Answer: 3/2
Step-by-step explanation:
the formula to find slope is:
m = rise/run = 
In this case:
m = 3/2 = 
5 consecutive even integers: 2n, 2n+2, 2n+4, 2n+6, 2n+8 for any integer n The sum of the two smallest of five consecutive even integers is 50 less than the sum of the other three integers: 2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 504n + 2 = 6n -3234 = 2nn = 17 <span>The smallest integer in our sequence is 2n = 2*17 = 34</span> Check:2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 5034 + 36 = 38 + 40 + 42 - 5070 = 120 - 5070 = 70
Answer:
10/48 - 2/24 - 1/12
Step-by-step explanation:
answer is 1/12
(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))
Recall that
tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)
so cos²(<em>θ</em>) cancels with the cos²(<em>θ</em>) in the tan²(<em>θ</em>) term:
(sin²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>))
Recall the double angle identity for cosine,
cos(2<em>θ</em>) = 2 cos²(<em>θ</em>) - 1
so the 1 in the denominator also vanishes:
(sin²(<em>θ</em>) - 1) / (2 cos²(<em>θ</em>))
Recall the Pythagorean identity,
cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1
which means
sin²(<em>θ</em>) - 1 = -cos²(<em>θ</em>):
-cos²(<em>θ</em>) / (2 cos²(<em>θ</em>))
Cancel the cos²(<em>θ</em>) terms to end up with
(tan²(<em>θ</em>) cos²(<em>θ</em>) - 1) / (1 + cos(2<em>θ</em>)) = -1/2
It would go by (-4,2) then (0,-2) next would be (2,4)
