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3 years ago

PLEASE HELP ASAP!!!! IS THIS CORRECT?

Mathematics

2 answers:

xz_007 [3.2K]3 years ago

7 0

From what I am looking at, it seems as if it is. :)

stealth61 [152]3 years ago

6 0

No you got it wrong sorry and good luck :}

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Given the expression below, what are the missing values, respectively?

asambeis [7]

Answer:

Step-by-step explanation:

20=4*5

9=4+5

7 0

2 years ago

Part 3 - Discussion/Explanation Question

SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

$\frac{1}{x - 5}$

Set the expression in the denominator equal to 0, because you can't divide by 0.

$x - 5 = 0$

$x = 5$

So the vertical asymptote is x=5.

Disclaimer if you see something like this

$\frac{(x - 5)(x + 3)}{(x - 5)}$

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

$\frac{1}{x}$

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

$a \times {x}^{0}$

For example,

$1 = 1 \times {x}^{0}$

$2 = 2 \times {x}^{0}$

And so on,

and

$x = {x}^{1}$

So our equation is basically

$\frac{1 \times {x}^{0} }{ {x}^{1} }$

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

$\frac{3 {x}^{2} }{ {x}^{2} + 1}$

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

$x = \frac{3}{1}$

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

8 0

1 year ago

In ΔSTU, the measure of ∠U=90°, SU = 7, UT = 24, and TS = 25. What ratio represents the tangent of ∠T?

Vinvika [58]

Answer:

Tan T = 7/24

Step-by-step explanation:

Firstly, please check attachment to have a picture of the triangle we are solving.

Now, we are concerned with calculating the ratio that represents the tangent of angle T.

Mathematically, the tangent of an angle is the ratio of the length of the opposite to the length of the adjacent.

In this question, our opposite is 7 while the adjacent is 24.

Thus Tan T = 7/24

6 0

3 years ago

54% in simplest form

Papessa [141]

As a fraction, 54% is 27/50.

4 0

2 years ago

Read 2 more answers

If X and Y are independent continuous positive random

Leni [432]

a) $Z=\frac XY$ has CDF

$F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy$

where the last equality follows from independence of $X,Y$ . In terms of the distribution and density functions of $X,Y$ , this is

$F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy$

Then the density is obtained by differentiating with respect to $z$ ,

$f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy$

b) $Z=XY$ can be computed in the same way; it has CDF

$F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Differentiating gives the associated PDF,

$f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Assuming $X\sim\mathrm{Exp}(\lambda_x)$ and $Y\sim\mathrm{Exp}(\lambda_y)$ , we have

$f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}$

and

$f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy$

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

6 0

3 years ago