<img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B4%7D%20-%2012%20%5C%5C%20x%20-%202" id="TexFormula1" title=" {x}^{4}

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3 years ago

- 2" alt=" {x}^{4} - 12 \\ x - 2" align="absmiddle" class="latex-formula">
find the quotient showing the long division process

Mathematics

1 answer:

Natalka [10]3 years ago

5 0

Hope this helps you

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Which fraction us less than a half?5/6 3/4 or 2/5

Olin [163]

Answer:

That's easy 2/5

Step-by-step explanation:

Half of 5 is 2.5 and well ovbi 2 is less than 2.5 therefore 2/5 is lees than a half

Hope this helps!

8 0

3 years ago

1. Determine whether or not each of the following is a Binomial random variable. If it is not a Binomial random variable, provid

satela [25.4K]

Answer:

a) You roll a die 25 times. Let X be the number of times that you roll "3".

* Is a Binomial

b) The SJSU football team will play 12 games next season. They have a 70% chance of winning each conference game and a 60% chance of winning each nonconference game. Let X be the number of games the SJSU football team wins next season.

* Not a Binomial because probability of success is not constant

c) Evelyn has a 75% chance of getting an A in each class she takes. Let X be the number of times Evelyn earns an A during four years at SJSU.

* Is a Binomial

8 0

3 years ago

How do you write 0.326 and 0.924 in word form?

Sergeeva-Olga [200]

three hundred twenty-six thousandths = .326
nine hundred twenty-four thousandths = .924

Hope this helped!! :D

4 0

3 years ago

Read 2 more answers

Write the expression as a square of a monomial.

frozen [14]

Answer:

The square of a monomial is $(9x^2)^2$

Step-by-step explanation:

Consider the provided monomial.

$81x^4$

We need to Write the expression as a square of a monomial.

The above expression can be written as:

$9\times 9\times x^2\times x^2$

$9^2(x^2)^2$

$(9x^2)^2$

Hence, the square of a monomial is $(9x^2)^2$

8 0

3 years ago

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

3 years ago