If the point ( x , √3/2) is on the unit circle, what is x?

A. √3/2

B. 1/2

C. - √3/2

D. 2/√3

B. 1/2

Step-by-step explanation:

This problem involves the use of the Pythagorean theorem. In the unit circle, the hypotenuse of any right triangle formed is 1 while the coordinates of the point are then the two legs that make up the triangle.

a = 1 (given)

b = √3/2 (given)

a² = b² + c² (Pythagorean Theorem)

1² = (√3/2)² + c² (Substitute information)

Now, we need to solve for c

1 = (3/4) + c² (square the hypotenuse and one of the legs)

1 - (3/4) = (3/4) + c² - (3/4) (subtract 3/4 on both sides)

1/4 = c² (combine like terms)

√(1/4) = √(c²) (square root both sides)

√1 / √4 = c (square root both sides)

1/2 = c (final answer)

Therefore the other leg of the right triangle is 1/2, this also means that the other coordinate, or x, is 1/2, so the answer is B. 1/2.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

300pi

step-by-step explanation: