I need points and i’m using this bc you didn’t even give an equation
Answer:
I think c 430 and if it is wrong I'm so sorry
J = 14
J = 1/2S + 4 1/2
14 = 1/2S + 9/2....multiply everything by 2 to get rid of fractions
28 = S + 9
28 - 9 = S
19 = S
check...
14 = 1/2S + 9/2......S = 19
14 = 1/2(19) + 9/2
14 = 9 1/2 + 9/2
14 = 19/2+ 9/2
14 = 28/2
14 = 14 (correct)
so the sister is 19 <===
The Equation of a Line
The slope-intercept form of a line can be written as:
y = mx + b
Where m is the slope of the graph of the line and b is the y-intercept.
In the specific case where the line passes through the origin (0,0), we can find the value of b by substituting x=0 and y=0:
0 = m(0) + b
Solving for b:
b = 0.
Thus, the equation of the line reduces to:
y = mx
We only need to find the value of the slope.
That is where we need the second data. Our line is perpendicular to the line of equation 4x + 3y = 6.
Solving for y:

The slope of the second line is -4/3.
We must recall that if two lines of slopes m1 and m2 are perpendicular, then:

Substituting the value of m1 and solving for m2:

The slope of our line is 3/4 and the required equation is:

From this last equation, we need to find the general form of the line.
Multiply both sides of the equation by 4:
4y = 3x
Subtract 3x on both sides:
4y - 3x = 0
Reorder:
-3x + 4y = 0
The axis of symmetry can be found by finding the average of the zeros, a derivation from the conservation of energy :P, or by finding the point when the velocity is equal to zero.
df/dx=-6x+12 so df/dx, velocity, equals zero when:
-6x+12=0
6x=12
x=2 so the axis of symmetry is the vertical line x=2
....
average of zeros...
3x^2-12x+6=0
x^2-4x+2=0
x^2-4x=-2
x^2-4x+4=2
(x-2)^2=2
x-2=±√2
x=2±√2 so the average of the zeros is obviously 2.
....
conservation of energy
vf-vi=at When vf=0, this is the maximum value for f(x)...
-vi=at, vi=b and a(acceleration)=2a(from quadratic) and t=x
-b=2ax
x=-b/(2a) in this case
x=-12/(2(-3))
x=-12/-6
x=2