Answer:
Option A
Step-by-step explanation:
Here is how to approach the problem:
We see that all our restrictions for all four answer choices are relatively the same with a couple of changes here and there.
One way to eliminate choices would be to look at which restrictions don't match the graph.
At x<-5, there is a linear function that does have a -2 slope and will intersect the x axis at -7. The line ends with an open circle, so any answer choice with a linear restriction of x less than or equal to -5 is wrong. This cancels out choices C and D.
Now we have two choices left.
For the quadratic in the middle, the vertex is at (-2,6) and the vertex is a maximum, meaning our graph needs to have a negative sign in front of the highest degree term. In our case, none of our quadratics left are in standard form, and instead are in vertex form.
Vertext form is f(x) = a(x-h)^2 + k.
h being the x-coordinate of the vertex and k being the y-coordinate.
We know that the opposite of h will be the actual x-coordinate of the vertex, so if our vertex is -2, we will see x+2 inside the parenthesis. This leaves option A as the only correct choice.
Answer:
<h2>3(cos 336 + i sin 336)</h2>
Step-by-step explanation:
Fifth root of 243 = 3,
Suppose r( cos Ф + i sinФ) is the fifth root of 243(cos 240 + i sin 240),
then r^5( cos Ф + i sin Ф )^5 = 243(cos 240 + i sin 240).
Equating equal parts and using de Moivre's theorem:
r^5 =243 and cos 5Ф + i sin 5Ф = cos 240 + i sin 240
r = 3 and 5Ф = 240 +360p so Ф = 48 + 72p
So Ф = 48, 120, 192, 264, 336 for 48 ≤ Ф < 360
So there are 5 distinct solutions given by:
3(cos 48 + i sin 48),
3(cos 120 + i sin 120),
3(cos 192 + i sin 192),
3(cos 264 + i sin 264),
3(cos 336 + i sin 336)
That’s true!!!! They totally should do that!
Answer:
(20, 16)
Step-by-step explanation:
6 hours and 40 mins which would be 6:45
