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3 years ago

8

What's 5m²+5m+9m+8m²-m²= plss

2 answers:

lana66690 [7]3 years ago

8 0

It is 26m hope you have a wonderful day.

NISA [10]3 years ago

6 0

Answer:

12m^2+14m

Step-by-step explanation:

The "^means exponet" Hope I made your day

:)

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Is 36, 12, 4, 4/3, 4/9 a geometric sequence?

Mars2501 [29]

Answer:

Step-by-step explanation:

36 : 3 = 12

12 : 3 = 4

...

4/3 : 3 = 4/

6 0

3 years ago

Write the related facts for the set of numbers.<br> 3,7,21

STALIN [3.7K]

They all have either a common factor, common multiple, LCF, GCF, LCM, or GCM. The G and L stand for greater and least.

4 0

2 years ago

Read 2 more answers

Plz help me with this

Aleks04 [339]

Answer: FALSE

<u>Step-by-step explanation:</u>

x⁵ + 0x⁴ - 4x³ + 2x² - 4x + 1

If x - 2 is a factor, then x =2 should result in a remainder of 0 when using synthetic division.

2 | 1 0 -4 2 -4 1

<u>| ↓ 2 4 0 4 0 </u>

1 2 0 2 0 1 ← Remainder ≠ 0 so x - 2 is NOT a factor

3 0

3 years ago

Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

$y_p=a\sin x+b\cos x$

${y_p}'=a\cos x-b\sin x$

${y_p}''=-a\sin x-b\cos x$

Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

$y''+y=x\cos(2x)$

$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

$-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)$

$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

7 0

2 years ago

Suppose that the functions r and s are defined for all real numbers x as follows.<br> HELP PLEASE

Bas_tet [7]

Answer:

-x-3, 7x-3, -15

Step-by-step explanation:

For first 2 just add/subtract their equations. For the last one (r.s)(-1) is the same as r(s(-1)) so s(-1)=4(-1)=-4 and r(-4)=3(-4)-3=-12-3=-15

3 0

3 years ago