Answer:
1/sqrt10
Step-by-step explanation:
1) Find out cosA using formula (cosA)^2+(sinA)^2=1
The module of cosA= sqrt (1- (-3/5)^2)= sqrt 16/25=4/5
So cosA=-4/5 or cosA=4/5.
Due to the condition 270degrees< A<360 degrees, 0<cosA<1 that's why cosA=4/5.
2) Find sinA/2 using a formula cosA= 1-2sinA/2*sinA/2 where cosA=4/5.
(sinA/2)^2= 0.1
sinA= sqrt 0.1= 1/ sqrt10 or sinA= - sqrt 0.1= -1/sqrt10
But 270°< A< 360°, then 270/2°<A/2<360/2°
135°<A/2<180°, so sinA/2 must be positive and the only correct answer is
sin A/2= 1/sqrt10
- Perpendicular=P=4
- Base=B=4
Hypotenuse be H
Apply Pythagorean theorem
Answer:
Are there multiple choice options? If so, please include them. That way I will actually be able to help you :)
Step-by-step explanation:
Answer:
f(-2) = 1/9
Step-by-step explanation:
Plug in -2 to x in the equation:
f(x) = 3^x
f(-2) = 3^(-2)
Solve. Note that if there is a negative in the power, you must flip the "fraction" and set it over 1. The answer itself does not become negative:
f(-2) = 1/(3^2)
f(-2) = 1/(9)
f(-2) = 1/9 is your answer.
~
1/7x - 6 = 8(11-3)
1/7x - 6 = 8(8)
1/7x - 6 = 64
1/7x = 70
1/7 = 70x
1 = 490x
1/490 = x
