When it hits the ground, y = 0...so....
We just need to solve this
-4.9t^2 + 42t + 18.9 = 0
Using the quad fromula we get
[ -42 ± sqrt (42^2 - 4 ( 18.9) ( -4.9) )] / ( 2 * - 4.9)
Taking the positive value.....t = 9
( It hits the ground at 9 sec )
q must be negative ( if it were positive, we would have factors of ( x + m) ( x + n) = 0, and the roots m, n would both be negative)
So......any negative for q would work because the discriminant
3^2 - 4(q)(1) will be positive
So
q < 0
1. (x + 2) / ( (2x + 2) = (4x + 3) / ( 7x + 3) cross-multiply
(x + 2) ( 7x + 3) = ( 2x + 2) ( 4x + 3) simplify
7x^2 + 17x + 6 = 8x^2 + 14x + 6
7x^2 + 17x = 8x^2 + 14x rearrange as
x^2 - 3x = 0
x ( x - 3) = 0
x = 0 is one solution.....so....the product of all real solutions must be 0
See here : https://web2.0calc.com/questions/i-m-a-little-confused
No prob !!!
The denomiantor cannot = 0
So...we can guarantee no reals will make the denominator = 0 if we solve this :
b^2 - 4 (8)(1) < 0
b^2 - 32 < 0
b^2 < 32 take the positive root
b < sqrt 32
b < 5.6
So 5 is the greatest integer
The common difference is -1/3 - 1/4 = -7/12
We need to evaluate this
-1/3 + (7/12) ( 10 - 2)
-1/3 + (7/12) ( 8)
-1/3 + 56/12
-1/3 + 14/3 =
13 / 3 = the second term
Slope between the two points = [ 5 - - 3 ] / [ 4 - 2 ] = 8/2 = 4
Equation of the line using ( 4, 5)
y = 4 ( x - 4) + 5
y = 4x - 16 + 5
y = 4x - 11 rearrange as
4x - y = 11
The common difference is 10 - -8 = 18
10 - 18 ( 17 - 2) =
10 - 18 * 15 =
10 - 270 =
-260 = 2nd term
Welcome.....
