The question.

Please help whoever can with #13 and thank you

bixtya [17]

These are so great! They are a perfect combination of Physics and pre-calculus! Your max height of that projectile is going to occur at the max value of the parabola, or at its vertex. So we need to find the vertex. The coordinates of the vertex will give us the x value, which is the time in seconds it takes to reach y which is the max height. Do this by completing the square. Begin by setting the equation equal to 0 and then moving the 80 over to the other side. Then factor out the -16. This is all that: $-16( t^{2}-4t)=-80$ . Take half the linear term which is 4 and square it and add it in to both sides. Half of 4 is 2, 2 squared is 4, so add 4 into the set of parenthesis and to the -80. $-16( t^{2}-4t+4)=-80-64$ . The -64 on the right comes from the fact that when you added 4 into the parenthesis, you had the -16 out in front which is a multiplier. -16 * 4 - -64. So what you really added in was -64. Now the perfect square binomial we created in that process was $-16(x-2) ^{2} =-144$ . When we move the 144 back over by addition we find that the vertex of the polynomial is (2, 144). And that tells us that it takes 2 seconds for the projectile to reach its max height of 144 feet. To find the time interval in which the object's height decreases occurs from its max height of 144 to where the graph of the parabola goes through the x-axis to the right of the max. To find where the graph goes through the x-axis, or the zeroes of the graph, you factor the polynomial. When you do that using the quadratic formula you get that x = -1 and 5. So at its max height it is at 2 seconds, and by 5 seconds it hits the ground. So the time interval of its height decreasing is from 2 seconds to 5 seconds, or a total of 3 seconds. I think you need the 2 and 5, from the wording of your problem.

8 0

4 years ago

Which of the following graphs shows the solution to the system of equations? <br> y=5x-1<br> y=x+3

raketka [301]

Answer:

sorry,but there is no graph ...its difficutlt to say u the ans

7 0

3 years ago

pls There are (4^9)^5 ⋅ 4^0 books at the library. What is the total number of books at the library? A.4^9 B. 4^45 C. 4^46 D. 1

jekas [21]

We know that there are (4^9)^5 ⋅ 4^0 at the library, so we just need to simplify this to get the answer.
(4^9)^5 ⋅ 4^0
=(4^9)^5×1
=(4^9)^5
=4^9×5
=4^45. As a result, the total number of books at the library is 4^45 books at the library or B is the final answer. Hope it help!

6 0

3 years ago

From Euler’s relation eiθ = cosθ + isinθ,

adelina 88 [10]

Answer:

(a) The graphic representation is in the attached figure.

(b) $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$ .

(c) $\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ .

Step-by-step explanation:

(a) Given a complex number $e^{i\theta}$ we know, from Euler's formula that $e^{i\theta} = \cos(\theta)+i\sin(\theta)$ . So, it is not difficult to notice that

$|e^{i\theta}|^2 = \cos^2(\theta)+\sin^2(\theta) =1$

so it is on the unit circumference. Also, notice that the Cartesian representation of the complex number is $(\cos(\theta), \sin(\theta))$ .

Now,

$e^{-i\theta} = \cos(\theta)+i\sin(-\theta) = \cos(\theta)-i\sin(\theta)$ .

Notice that $e^{-i\theta}$ has the same modulus that $e^{i\theta}$ , so it is on the unit circumference. Beside, its Cartesian representation is $(\cos(\theta), -\sin(\theta))$ .

So, the points $(\cos(\theta), \sin(\theta))$ and $(\cos(\theta), -\sin(\theta))$ are symmetric with respect to the X-axis. All this can be checked in the attached figure.