What is probality of class 9th in math ?

Natalka [10]

.33 with a line over the .33 because it is not able to be simplified into a whole decimal when ever both numbers are a factor of 3 besides when it is half that will happen

4 0

3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

2 years ago

If x is a positive integer and y is a negative integer than x + y is a rational number.

romanna [79]

Answer:

Yes

Step-by-step explanation:

An integer is a whole number so if x is positive and y is negative then the equation would just be x plus negative y. But x+-y is an odd equation so the simplified version is x-y so it is a rational number, since a rational number is a number that can be rewritten as a simple fraction.

5 0

3 years ago

Which property was used to simply the expression 3c+9+4c=3c+4c+9

34kurt

<span> commutative property</span>

7 0

2 years ago

HELP RIGHT NOW PIZZZ = FASTEST ANSWER BRAINLIEST AND I WILL THANK U

Nuetrik [128]

Answer:

1) The slope of the function $g(x)$ is $0$ and the slope of the function $f(x)$ is $-1$ .

2) The negative slope of the function $f(x)$ shows that it is the line is increasing and the slope $0$ of the function $g(x)$ shows that the line will always have the same y-coordinate.

3) The slope of the function is $f(x)$ is greater than the slope of the function $g(x)$ .

Step-by-step explanation:

For this exercise you need to know that the slope of any horizontal line is zero ( $m=0$ )

The slope of a line can be found with the following formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

You can observe in the graph of the function $g(x)$ given in the exercise, that this is an horizontal line. Then, you can conclude that its slope is:

$m=0$

The steps to find the slope of the function $f(x)$ shown in the table attached, are the following:

- Choose two points, from the table:

$(0,3)$ and $(4,-1)$

- You can say that:

$y_2=-1\\y_1=3\\\\x_2=4\\x_1=0$

- Substitute values into the formula $m=\frac{y_2-y_1}{x_2-x_1}$ :

$m=\frac{-1-3}{4-0}$

- Finally, evaluating, you get:

$m=\frac{-4}{4}\\\\m=-1$

Therefore:

1) The slope of the function $g(x)$ is $0$ and the slope of the function $f(x)$ is $-1$ .

2) The negative slope of the function $f(x)$ shows that it is the line is increasing and the slope $0$ of the function $g(x)$ shows that the line will always have the same y-coordinate.

3) The slope of the function is $f(x)$ is greater than the slope of the function $g(x)$ .

8 0

3 years ago