We have:
(30x²+23x+16)/(cx+3) - 13/(cx+3) = 6x+1
(30x²+23x+16 - 13)/(cx+3) = 6x+1
(30x²+23x+3)/(cx+3) = 6x+1
30x²+23x+3 = (cx+3)(6x+1)
30x²+23x+3 = 6cx²+cx+18x+3
30x² + 23x + 3 - 6cx² - cx - 18x - 3 = 0
(30 - 6c)x² +(5 - c)x = 0
6(5 - c)x² +(5 - c)x = 0
(5 - c)(6x² +x) = 0, and x∈ R\ {3/c} ⇒ 5 - c = 0 ⇒ c = 5.
The neq coordinates would be -8,8 and 8,-8
Answer:
52°
Step-by-step explanation:
(I'm assuming lines JK and HI are parallel)
∠JKA and ∠HIA are corresponding so they are congruent.
Therefore, since ∠JKA is 62, so is ∠HIA. Then knowing the sum of interior angles of a triangle always must be 180, you can just do the math to get 52° as the remaining angle.
66° + 62° + ∠x = 180
∠x = 52°
The answer is 2.11.
For negtives i always subtract the numbers: 51.64-46.09
That always helps me
