Ok here is what I think.
Let us first number these statements, as #1, and #2.
First statement: 3x + 8y = 12 (1)
Second Statement: 2x + 2y = 3 (2)
Now, we can work from this.
We want to make one of the equations be equal to 0 so that at the end when we check they can be equal to each other.
Let us use 4.
3x+8y=12 1-8x-8y=-12 2
This gives us:-5x = 0
Now we should try and isolate x so we can substitute it into one of the equations.
We have -5x=0
and x=0
3(0)+8y=12
8y=12
y=12/8
y=3/2
Plug in these new equations
y=3/2 and y=0 into any of the first equations
3x+8y=12 3(0)+8(3/2)= 12 8(3/2)=12 4(3)=12 12=12
Now we know it works, thats our check^^
If 22% are sixth graders, then the rest must not sixths graders.
100% of the school - 22% sixth grades = 78% are not sixths graders
This involvea factoring the polynomial into binomials and solving.
So make it (3x+1)(x-7)
X equals 7 and -1/3
The middle one is the answer.
The -1 moves the graph down by one unit.
if it was |x-1| that would move it one unit right.
The equation for a circle is (x-h)^2 +(y-k)^2=r^2
You just have to plug in your coordinates h is the x and k is the y.
The r is radius squared.
so, (x-2)^2+(y-(-5)^2=144
You have to be careful of those tricky negative values. When you distribute the y-(-5) it turns into y+5.
So your answer is D.
