Answer:
· This image may result from the construction of <em>an angle congruent to a given angle</em>.
· The next step in this construction is to set the compass width to <em>arc JK and draw an arc centered at L intersecting the existing arc through L</em>.
Step-by-step explanation:
If the step shown above results in point of intersection P, and the construction is completed by drawing ray DP, then this construction produces triangle DLP congruent to triangle BKJ. The angle at D will be congruent to the angle at B because CPCTC. Hence an angle congruent to a given angle will have been constructed.
The angle would be 70 degrees because 180-110=70
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I think you mean "if the points <span>(2,5), (3,2) and (4,5) satisfy an unknown 3rd degree polynomial, what is the polynomial?"
Since 3 roots {2, 3, 4} are known, we might begin by assuming that this poly would have the form y = ax^3 + bx^2 + cx + d (which has three factors). Unfortunately, three roots are not enough to determine all four constants {a, b, c, d}.
So, let's assume, instead, that the poly would have the form y = ax^2 + bx + c. Three given points should make it possible to determine {a, b, c}.
(2,5): 5 = a(2)^2 + b(2) + c => 5 = 4a + 2b + c
(3,2): 2 = a(3)^2 + b(3) + c => 2 = 9a + 3b + 5 - 4a - 2b
(4,5): 5 = a(4)^2 + b(4) + c => 5 = 16a + 4b + 5 - 4a - 2b
Now we have two equations in a and b alone, which enables us to solve for a and b:
</span>2 = 9a + 3b + 5 - 4a - 2b becomes -3 = 5a + b
<span>and
</span>5 = 16a + 4b + 5 - 4a - 2b becomes 0 = 12a + 2b, or 0 = 6a + b, or 0=-6a-b
<span>
Adding this result to -3 = 5a + b, we get -3 = -a, so a =3.
Thus, since -3 = 5a + b, -3 = 5(3) + b, so b = -18
All we have to do now is to find c. Let's do this using </span>5 = 4a + 2b + c.
We know that a = 3 and b = -18, so this becomes 5 = 4(3) + 2(-18) + c.
Thus, 5 = 12 - 36 + c, or c = 29.
With a, b and c now known, we can write the poly as y = 3x^2 - 18x + 29.
Now the only thing to do remaining is to verify that each of the three given points satsifies y = 3x^2 - 18x + 29. Try this, please.
