Algebra 1 is a high school course. You learn some of the same things but in algebra it's more complicated.
( 'x' is not 144 .)
The supplement of an angle is (180 - x) .
The problem says that (2/3) of 'x' is equal to (180 - x) .
180 - x = 2/3 x
Multiply each side by 3 :
( Note: 3 x 180 = 540 .)
540 - 3x = 2x
Add 3x to each side:
540 = 5x
Divide each side by 5 :
<u>x = 108°</u> .
<h3>
Answer: 1</h3>
Explanation:
The original expression is the same as
since 
The degree of any polynomial is always the largest exponent. This applies to single variable polynomials only.
Therefore, the degree of
is 1. This is a linear polynomial, and it's also a binomial since it has 2 terms 17x and 4.
Given the points A and B
The coordinates of point A = ( 3 , 1 )
The coordinates of point B = (-1 , -1)
The midpoint of AB, is the point C
C will be calculated as following :

so, the midpoint of AB = (1 , 0 )
<h3>E
xplanation:</h3>
Replace cos^2(θ) with 1-sin^2(θ), and cot(θ) with cos(θ)/sin(θ).
cos^2(θ)cot^2(θ) = cot^2(θ) - cos^2(θ)
(1 -sin^2(θ))cot^2(θ) = . . . . . replace cos^2 with 1-sin^2
cot^2(θ) -sin^2(θ)·cos^2(θ)/sin^2(θ) = . . . . . replace cot with cos/sin
cot^2(θ) -cos^2(θ) = cot^2(θ) -cos^2(θ) . . . as desired