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3 years ago

ANSWER ASAP !!

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

3 0

14 units2 is the answer

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How do I solve this algebra two btw

pychu [463]

$\bf \begin{cases} x^2+y^2=400\\ \boxed{y}=x-28 \end{cases}\qquad \stackrel{\textit{substituting on the 1st equation}}{x^2+\left( \boxed{x-28} \right)^2=400} \\\\\\ x^2+(x^2-56x+28^2)=400\implies x^2+x^2-56x+28^2=400 \\\\\\ 2x^2-56x+784=400\implies 2x^2-56x+384=0 \\\\\\ 2(x^2-28x+192)=0\implies x^2-28x+192=0 \\\\\\ (x-16)(x-12)=0\implies x = \begin{cases} 16\\ 12 \end{cases}$

now that we know what are the x-values, what are the y-values? well, we can just use the 2nd equation, since we know that y = x - 28, then

$\bf y = x - 28\implies \stackrel{\textit{when x = 16}}{y = 16 - 28}\implies y = -12 \\\\\\ y = x - 28\implies \stackrel{\textit{when x = 12}}{y = 12 - 28}\implies y = -16 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{x}{16}~~,~~\stackrel{y}{-12})\qquad,\qquad (\stackrel{x}{12}~~,~~\stackrel{y}{-16})~\hfill$

7 0

3 years ago

Evaluate the logarithmic equation:

evablogger [386]

Answer

simplified is log5 (2)

the 5 is suposed to be the subscript one.

6 0

2 years ago

Complete the area model to identify the factored

beks73 [17]

I think it’s C but I am not sure sorry dude

5 0

3 years ago

Which is the graph of the function f(x) = one-halfx2 + 2x – 6?

guapka [62]

Answer:

1/2 x^2 + 2x - 6 = 0

(1/2 x - 1)(x + 6) = 0

zeroes are 2 and -6

so the graph intersects x axis at -6 and 2

The only one to do that is diagram A

Step-by-step explanation:

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 $U\sim N(n,1)$ :

$E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1$

so we end up with

$E[Y^n]=e^{\frac12n^2}$

3 0

3 years ago