Write a number that has 1/10 the value of 9 in 968

kifflom [539]

Answer:

The answers to the questions about problem 2 are:

The fraction of the length after Priya stops two times is 3/4.
The fraction of the length after Priya stops four times is 15/16.
Priya will never reach the end of the hallway.

Step-by-step explanation:

The explanation about each answer is below:

1. In the first stop, Priya has walked half of the total length, and there would still be half the hallway, however, when she stops the second time, she has walked the half of the half, I mean:

$\frac{1}{2}/2=\frac{1}{4}$

If you add the two values you obtain the total distance traveled by Priya:

$\frac{1}{2}+\frac{1}{4}= \frac{3}{4}$

And the distance traveled in two stops is 3/4 of the total length of the hallway.

2. With four stops the system is the same, we know with two stops Priya travels 3/4 of the hallway, now with the third stop, she will travel half of the remaining distance:

$\frac{1}{4}/2=\frac{1}{8}$

And in the fourth stop she will travel half of the remaining, I mean:

$\frac{1}{8}/2=\frac{1}{16}$

Now, we add all the values of the distances obtained:

$\frac{3}{4}+\frac{1}{8}+\frac{1}{16}= \frac{15}{16}$

So, the distance traveled by Priya in the fourth stop is 15/16 of the total length of the hallway.

3. How we know, the numbers are infinite, in the same forms the distances, by this reason, how the problem says that Priya walks just half of the distance, she will never reach the end because despite she has very near of the end, she will continue walking just half and ever smaller distances.

3 0

3 years ago

Quadrilateral ABCD has the following vertices: A(-4, -3), B(2, -3), C(4, -6), and D(-4, -6). Quadrilateral ABCD is a

seropon [69]

In order to get the figure, you have to plot it.

Quadlirateral ABCDwith <span>vertices: A(-4, -3), B(2, -3), C(4, -6), and D(-4, -6) is </span>s A. TRAPEZOID

6 0

3 years ago

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PLEASE HELP I DO NOT UNDERSTAND AT ALL ITS PRECALC PLEASE SERIOUS ANSWERS

Elina [12.6K]

You want to end up with $A\sin(\omega t+\phi)$ . Expand this using the angle sum identity for sine:

$A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi$

We want this to line up with $2\sin(4\pi t)+5\cos(4\pi t)$ . Right away, we know $\omega=4\pi$ .

We also need to have

$\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}$

Recall that $\sin^2x+\cos^2x=1$ for all $x$ ; this means

$(A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}$

Then

$\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)$