Select all statements that are true about the linear equation.
1 answer:
<h2>Linear equations</h2><h3>Concept</h3>
Linear equations, or a straight line on a graph, generally consist of two variables, x, and y - as well as slope.
<h3>Utilization</h3>You have been provided a linear equation in slope-intercept form (y=mx+b). Let's look at the true and false statements now:
The graph of the equation is the set of all points that are solutions to the equation, TRUE or FALSE. The answer to this is true, because any point on the line should provide a valid solution to the problem. TRUE
The point (10, 3) is on the graph of the equation. We have already determined that in order for a point to be on the line - it must be a solution... so, plug it in and test:
3 is not equal to 118, so the point (10,3) is not on this line. FALSE
The point (−2,−1) is on the graph of the equation. Use the same setup as before:
-1 is not equal to -25, so this point is once again, not on the line. FALSE
The graph of the equation is a single point representing one solution to the equation. This is nearly the inverse of the first statement, so we can rule this out as well. FALSE
<h3>Answer</h3>The only true statement here seems to be The graph of the equation is the set of all points that are solutions to the equation. Hope this helps you!
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<h3>Answer: Venn Diagram is below</h3>
The rectangle represents the universal set (marked as "set U"). Everything is contained in this rectangle.
Circle A represents set A = {a, b, e, f, h, j}
Circle B represents set B = {b, c, d, f, j}
The common elements between the two sets are {b, f, j}, so we'll write these three values in the overlapping portion of the two circles.
The items {a,e,h} will go in circle A, but not in the overlapping portion since they are found only in set A.
Similarly we'll have {c,d} go in circle B but not in the overlapping portion since these items are only found in set B.
The remaining terms {g, i, k} go outside both circles, but inside the rectangle, because neither of these elements are in set A or set B.
Answer:
The polynomial function of the lowest degree that has zeroes at -1, 0 and 6 and with a leading coefficient of one is .
Step-by-step explanation:
From Fundamental Theorem of Algebra, we remember that the degree of the polynomials determine the number of roots within. Since we know three roots, then the factorized form of the polynomial function with the lowest degree is:
(1)
Where ,
and
are the roots of the polynomial.
If we know that ,
and
, then the polynomial function in factorized form is:
(2)
And by Algebra we get the standard form of the function:
(3)
The polynomial function of the lowest degree that has zeroes at -1, 0 and 6 and with a leading coefficient of one is .