Answer: No, it is not a solution
Work Shown:
-2 ≤ k/3
-2 ≤ -9/3
-2 ≤ -3
The last inequality is false because -3 should be smaller than -2 (not the other way around). Use a number line to help see this.
Since the last inequality is false, the original inequality must also be false for that particular k value. Therefore, k = -9 is not a solution.
Just solve the square roots and compare all four to see which one is the normalest
Let's examine the given function first:
f(x) = x^2 + 1 is the same as f(x) = 1(x-0)^2 + 1.
The vertex of the graph of this function is at (0, 1).
Let x=0 to find the y-intercept: f(0)=0^2+1 = 1; y-int. is at (0,1) (which happens to be the vertex also)
Comparing f(x) = x^2 + 1 to y = x^2, we see that the only difference is that f(x) has a vertical offset of 1. So: Graph y=x^2. Then translate the whole graph UP by 1 unit. That's it. Note (again) that the vertex will be at (0,1), and (0,1) is also the y-intercept.
Answer:
for what??
Step-by-step explanation:
2/3=x/12
Use cross multiply
2*12=3*x
24=3x
Divided both side by 3
24/3=3x/3
x=8
Check:
2/3=x/12
Substitute x with 8
2/3=8/12
2/3=(8/4)/(12/4)
2/3=2/3. As a result, x=8. Hope it help!
