The earliest train Brian can catch

ale4655 [162]

Charles law:
V1/T1=V2/T2
where T is in deg. K.
75°F=297.04°K
55°F=285.92°K
=>
V2=V1/T1*T2
=3000*(285.92/297.4)
=2887.8 c.f.

4 0

3 years ago

PLZ HELP 50B please Please Please

Marysya12 [62]

Answer:

$40

$32

Step-by-step explanation:

It says that both your friend's monthly allowance and your monthly allowance together is $72. To find out how much one box is worth, we need to divide $72 by 9 because there are 9 boxes. 72 ÷ 9 = 8.

We know that each box is equal to $8.00, so to find out how much your monthly allowance is, we multiply 8 by 5 because you have five squares. 8 x 5 = 40.

To find out how much money your friend's monthly allowance is, we multiply 8 by 4 because your friend has four squares. 8 x 4 = 32.

We can check our work by adding 40 and 32 together. 40 + 32 = 72. Which means that it is correct. :)

I hope that this helps!

7 0

3 years ago

Use the method of "undetermined coefficients" to find a particular solution of the differential equation. (The solution found ma

Naddika [18.5K]

Answer:

The particular solution of the differential equation

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$ + $\frac{1}{37}185e^{6x})$

Step-by-step explanation:

Given differential equation y''(x) − 10y'(x) + 61y(x) = −3796 cos(5x) + 185e6x

The differential operator form $(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

<u>Rules for finding particular integral in some special cases:-</u>

let f(D)y = $e^{ax}$ then

the particular integral $\frac{1}{f(D)} (e^{ax} ) = \frac{1}{f(a)} (e^{ax} ) if f(a)$ ≠ 0

let f(D)y = cos (ax ) then

the particular integral $\frac{1}{f(D)} (cosax ) = \frac{1}{f(D^2)} (cosax ) =\frac{cosax}{f(-a^2)}$ f(-a^2) ≠ 0

Given problem

$(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

P<u>articular integral</u>:-

$P.I = \frac{1}{f(D)}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + \frac{1}{D^2-10D+61}185e^{6x})$

P.I = $I_{1} +I_{2}$

we will apply above two conditions, we get

$I_{1}$ =

$\frac{1}{D^2-10D+61}( −3796 cos(5x) = \frac{1}{(-25)-10D+61}( −3796 cos(5x) ( since D^2 = - 5^2)$ $= \frac{1}{(36-10D}( −3796 cos(5x) \\= \frac{1}{(36-10D}X\frac{36+10D}{36+10D} ( −3796 cos(5x)$

on simplification we get

= $\frac{1}{(36^2-(10D)^2}36+10D( −3796 cos(5x)$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{1296-100(-25)}$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$

$I_{2}$ =

$\frac{1}{D^2-10D+61}185e^{6x}) = \frac{1}{6^2-10(6)+61}185e^{6x})$

$\frac{1}{37}185e^{6x})$

Now particular solution

P.I = $I_{1} +I_{2}$

P.I = $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$ + $\frac{1}{37}185e^{6x})$

8 0

3 years ago

Write the equation for a line that has an initial value of 3 and 3/4 as it’s rate of change

frutty [35]

the slope goes by several names

• average rate of change

• rate of change

• deltaY over deltaX

• Δy over Δx

• rise over run

• gradient

• constant of proportionality

however, is the same cat wearing different costumes.

initial value of 3, namely when x = 0, y = 3, so we have the point (0 , 3) and it has a rate or slope of 3/4.

$(\stackrel{x_1}{0}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{3}{4} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{3}=\stackrel{m}{\cfrac{3}{4}}(x-\stackrel{x_1}{0})\implies y=\cfrac{3}{4}x+3$

3 0

3 years ago

Simplify the radical expression.

GREYUIT [131]

Answer:

c

Step-by-step explanation:

160 = 16 * 10 = 4 * 4 * 2 * 5 = 2 * 2 * 2 * 2 * 2 *5

$-4\sqrt{160} = -4\sqrt{2*2*2*2*2*5}\\\\= -4*2*2\sqrt{2*5}\\\\= -16\sqrt{10}$

5 0

3 years ago

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