Plz answer 7-12!!!!!!!!!!!!!!!
2 answers:
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<h3>Answer:</h3>
See the attached
<h3>Step-by-step explanation:</h3>When you square the binomial (a -b), you get ...
... (a -b)² = a² -2ab +b²
That is, both the a² and b² terms have positive signs, and the middle term is twice the product of the roots of the squared terms.
The last two selections have negative signs on the constant, so cannot be perfect square trinomials.
The first selection has a middle term that is -ab, not -2ab, so it is not a perfect square trinomial, either.
The second selection is the correct one:
... 4a² -20a +25 = (2a +5)²
Explanation: To graph the equation y = x - 2, the idea is simple.
We pick three values for <em>x</em>, plug them into the
equation, and find the corresponding values for <em>y</em>.
In order to do this in an organized way, we set up a chart.
Down the left side we make a column for our x's, down the right side we make a column for our y's, and in the middle we make a column for the side of the equation that contains the <em>x</em>, in this case, x - 2.
So reading the chart from left to right, we have our <em>x</em> or what we plug into the equation, then we have the x - 2 or what happens to the x when we plug it into the equation, and finally we have our <em>y</em> or what we end up with after plugging <em>x</em> into the equation.
So our first step in filling out the chart is choosing
the x's that we want to plug into the equation.
When graphing lines, I always choose three x's,
a positive number, zero, and a negative number.
The easiest numbers to pick are usually 1, 0, and -1.
So for 1, we plug a 1 into the equation for <em>x</em>
and we have y = (1) - 2 or -1
For 0 we plug a 0 into the equation for <em>x</em>
and we have y = (0) - 2 or -2.
For -1 we plug a -1 into the equation for <em>x</em>
and we have y = (-1) - 2 or -3.
So our points are (1, -1), (0, -2), and (-1, -3).
I looked at the x and y column to find those points.
Since our domain is all real numbers, we can connect these points based on the pattern that they are forming on the graph which is a line.
