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3 years ago

How many lines of symmetry does a regular polygon with x sides have

Mathematics

2 answers:

anastassius [24]3 years ago

6 0

There are actually 2 variations in this case.

> First case: the number of sides is an odd number

<span> The number of lines of symmetry is simply equal to number of sides</span>

> Second case: the number of sides is an even number

<span> The number of lines of symmetry is equal to twice the number of sides</span>

BaLLatris [955]3 years ago

6 0

X sides because every regular polygon has the same amount of lines of symmetry as the amount of sides.

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Janet can plant nine feet of carrots in 15 minutes while her daughter Amy can plant 17 feet of carrots in half an hour.

Lena [83]

Answer:

Step-by-step explanation:

In order to find who is more efficient at planting carrots, we want to put both ratios in terms of the same amount of time.

For simplicity, lets see how many carrots each can make in 1 hour.

$Janet:\\9ft \: in \: 15 \:minutes$

$\frac{9ft}{15min} = \frac{xft}{60min}\\x = \frac{60*9}{15} = 36$

Janet can plant 36 feet of carrots in one hour

$Amy:\\\frac{17ft}{30min} = \frac{xft}{60min}\\x = \frac{17*60}{30} = 34$

Amy can plant 34 feet of carrots in one hour

So, Janet can plant carrots more quickly

6 0

3 years ago

you invest 2,600 in an account that pays an interest rate of 8.5% compounded continuously. Calculate the balance of your account

soldier1979 [14.2K]

The formula is
A=p e^rt
A future value?
P present value 2600
R interest rate 0.085
T time 5years
E constant
A=2,600×e^(0.085×5)
A=3,976.94

6 0

3 years ago

Will someone please answer this and remind me how to solve this

Veseljchak [2.6K]

Answer:

Step-by-step explanation:

Since the two triangles are similar, you can use proportions to solve for the value of $x$

$\frac{x}{12} = \frac{4}{3}$

Now, just multiply 4 and 12.

$4 * 12 = 48$

Then divide that by 3.

$48 / 3 = 16$

Therefore, $x = 16$ .

5 0

3 years ago

Read 2 more answers

Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

Cloud [144]

If $F_Y(y)$ is the cumulative distribution function for $Y$ , then

$F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)$

Then the probability density function for $Y$ is $f_Y(y)={F_Y}'(y)$ :

$f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}$

The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :