-1 is less than 0, so you use the first equation:
3(-1) +2 = -3+2 = -1
f(-1) = -1
For 0 use the 2nd equation:
3(0) + 4 = 0+4 = 4
f(0) = 4
For 2 use the 2nd equation:
3(2) + 4 = 6+4 = 10
f(2) = 10
Answer:
rhombus
Step-by-step explanation:
It is helpful to plot the points on a graph. There, you can see that the rise and run of each segment are 1 and 4 (or vice versa). This tells you they are the same length, but not perpendicular.
A quadrilateral with all sides the same length, but not perpendicular, is a rhombus.
Factor out the gcf first.
3(4y^2-9)
The binomial inside is a perfect square so keep factoring.
3(2y+3)(2y-3)
The pair of binomials cannot be further factored so stop there.
1) Solving in terms of h
V = lwh <em>Divide both sides by h</em>
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So rearranging that equation we can find h, in terms of V, and l and w.
If we want to solve in terms of l, or w, we'll proceed similarly to isolate the variable we want on the left side, and the other terms on the right side.
The equation would be 3(x + 5)