From the law of sines, we have

where
is the leg we're interested in,
is the hypothenuse,
is the angle opposite to
, and
is the angle between
and
.
So, in the first case, we can use

And in the second excercise, we use

Answer:
x=1
Step-by-step explanation:
According to your typed question,
f(x)=2x-3
Let f(x)=y
y=2x-3
Now,
Interchanging the positions of x and y
x=2y-3
x+3=2y
x+3/2=y
f'(x)=x+3/2
Then,
f(x)=f'(x)
2x-3=x+3/2
2(2x-3)=x+3
4x-3=x+3
4x-x=3+3
3x=3
x=3/3
x=1
According to your image question,
f(x)=x/2x-3
f(x)=f'(x)
Now,
Let y=f(x)
y=x/2x-3
y(2x-3)=x
2xy-3=x
2xy-x=3
x(2y-1)=3
2y-1=3x
2y=3x+1
y=3x+1/2
f'(x)=3x+1/2
Then,
f(x)=f'(x)
x/2x-3=3x+1/2
2x=6xsq+2x-9x+3
2x=6xsq+7x+3
solve for x ok
The answer is 12. Thanks (;
Answer:
it would be 12
Step-by-step explanation:
Answer:
Unforgettable and Chanaynay by wewantwraiths
Step-by-step explanation: