To put an equation into (x+c)^2, we need to see if the trinomial is a perfect square.
General form of a trinomial: ax^2+bx+c
If c is a perfect square, for example (1)^2=1, 2^2=4, that's a good indicator that it's a perfect square trinomial.
Here, it is, because 1 is a perfect square.
To ensure that it's a perfect square trinomial, let's look at b, which in this case is 2.
It has to be double what c is.
2 is the double of 1, therefore this is a perfect square trinomial.
Knowing this, we can easily put it into the form (x+c)^2.
And the answer is: (x+1)^2.
To do it the long way:
x^2+2x+1
Find 2 numbers that add to 2 and multiply to 1.
They are both 1.
x^2+x+x+1
x(x+1)+1(x+1)
Gather like terms
(x+1)(x+1)
or (x+1)^2.
-38 due to the fact the lower the number, the colder it is.
For the information given, a line segment of 4 inches can be drawn
<h3>How to illustrate the segment?</h3>
For the 'copy', you would measure the line drawn, and duplicate it to the right, measuring it to make sure it is the exact length of the first line.
To bisect the original line segment, measure the line with a ruler or other device, and calculate 1/2 distance from one end, mark it, measure from left end to mark, make sure right section is equal in length.
Draw an angle of 90 degree and label it to avoid confusion. Now use a scale to draw a ray and label the vertex point so we can know the point where to place compass. Use compass to draw an arc on angle. This angle will help you measure where to put the new point on our new angle.
Similarly draw the same arc on the new ray. This is the first step in finding the new point. Then, use compass to measure the difference between the two places where the arc meets the angle on angle.
Furthermore, use the measurements from the previous step to draw another arc. This step is to help establish a point where the two arcs meet. Now draw a point where the two arcs met and drew a line through it.
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Answer: 1 and 7/15
Step-by-step explanation:
I made the common donominator 15 and then did the problem but the donominator stays the same
