }{1 - \tan(x) \tan(y) }](https://tex.z-dn.net/?f=%20%5Ctan%28x%20%2B%20y%29%20%20%3D%20%20%5Cfrac%7B%20%5Ctan%28x%29%20%20%2B%20%20%5Ctan%28y%29%20%7D%7B1%20-%20%20%5Ctan%28x%29%20%5Ctan%28y%29%20%20%7D%20)
![\frac{ \tan(x) + \tan( \frac{\pi}{6} ) }{1 - \tan(x) \tan( \frac{\pi}{6} ) }](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Ctan%28x%29%20%2B%20%20%5Ctan%28%20%5Cfrac%7B%5Cpi%7D%7B6%7D%20%29%20%20%7D%7B1%20-%20%20%5Ctan%28x%29%20%5Ctan%28%20%5Cfrac%7B%5Cpi%7D%7B6%7D%20%29%20%20%7D%20)
Tan if pi/6 is sqr root of 3/3
![\frac{ \tan(x) + ( \frac{ \sqrt{3} }{3} ) }{1 - \tan(x) (\frac{ \sqrt{3} }{3} ) }](https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Ctan%28x%29%20%2B%20%20%28%20%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B3%7D%20%29%20%20%7D%7B1%20-%20%20%5Ctan%28x%29%20%20%28%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B3%7D%20%29%20%20%7D%20)
Since my phone about to die if you later simplify that,
you'll get
![\frac{(3 \tan(x) + \sqrt{3} )(3 + \sqrt{3} \tan(x) }{3(3 - \tan {}^{2} (x) }](https://tex.z-dn.net/?f=%20%5Cfrac%7B%283%20%5Ctan%28x%29%20%2B%20%20%5Csqrt%7B3%7D%20%29%283%20%2B%20%20%5Csqrt%7B3%7D%20%20%5Ctan%28x%29%20%20%7D%7B3%283%20-%20%20%5Ctan%20%7B%7D%5E%7B2%7D%20%28x%29%20%7D%20)
Replace theta with X.
True? I dont know what you're trying to ask. if you're asking whether or not it's true, it is.
Answer: 42.25 feet
Step-by-step explanation:
We know that after "t" seconds, its height "h" in feet is given by this function:
![h(t) = 52t -16t^2](https://tex.z-dn.net/?f=h%28t%29%20%3D%2052t%20-16t%5E2)
The maximum height is the y-coordinate of the vertex of the parabola. Then, we can use the following formula to find the corresponding value of "t" (which is the x-coordinate of the vertex):
![x=t=\frac{-b}{2a}](https://tex.z-dn.net/?f=x%3Dt%3D%5Cfrac%7B-b%7D%7B2a%7D)
In this case:
![a=-16\\b=52](https://tex.z-dn.net/?f=a%3D-16%5C%5Cb%3D52)
Substituting values, we get :
![t=\frac{-52}{2(-16)}\\\\t=1.625](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-52%7D%7B2%28-16%29%7D%5C%5C%5C%5Ct%3D1.625)
Substituting this value into the function to find the maximum height the ball will reach, we get:
![h(1.625) = 52(1.625) -16(1.625)^2\\\\h(1.625) =42.25\ ft](https://tex.z-dn.net/?f=h%281.625%29%20%3D%2052%281.625%29%20-16%281.625%29%5E2%5C%5C%5C%5Ch%281.625%29%20%3D42.25%5C%20ft)
Answer:
<u>13 - 2x </u>
Step-by-step explanation:
(x +2)² + (x - 3)² - 2x²
<em>Expand.</em>
(x + 2)(x + 2) + (x - 3)(x - 3) - 2x²
x² + 2x + 2x + 4 + x² - 3x - 3x + 9 - 2x²
<em>Bring all the like terms together then simply.</em>
x² + x² - 2x² + 2x + 2x - 3x - 3x + 4 + 9
2x² - 2x² + 4x - 6x + 13
= <u>13 - 2x</u> (or <u>-2x + 13</u> )