Answer:
The answer for 10 is y= -4x+2
If the slope is exactly the same but the y intercept is different then it will be parallel
The answer for 11 is y-5=-3/2 (x-1)
Step-by-step explanation:
Let x be the 1st odd number, and x+2 the second odd consecutive number:
(x)(x + 2) = 6[((x) + (x+2)] -1
x² + 2x = 6(2x + 2) - 1
x² + 2x = 12x +12 - 1
And x² - 10x - 11=0
Solve this quadratic expression:
x' = [+10 +√(10²- 4.(1)(-11)]/2 and x" = [+10 -√(10²- 4.(1)(-11)]/2
x' = [10 + √144]/2 and x" = [10 - √64]/2
x' = (10+12)/2 and x" = (10-12)/2
x = 11 and x = -1
We have 2 solutions that satisfy the problem:
1st for x = 11, the numbers at 11 and 13
2nd for x = - 1 , the numbers are -1 and +1
If you plug each one in the original equation :(x)(x + 2) = 6[((x) + (x+2)] -1
you will find that both generates an equlity
Solve for b.
b^2= x+4ac
So b=sq root of x+4ac
The list of choices you included doesn't have any correct expressions on it.
-- If the rocket went straight up, and then dropped back at the same place
it was launched from, then it must have dropped straight down.
-- The route it followed on the way down was exactly the same as the
route it followed on the way up.
-- Since it reached a maximum height of 150-ft from the ground
and the route was straight, the length of the route was 150-ft.
-- (150-ft going up) plus (150-ft coming down) = 300-ft traveled all together.
Answer:
step one needs to be fixed... you are supposed to subtract 10 from both sides of the equation.
Step-by-step explanation:
STEP 1: Move all terms not containing n to the right side of the equation.
STEP 2: Divide each term by 2 and simplify.
