Answer:
<h2>
</h2>
Step-by-step explanation:
<h3>
</h3>
To find f(½) substitute the value of x that's ½ into f(x). That is for every x in f(x) replace it with 1/2
That's
<h3>
</h3>
<u>Find the LCM</u>
The LCM is 3
We have
<h3>
</h3>
We have the final answer as
<h3>
</h3>
Hope this helps you
Answer:
a + a + a = 3a
Step-by-step explanation:
let me know if there is more to the question
Answer:
B. 5√3 is your answer.
Step-by-step explanation:
This is an equilateral triangle, which means that all external sides (that make up the triangle) have the same measurement.
This means that line segment AC = 10.
Solve for the measurement of line segment AB. Line Segment AB is 1/2 of AC, making it: 10/2 = 5
Line AB = 5
Line AD = 10
This is a 1, √3, 2 triangle, in which:
1 = 5
2 = 10
Solve for √3. Multiply √3 with the 1:
5 x √3 = 5√3
B. 5√3 is your answer.
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Answer:
The answer is below
Step-by-step explanation:
Show that f(x) f(y) = f(x+y)
From trigonometric:
sin(x + y) = sinxcosy + cosxsiny
sin(x - y) = sinxcosy - cosxsiny
cos(x + y) = cosxcosy - sinxsiny
cos(x - y) = cosxcosy + sinxsiny
![f(x)=\left[\begin{array}{ccc}cosx&-sinx&0\\sinx&cosx&0\\0&0&1\end{array}\right] ,f(y)=\left[\begin{array}{ccc}cosy&-siny&0\\siny&cosy&0\\0&0&1\end{array}\right] \\\\\\f(x)f(y)=\left[\begin{array}{ccc}cosxcosy-sinxsiny&-cosxsiny-sinxcosy&0\\sinxcosy+cosxsiny&-sinxsiny+cosxcosy&0\\0&0&1\end{array}\right] \\\\\\f(x)f(y)=\left[\begin{array}{ccc}cos(x+y)&-sin(x+y)&0\\sin(x+y)&cos(x+y)&0\\0&0&1\end{array}\right] \\\\\\](https://tex.z-dn.net/?f=f%28x%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcosx%26-sinx%260%5C%5Csinx%26cosx%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%2Cf%28y%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcosy%26-siny%260%5C%5Csiny%26cosy%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5Cf%28x%29f%28y%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcosxcosy-sinxsiny%26-cosxsiny-sinxcosy%260%5C%5Csinxcosy%2Bcosxsiny%26-sinxsiny%2Bcosxcosy%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5Cf%28x%29f%28y%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%28x%2By%29%26-sin%28x%2By%29%260%5C%5Csin%28x%2By%29%26cos%28x%2By%29%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C)
![f(x+y)=\left[\begin{array}{ccc}cos(x+y)&-sin(x+y)&0\\sin(x+y)&cos(x+y)&0\\0&0&1\end{array}\right] \\\\\\Therefore\ f(x)f(y)=f(x+y)](https://tex.z-dn.net/?f=f%28x%2By%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dcos%28x%2By%29%26-sin%28x%2By%29%260%5C%5Csin%28x%2By%29%26cos%28x%2By%29%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5CTherefore%5C%20f%28x%29f%28y%29%3Df%28x%2By%29)
The answer is C: first add 2 both sides then divide both sides by -5.
