This is a quadratic formula with a general form of a²x + bx + c = 0. For quadratic equations, we can solve for its two roots using the quadratic formula shown in the attached picture.
a = -3
b = -4
c = -4
x = [-(-4) + √(-4)² - 4(-3)(-4)]/2(-3) =
<em>2 + √-32/2</em>x = [-(-4) - √(-4)² - 4(-3)(-4)]/2(-3) =
<em> 2 - √-32/2</em>
4(2t+6)
8t+24 is the answer
For the division part:
But I think your question differs than this..
Am I right?
as your options do not include 1
parallel lines have the same slope
The slope-intercept form of a linear equatio is y=mx+b, where m stands for the "slope of the line" and b stands for the "y-intercept of the line"
They give you the equation y= -5/6x+3 Notice this is already on the slope-intercept form, so in this case the slope is -5/6 and the y-intercept is 3
You want an equation of the line that is parallel to the given line. The slopes must be the same, so m=-5/6
So far we have y=-5/6x + b
We don't have b yet but that can be found using the given point (6,-1) which tells you that "x is 6 when y is -1"
Replace that on the equation y=-5/6x + b and you get
-1 = (-5/6)(6) + b
-1 = -5 +b
4 = b
b = 4
We found b, or the y-intercept
Go back to the equation y = -5/6 x + b and replace this b with the b we just found
y = -5/6x + 4
Answer:
C. (-3,11)
Step-by-step explanation:
Tp is (-3,6) implies the quadratic could have been
f(x) = (x+3)²+6
(2/3)f(x) = (2/3)[(x+3)²+6]
= (2/3)(x+3)²+4
(2/3)f(x)+3 = (2/3)(x+3)²+4+3
= (2/3)(x+3)²+7
Tp at (-3,7)
Alternately,
No change in domain so x remains-3
(2/3)f(x) changes y from 6 to 4 (6×2/3)
+3 increases the y by 3
i.e 4+3 = 7
So, (-3,7)
