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3 years ago

Can I get help on this please

Mathematics

1 answer:

USPshnik [31]3 years ago

6 0

1. dismal 2. wager 3. peril 4. recline 5. shriek 6. sinister 7. conceal 8. inhabit 9. frigid 10. numb 11. corpse 12. tempt

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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

Gavin, Colin and Dan share some sweets in the ratio 1:4:1. Gavin gets 11 sweets. How many did Colin get?

djverab [1.8K]

If Gavin got 11 sweets, then the new ratio would be: 11:44:11

Colin got 44 sweets.

4 0

3 years ago

Using only whole numbers Nikki wrote as many multiplication equations as she could with 12 as the product what where her equatio

erastovalidia [21]

1 times 12,2 times 6,and 3times 4

4 0

2 years ago

Helpppp now please helpppp

algol [13]

The correct answer is: [B]: " $\frac{(-3)(4)}{(6)}$ " .
__________________________________________________________
Consider choice [A]: $\frac{(3)(4)}{(6)}$ ;
= $\frac{12}{6}$ ;
= 2 ; "2 ≠ -2" ; so we can rule out: "Choice [A]" .
__________________________________________________________
Consider choice [B]: $\frac{(-3)(4)}{(6)}$ ;
= $\frac{-12}{6}$ ;
= "-2" ; Yes!
→ Let us proceed with the final answer choice ;
__________________________________________________________
Consider choice [C]: $\frac{(-3)(-4)}{(6)}$ ;
= $\frac{12}{6}$ ;
= 2 ; "2 ≠ -2" ; so we can rule out: "Choice [C]" .
__________________________________________________________
The correct answer is: [B]: " $\frac{(-3)(4)}{(6)}$ " .
__________________________________________________________

7 0

3 years ago

What is the volume of the solid figure below? no fake answers

hram777 [196]

You can seperate the prisms then use the formula l*w*h which for the one on top u get V=20 and for the other one is 48 so 20+48= 68 so the volume of the prism is

(☞ﾟヮﾟ)☞ 68 ☜(ﾟヮﾟ☜)

8 0

2 years ago

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