Answer:
acute
Step-by-step explanation:
Evaluate the angle C = (11x + 85)/3 at x = 13:
C(13) = (143 + 85)/3, or 76 degrees. This is an acute angle (more than 0 and less than 90 degrees).
Answer:5/48,3/16,0.5,0.75
Step-by-step explanation:
9514 1404 393
Answer:
A. 5
Step-by-step explanation:
The base AB is twice the length of the midsegment CD.
8n +10 = 2(5n)
8n +10 = 10n . . . . . eliminate parentheses
10 = 2n . . . . . . . . . subtract 8n
n = 5 . . . . . . . . . . . divide by 2
Thet contradict each other, that's why both of them are incorrect.
<span>Suppose that a polynomial has four roots: s, t, u, and v. If the polynomial were evaluated at any of these values, it would have to be zero. Therefore, the polynomial can be written in this form.
p(x)(x - s)(x - t)(x - u)(x - v), where p(x) is some non-zero polynomial
This polynomial has a degree of at least 4. It therefore cannot be cubic.
Now prove Kelsey correct. We have already proved that there can be no more than three roots. To prove that a cubic polynomial with three roots is possible, all we have to do is offer a single example of that. This one will do.
(x - 1)(x - 2)(x - 3)
This is a cubic polynomial with three roots, and four or more roots are not possible for a cubic polynomial. Kelsey is correct.
Incidentally, if this is a roller coaster we are discussing, then a cubic polynomial is not such a good idea, either for a vertical curve or a horizontal curve. I hope this helps</span><span>
</span>
I'll talk you through it so you can see why it's true, and then
you can set up the 2-column proof on your own:
Look at the two pointy triangles, hanging down like moth-wings
on each side of 'OC'.
-- Their long sides are equal, OA = OB, because both of those lines
are radii of the big circle.
-- Their short sides are equal, OC = OC, because they're both the same line.
-- The angle between their long side and short side ... the two angles up at 'O',
are equal, because OC is the bisector of the whole angle there.
-- So now you have what I think you call 'SAS' ... two sides and the included angle of one triangle equal to two sides and the included angle of another triangle.
(When I was in high school geometry, this was not called 'SAS' ... the alphabet
did not extend as far as 'S' yet, and we had to call this congruence theorem
"broken arrow".)
These triangles are not congruent the way they are now, because one is
the mirror image of the other one. But if you folded the paper along 'OC',
or if you cut one triangle out and turn it over, it would exactly lie on top of
the other one, and they would be congruent.
So their angles at 'A' and at 'B' are also equal ... those are the angles that
you need to prove equal.
