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2 years ago

PLS HELP ME I ONLY HAVE 5 MINUTES PLS

Mathematics

1 answer:

Viefleur [7K]2 years ago

3 0

Answer: 71 ft

i hope its the right answeer, do well

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What is the factored form of x^2y^3 -2y^3 - 2x^2 +4

Law Incorporation [45]

(-x^2+4)

x^2y^3-2y^3-2x^2+4=x^2-2x^2+4=

(-x^2+4)

8 0

3 years ago

2/5 of 1 meter=____cm<br> Plz help thx

Crazy boy [7]

Answer:

40 cm

Explanation:

2/5 = 0.4

1 meter = 100 cm

so,

100 x 0.4= 40 cm

4 0

3 years ago

If a triangle has sides 4, 6 and 10 and a 2nd similar triangle has a perimeter of 60, find the 2nd largest side of the 2nd trian

algol13

The first triangle has a perimeter of 4+6+10 = 20.

The 2nd longest side is 6/20 = 3/10 of the perimeter. In the larger triangle, the 2nd longest side will have length (3/10)*60 = 18.

_____
These side lengths mean the "triangle" is a line segment. That is, it has zero area.

8 0

3 years ago

4sin²<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B2%7D" id="TexFormula1" title="\frac{x}{2}" alt="\frac{x}{2}" align="absm

raketka [301]

Answer:

$\displaystyle x=\left \{\frac{2\pi}{3}+2\pi k,\frac{4\pi}{3}+2\pi k, \frac{8\pi}{3}+2\pi k, \frac{10\pi}{3}+2\pi k\right \}k\in \mathbb{Z}$

Step-by-step explanation:

Hi there!

We want to solve for $x$ in:

$4\sin^2(\frac{x}{2})=3$

Since $x$ is in the argument of $\sin^2$ , let's first isolate $\sin^2$ by dividing both sides by 4:

$\displaystyle \sin^2\left(\frac{x}{2}\right)=\frac{3}{4}$

Next, recall that $\sin^2x$ is just shorthand notation for $(\sin x)^2$ . Therefore, take the square root of both sides:

$\displaystyle \sqrt{\sin^2\left(\frac{x}{2}\right)}=\sqrt{\frac{3}{4}},\\\sin\left(\frac{x}{2}\right)=\pm \sqrt{\frac{3}{4}}$

Simplify using $\displaystyle \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ :

$\displaystyle \sin\left(\frac{x}{2}\right)=\pm \sqrt{\frac{3}{4}},\\\sin\left(\frac{x}{2}\right)=\pm \frac{\sqrt{3}}{\sqrt{4}}=\pm \frac{\sqrt{3}}{2}$

Let $\phi = \frac{x}{2}$ .

<h3><u>Case 1 (positive root):</u></h3>

$\displaystyle \sin(\phi)=\frac{\sqrt{3}}{2},\\\phi = \frac{\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\phi =\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z}$

Therefore, we have:

$\displaystyle \frac{x}{2}=\phi = \frac{\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\frac{x}{2}=\phi =\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z},\\\\\begin{cases}x=\boxed{\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z}},\\x=\boxed{\frac{4\pi}{3}+2\pi k , k \in \mathbb{Z}}\end{cases}$

<h3><u>Case 2 (negative root):</u></h3>

$\displaystyle \sin(\phi)=-\frac{\sqrt{3}}{2},\\\phi = \frac{4\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\phi =\frac{5\pi}{3}+2\pi k, k\in \mathbb{Z},\\\begin{cases}x=\boxed{\frac{8\pi}{3}+2\pi k, k\in \mathbb{Z}},\\x=\boxed{\frac{10\pi}{3}+2\pi k , k \in \mathbb{Z}}\end{cases}$

8 0

2 years ago

The computer center at Dong-A University has been experiencing computer down time. Let us assume that the trials of an associate

Schach [20]

Answer:

(a)0.16

(b)0.588

Step-by-step explanation:

The matrix below shows the transition probabilities of the state of the system.

$\left(\begin{array}{c|cc}&$Running&$Down\\---&---&---\\$Running&0.90&0.10\\$Down&0.30&0.70\end{array}\right)$

(a)To determine the probability of the system being down or running after any k hours, we determine the kth state matrix $P^k$ .

(a)

$P^1=\left(\begin{array}{c|cc}&$Running&$Down\\---&---&---\\$Running&0.90&0.10\\$Down&0.30&0.70\end{array}\right)$

$P^2=\begin{pmatrix}0.84&0.16\\ 0.48&0.52\end{pmatrix}$

If the system is initially running, the probability of the system being down in the next hour of operation is the $(a_{12})th$ entry of the P^2$ matrix.$

The probability of the system being down in the next hour of operation = 0.16

(b)After two(periods) hours, the transition matrix is:

$P^3=\begin{pmatrix}0.804&0.196\\ 0.588&0.412\end{pmatrix}$

Therefore, the probability that a system initially in the down-state is running

is 0.588.

(c)The steady-state probability of a Markov Chain is a matrix S such that SP=S.

Since we have two states, $S=[s_1$ s_2]$

$[s_1$ s_2]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[s_1$ s_2]$

Using a calculator to raise matrix P to large numbers, we find that the value of $P^k$ approaches [0.75 0.25]:

Furthermore,

$[0.75$ 0.25]\left(\begin{array}{ccc}0.90&0.10\\0.30&0.70\end{array}\right)=[0.75$ 0.25]$

The steady-state probabilities of the system being in the running state and in the down-state is therefore:

$[s_1$ s_2]=[0.75$ 0.25]$

4 0

3 years ago