Answer:
Step-by-step explanation:
The complete question in the attached figure
we have
we know that
(ab)(x) is equal to the product of a(x) and b(x)
so
Applying distributive property
This is a quadratic equation
The other expressions not produce a quadratic equation
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
Answer:
y = 3x-9
Step-by-step explanation:
y = -1/3 x +2
The slope is -1/3
We want a line perpendicular to it
Perpendicular lines have negative reciprocal slopes
The new slope is -(-3/1) = 3
We have a slope m=3 and a point (5,6)
We will use point slope form to find the line
y-y-1 = m(x-x1)
y-6 = 3(x-5)
In slope intercept form
Distribute
y-6 = 3x-15
Add 6 to each side
y-6+6 = 3x-15+6
y = 3x-9
