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Mathematics

1 answer:

ololo11 [35]2 years ago

8 0

If Carla would have bought 4 plants it would have been 13 dollars from green touch

If she would have bought them for the grocery store it would have been 14 dollars

So in conclusion she saved 1 dollar
Option 3

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Help me please..................

Rudiy27

253. First, I squared 8, to get 64, then multiplied that by 12. I subtracted 9 from this because 3x3 is 9 to get 759. Then, divide this by 3 to get 253. Hope this helped!

8 0

2 years ago

Read 2 more answers

Find the common difference for the sequence below. 1/4, 5/16, 3/8,..

vredina [299]

Answer:

the common difference of the sequence is $(\frac{1}{16})$

Step-by-step explanation:

A sequence of numbers is said to be in arithmetic progression of each term of the sequence has same common difference.

Now, here first term a1 = 1/4

Second term a2 = 5/16

Third term a3 = 3/8

Now, Common difference (d) = a2 - a1 = a3 - a2

Now, a2 - a1 = $\frac{5}{16} - \frac{1}{4} = \frac{5 - 4}{16} = \frac{1}{16}$

Now, a3 - a2 = $\frac{3}{8} - \frac{5}{16} = \frac{6 - 5}{16} = \frac{1}{16}$

Hence, the common difference of the sequence is $(\frac{1}{16})$

5 0

2 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

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