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In-s [12.5K]
3 years ago
7

What is the value of x

Mathematics
1 answer:
kozerog [31]3 years ago
6 0

Answer:


Step-by-step explanation:

If the upper triangle has a right angle at Island C, we need only use the Pythagorean Theorem to find x:

(10 mi)^2 + (20 mi)^2 = x^2, so that

x^2 = 100 mi^2 + 400 mi^2 = 500 mi^2, which reduces to

x = +10√5 mi

Note that the shortest side (10 mi) is adjacent to a 30 degree angle.  Thus, the upper triangle is a 30-60-90 triangle, meaning that the Pythagorean Theorem DOES apply here.

Next time, please incude ALL of the verbal instructions.  Thank you.

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RideAnS [48]

Answer:

X.

Step-by-step explanation:

Hope this helps.

3 0
3 years ago
Let Y be a random variable with a density function given by
Neporo4naja [7]

From the given density function we find the distribution function,

F_Y(y)=P(Y\le y)=\displaystyle\int_{-\infty}^y f_Y(t)\,\mathrm dt=\begin{cases}0&\text{for }y

(a)

F_{U_1}(u_1)=P(U_1\le u_1)=P(3Y\le u_1)=P\left(Y\le\dfrac{u_1}3\right)=F_Y\left(\dfrac{u_1}3\right)

\implies F_{U_1}(u_1)=\begin{cases}0&\text{for }u_1

\implies f_{U_1}(u_1)=\begin{cases}\frac{{u_1}^2}{18}&\text{for }-3\le u_1\le3\\0&\text{otherwise}\end{cases}

(b)

F_{U_2}(u_2)=P(3-Y\le u_2)=P(Y\ge3-u_2)=1-P(Y

\implies F_{U_2}(u_2)=\begin{cases}0&\text{for }u_2

\implies f_{U_2}(u_2)=\begin{cases}\frac32(u_2-3)^2&\text{for }2\le u_2\le4\\0&\text{otherwise}\end{cases}

(c)

F_{U_3}(u_3)=P(Y^2\le u_3)=P(-\sqrt{u_3}\le Y\le\sqrt{u_3})=F_Y(\sqrt{u_3})-F_y(\sqrt{u_3})

\implies F_{U_3}(u_3)=\begin{cases}0&\text{for }u_3

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5 0
3 years ago
Save me the headache
maxonik [38]

(9\sin2x+9\cos2x)^2=81

Taking the square root of both sides gives two possible cases,

9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1

or

9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1

Recall that

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

If \alpha=2x and \beta=\dfrac\pi4, we have

\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}

so in the equations above, we can write

\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1

Then in the first case,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}

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(where n is any integer)

\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi

\implies x=n\pi\text{ or }\dfrac\pi4+n\pi

and in the second,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi

\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi

\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi

Then the solutions that fall in the interval [0,2\pi) are

x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4

5 0
3 years ago
Read 2 more answers
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EleoNora [17]

Product is multiplication.

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3 years ago
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vovangra [49]

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Given Data

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y = 57x+ 32 -----------------1

Company B

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Equating 1 and 2 we have

57x+ 32 = 62x+ 0

Solving for x we have

62x-57x = 32

5x = 32

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x = 32/5

x = 6.4 days

Learn more about word problem here:

brainly.com/question/13818690

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2 years ago
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