Evaluate the equation.

A. h=
B. h=
C. h=
D. h=

where is the equation?

Explanation:

For h(w) = w2 – 3, find h(2d + 1).

For every instance of the variable w, I'll need to plug in the expression 2d + 1. I'll use parentheses to make the replacements clear for my next step.

h(2d + 1) = (2d + 1)2 – 3

I need to multiply out the squared binomial next, and then simplify:

(2d + 1)2 – 3

= (4d 2 + 4d + 1) – 3

= 4d 2 + 4d + 1 – 3

= 4d 2 + 4d – 2

I've simplified as much as I can. My answer is:

h(2d + 1) = 4d 2 + 4d – 2

Notice that, while doing the squaring halfway through my work above, I didn't ignore (and then maybe forget) the "minus three" that was along for the ride. In whatever manner you do your work, make sure that you're doing it so you don't get so busy with one part of an exercise that you find yourself in danger of losing track of other parts.

Explanation:

The letter Y can be regarded as both a vowel and a consonant

Explanation: