Please answer this correctly

Mathematics

1 answer:

nikklg [1K]3 years ago

8 0

Answer:

20-29: Make it 1 unit tall

30-39: Make it 5 units tall

Step-by-step explanation:

20-29: 22 (1 number)

30-39: 31, 32, 32, 32, 32 (5 numbers)

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You are working on an experiment where you want to test the effects of caffeinated beverages on the behavior of cats. You will t

vichka [17]

For the experiment, you need 2L of cola. Your first option would be to purchase 1 2L bottle of cola for $2.25.

To calculate the second option, let's convert milliliters to liters first. There are 1,000 milliliters in 1 liter. With this, we know that there are 2,000 milliliters in 2 liters. Option 2 comes in 500-milliliter cans, which means that you would need 4 of them (2,000/500 = 4). 4 cans multiplied by $0.50 would cost you $2.00.

Let's check the cost of your answer options.
A. 4 cans - As seen above, this would cost $2.00.
B. 1 bottle - From the question, we know this would cost $2.25.
C. 2 bottles - This would be more soda than you need and would cost $4.50 ($2.25x2)
D. 1 can - This would be .5L and not enough soda for the experiment.
E. 5 cans - This would cost $2.50, but would be an extra 500mL of soda.
F. 2 cans - This would only be 1L of soda and not enough for the experiment.
G. 3 cans - This would be 1.5L of soda and not enough for the experiment either.

For the best price option, you would choose A (four cans of soda). This would give you the amount of soda that you need at the lowest price.

6 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

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