Answer:
see explanation
Step-by-step explanation:
The equation of a parabola in vertex form is
y = a(x - h)² + k
where (h, k) are the coordinates of the vertex and a is a multiplier
Given
x² + x + 1 ← to obtain vertex form use metod of completing the square
To complete the square
add/ subtract (half the coefficient of the x- term )² to x² + x
x² + 2()x + - + 1
f(x) = (x + )² + ← in vertex form
Yes it can, it has to be a fraction
If we know your Pythagorean Triples we can immediately recognize that the last choice is a right triangle:
8² + 15² = 17²
If you don't know your Pythagorean Triples, it's worth learning the first few off the list because teachers use them in problems all the time. But for now let's just exhaustively check the Pythagorean Theorem for each triangle. We don't have to multiply everything out; we can analyze the common factors. If two have a common factor that the third one doesn't have, there's no way for the Pythagorean Theorem to add up.
Clearly 5²+15² is a multiple of 5 but 18² isn't so that one isn't a right triangle.
6²+12² is a multiple of 6, 16² isn't a multiple of 6, not an RT.
15²-5² is a multiple of 5, 13² isn't, no joy.
8²+15² = 64 + 225 = 289 = 17² -- that's a real right triangle, a valid Pythagorean Triple.
Answer:
Step-by-step explanation:
Identify the graph and describe the solution set of this system of inequalities.
x
ya- x -10
y> - x-2
3
'a' can't be uniquely found without some more information about the triangle. With only the given information, all we can say is that 5a must be at least 17. Is there a picture of the triangle ? Is there maybe possibly an angle given ? ?