x = 3/32
Option C is the right answer
Answer:
1/36
Step-by-step explanation:
When the coefficient is 1, the function has zeros at -3 and -5, one horizontal unit from the vertex. You want to move the zero to (2, 0), which is 6 units from the vertex. To achieve a horizontal stretch by a factor of 6, the value of x in the function must be replaced by x/6. That would make the coefficient of x^2 be (1/6)^2 = 1/36.
The coefficient of the squared term is 1/36.
Answer:
1st box: Asso. prop= m+(4+x)
2nd box: Comm. Prop= m+4=4+m
3rd box: iden. prop= m+0=m
4th box: Zero prop: m x 0=0
9514 1404 393
Answer:
focus: (0, -5.5)
directrix: y = 5.5
Step-by-step explanation:
The equation compares to the form ...
x² = 4py
where 4p = -22 ⇒ p = -22/4 = -5.5.
The value of p is the distance from the vertex to the focus. The negative value means the focus is below the vertex at (0, -5.5).
The directrix is above the vertex by the same amount, so is y = 5.5.
y=x+14 line 1
y=3x+2 line 2
These are both the equation of lines written in slope intercept form
y=mx+b where m is the slope and the point (0,b) is the y intercept.
The first line has a slope of m=1. The 2nd line has a slope of m=3
Since these lines have different slopes, they are not parallel, thus they will cross at some point. What you have to determine is where the lines cross, which will be a point (x,y) that is on both lines.
We already have y solved in terms of x from either equation so we can use substitution to solve the system.
Since y=x+14 from line 1, put x+14 in place of y in the equation of line 2.
x+14=3x+2
solve for x.
Subtract x from both sides...
14= 3x-x+2
14=2x+2
subtract 2 from both sides
14-2=2x
12=2x
divide both sides by 2
6=x
We now have the x value of the common point. Plug the value 6 in for x in one of the original equations and solve for y.
y=6+14
y=20
These two lines cross at the point (6,20) which is a point the two lines have in common.
Hope I helped (SharkieOwO)
