On Tuesday morning, the temperature was recorded

Mathematics

1 answer:

VashaNatasha [74]3 years ago

6 0

C. -2° :) hope this helps lmk!

You might be interested in

Hello! i need help! I Am a 5th grade scholar

stich3 [128]

0.4 is 10 times as much as 0.04

this is because 0.04 times 10 is 0.4

0.003 is 1/10 of 0.03

0.003 times 10 is 0.03

6 0

4 years ago

Indigo measures a 2-ounce sample for an experiment. How many grams are in 2 ounces? (1 oz = 28.35 g)

Nimfa-mama [501]

Answer:

56.70 grams

Step-by-step explanation:

Multiply: 28.35*2=56.70

(Picture included)

Sorry about my handwriting!

Hope this helped!! :)

Stay safe and have a wonderful day/night!!!!!

Brainliest?!?!

3 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$