How do you do this problem
1 answer:
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9514 1404 393
Answer:
"complete the square" to put in vertex form
Step-by-step explanation:
It may be helpful to consider the square of a binomial:
(x +a)² = x² +2ax +a²
The expression x² +x +1 is in the standard form of the expression on the right above. Comparing the coefficients of x, we see ...
2a = 1
a = 1/2
That means we can write ...
(x +1/2)² = x² +x +1/4
But we need x² +x +1, so we need to add 3/4 to the binomial square in order to make the expressions equal:
_____
Another way to consider this is ...
x² +bx +c
= x² +2(b/2)x +(b/2)² +c -(b/2)² . . . . . . rewrite bx, add and subtract (b/2)²*
= (x +b/2)² +(c -(b/2)²)
for b=1, c=1, this becomes ...
x² +x +1 = (x +1/2)² +(1 -(1/2)²)
= (x +1/2)² +3/4
_____
* This process, "rewrite bx, add and subtract (b/2)²," is called "completing the square"—especially when written as (x-h)² +k, a parabola with vertex (h, k).
A polygon is a two-dimensional closed object. The polygon that best describes a stop sign is a regular octagon.
<h3>What is a polygon?</h3>A polygon is a two-dimensional closed object with n number of straight sides that is flat or planar and the value of n is always greater than 2 (n>2).
<h3>What is an octagon?</h3>An octagon is a polygon that has 8 number of sides.
As we know that a stop sign has 8 sides, therefore, the polygon that is in the shape of the stop sign is an octagon.
An octagon has 8 sides, and as it is mentioned in the problem that sides and angles appear to be congruent, therefore, the polygon must be a regular polygon.
Hence, the polygon that best describes a stop sign is a regular octagon.
Learn more about Polygon:
Answer: y≤
Step-by-step explanation:
Since we are given the equation, all we have to do is to fix the inequality. We have four options: <, >, ≤, and ≥. The line in the graph is a solid line. This means the line reaches and is equal to the points on the line. We can automatically eliminate ≤ and ≥. Now, we know the answer is ≤. The shading shows that the answer is less than 7/3x+2. In other words, y≤ .