Answer:
Step-by-step explanation:
O
3x² - 11x - 4 = 0
a = 3 ; b = -11 , c = -4
W
8x² + 12x = -1
8x² + 12x + 1 = 0
a = 8 ; b = 12 ; c = 1
L
x² + x = 6
x² + x - 6 = 0
a = 1 ; b= 1 ; c = -6
G
x²- 2x - 2 =0
a = 1 ; b = -2 ; c = -2
E
4x - 5 = -4x²
4x² + 4x - 5 = 0
a = 4 ; b = 4 ; c = -5
B
12x² +x -6 = 0
a = 12 ; b = 1 ; c = -6
R
2x² + 5x - 25 = 0
a = 2 ; b = 5 ; c = -25
A
6x² -5x - 4 = 0
a = 6 ; b = -5 ; c = -4
![Sum \ of \ roots =\dfrac{-[-5]}{6}=\dfrac{5}{6}\\\\Product \ of \ roots = \dfrac{-4}{6}=\dfrac{-2}{3}](https://tex.z-dn.net/?f=Sum%20%5C%20of%20%5C%20roots%20%3D%5Cdfrac%7B-%5B-5%5D%7D%7B6%7D%3D%5Cdfrac%7B5%7D%7B6%7D%5C%5C%5C%5CProduct%20%5C%20of%20%5C%20roots%20%3D%20%5Cdfrac%7B-4%7D%7B6%7D%3D%5Cdfrac%7B-2%7D%7B3%7D)
It's D
Because a sphere is round, it's a 3D circle.
Hope this helps!
Answer:
the end by the finish line
Answer:
Checked on edg. , the answer is A. Thanks!
Step-by-step explanation:
Triangular sequence = n(n + 1)/2
If 630 is a triangular number, then:
n(n + 1)/2 = 630
Then n should be a positive whole number if 630 is a triangular number.
n(n + 1)/2 = 630
n(n + 1) = 2*630
n(n + 1) = 1260
n² + n = 1260
n² + n - 1260 = 0
By trial an error note that 1260 = 35 * 36
n² + n - 1260 = 0
Replace n with 36n - 35n
n² + 36n - 35n - 1260 = 0
n(n + 36) - 35(n + 36) = 0
(n + 36)(n - 35) = 0
n + 36 = 0 or n - 35 = 0
n = 0 - 36, or n = 0 + 35
n = -36, or 35
n can not be negative.
n = 35 is valid.
Since n is a positive whole number, that means 630 is a triangular number.
So the answer is True.
