We know that
if cos x is positive
and
sin x is negative
so
the angle x belong to the IV quadrant
cos x=5/13
we know that
sin²x+cos²x=1-------> sin²x=1-cos²x------> 1-(5/13)²---> 144/169
sin x=√(144/169)-------> sin x=12/13
but remember that x is on the IV quadrant
so
sin x=-12/13
Part A) <span>cos (x/2)
cos (x/2)=(+/-)</span>√[(1+cos x)/2]
cos (x/2)=(+/-)√[(1+5/13)/2]
cos (x/2)=(+/-)√[(18/13)/2]
cos (x/2)=(+/-)√[36/13]
cos (x/2)=(+/-)6/√13-------> cos (x/2)=(+/-)6√13/13
the angle (x/2) belong to the II quadrant
so
cos (x/2)=-6√√13/13
the answer Part A) is
cos (x/2)=-6√√13/13
Part B) sin (2x)
sin (2x)=2*sin x* cos x------> 2*[-12/13]*[5/13]----> -120/169
the answer Part B) is
sin(2x)=-120/169
Answer:
Step-by-step explanation:
The first one: The median is 58.
The second one: The maximum value is 58.
The third one: The upper quartile is 58.
The last one: The minimum value is 58.
Answer:
The graph of y = f(x) will shift to the right 9 units.
Step-by-step explanation:
Despite some lack of clarity, because usually this equality is in terms of x and y. Let's do it.
Taking f(x) as its parent function in other words
Testing some values:
x|y
0|0
1|1
2|2
Adding to the function -9 units is to shift it to the right
Testing some values to certify:
x|y
-9|0-9
-8|1-9
-7|2-9
Answer:
9) x = 5
10) x = 4
Step-by-step explanation:
a^2 + b^2 = c^2
9) 6^2 + x^2 = 8^2
- 36 + x^2 = 64
- 36 - 36
_______________
x^2 = 28
Square root them
Squareroot x^2 is x
Square root 28 is 5.29 round is 5
so x = 5
*Correct me if im wrong*
10) 3^2 + x^2 = 5^2
9 + x^2 = 25
- 9 - 9
____________________
x^2= 16
Square root them
square root of x ^2 is x and square root of 16 is 4
so x = 4
