HELP!!!!
1 answer:
<h3>3 Answers: Choice D, Choice E, Choice F</h3>
============================================================
Explanation:
The inequality 6x - 10y ≥ 9 solves to y ≤ (3/5)x - 9/10 when you isolate y.
Graph the line y = (3/5)x - 9/10 and make this a solid line. The boundary line is solid due to the "or equal to" as part of the inequality sign. We shade below the boundary line because of the "less than" after we isolated for y.
Now graph all of the points given as I've done so in the diagram below. The points in the blue shaded region, or on the boundary line, are part of the solution set. Those points are D, E and F.
We can verify this algebraically. For instance, if we weren't sure point E was a solution or not, we would plug the coordinates into the inequality to get...
6x - 10y ≥ 9
6(5) - 10(2) ≥ 9 .... plug in (x,y) = (5,2)
30 - 20 ≥ 9
10 ≥ 9 ... this is a true statement
Since we end up with a true statement, this verifies point E is one of the solutions. I'll let you check points D and F.
-----------
I'll show an example of something that doesn't work. Let's pick on point A.
We'll plug in (x,y) = (-1,1)
6x - 10y ≥ 9
6(-1) - 10(1) ≥ 9
-6 - 10 ≥ 9
-16 ≥ 9
The last inequality is false because -16 is smaller than 9. So this shows point A is not a solution. Choices B and C are non-solutions for similar reasons.
You might be interested in
Answer:
1)
Minimum is a 4th Degree
2)
Positive; Even
3)
Step-by-step explanation:
Part 1)
The minimum degree of our function will be 4.
Looking at the graph, we know that the graph crosses the x-axis at -4 and -1. Since it <em>crosses through</em> the x-axis at these two points, these two factors must have an odd multiplicity.
So, it can be anything 1, 3, 5, 7, etc.
However, we will choose the lowest one, 1.
Next, we know that the graph <em>bounces off</em> at 3.
So, it must have an even multiplicity. In other words, 2, 4, 6, 8, etc.
We choose the lowest one, 2.
Therefore, the minimum degree of our function will be 1+1+2 or 4.
Part 2)
The degree of our polynomial is (and will always be) even. Therefore, both ends of the graph will go in the same direction.
Recall the simplest even polynomial, the parent quadratic function. When the leading coefficient is positive, both of the ends go straight up.
This applies to all polynomials with even degrees.
Therefore, since the arms of the graph is going straight up towards positive infinity, the leading coefficient of our graph must be positive.
Part 3)
This is similar to Part 1.
We can see that the graph touches the x-axis at -4, -3, and 1. So, the zeros of the function is:
We know that it<em> passes through</em> . So, these two factors must have an odd multiplicity.
However, since the graph <em>bounces off</em> , this factor must have an even multiplicity.
And we're done!
The equation of a circle in standard form is
where (h, k) is the center of the circle, and r is the radius if the circle.
We need to find the radius and center of the circle.
We are given a diameter, so to find the center, we need the midpoint of the diameter.
M = ((-6 + 6)/2, (6 + (-2))/2) = (0, 2)
The center is (0, 2).
To find the radius, we find the length of the given diameter and divided by 2.
where (h, k) is the center of the circle, and r is the radius if the circle.
We need to find the radius and center of the circle.
We are given a diameter, so to find the center, we need the midpoint of the diameter.
M = ((-6 + 6)/2, (6 + (-2))/2) = (0, 2)
The center is (0, 2).
To find the radius, we find the length of the given diameter and divided by 2.