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3 years ago

12

What is the constant of proportionality?

1 answer:

devlian [24]3 years ago

4 0

Answer:

0.04 is the most common correct answer for the the formula

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A standard deck of playing cards contains 52 cards, equally divided among four suits (hearts, diamonds, clubs, and spades). Each

schepotkina [342]

If 3 of spades is drawn...and not replaced..

P(next card is spade) = 12/51
P(next card is a king) = 4/51

12/51 + 4/51 = 16/51

7 0

4 years ago

Read 2 more answers

Let U = {5, 10, 15, 20, 25, 30, 35, 40} A = {5, 10, 15, 20} V = {25, 30, 35, 40} C = {10, 20, 30, 40}. Find A C.

dusya [7]

The perimeter of the rectangle below is 202 units. Find the value of .

6 0

4 years ago

Find a second solution y2(x) of<br> x^2y"-3xy'+5y=0; y1=x^2cos(lnx)

rosijanka [135]

We can try reduction order and look for a solution $y_2(x)=y_1(x)v(x)$ . Then

$y_2=y_1v\implies{y_2}'=y_1v'+{y_1}'v\implies{y_2}''=y_1v''+2{y_1}'v+{y_1}''v$

Substituting these into the ODE gives

$x^2(y_1v''+2{y_1}'v+{y_1}''v)-3x(y_1v'+{y_1}'v)+5y_1v=0$

$x^2y_1v''+(2x^2{y_1}'-3xy_1)v'+(x^2{y_1}''-3x{y_1}'+5y_1)v=0$

$x^4\cos(\ln x)v''+x^3(\cos(\ln x)-2\sin(\ln x))v'=0$

which leaves us with an ODE linear in $w(x)=v'(x)$ :

$x^4\cos(\ln x)w'+x^3(\cos(\ln x)-2\sin(\ln x))w=0$

This ODE is separable; divide both sides by the coefficient of $w'(x)$ and separate the variables to get

$w'+\dfrac{\cos(\ln x)-2\sin(\ln x)}{x\cos(\ln x)}w=0$

$\dfrac{w'}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}$

$\dfrac{\mathrm dw}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}\,\mathrm dx$

Integrate both sides; on the right, substitute $u=\ln x$ so that $\mathrm du=\dfrac{\mathrm dx}x$ .

$\ln|w|=\displaystyle\int\frac{2\sin u-\cos u}{\cos u}\,\mathrm du=\int(2\tan u-1)\,\mathrm du$

Now solve for $w(u)$ ,

$\ln|w|=-2\ln(\cos u)-u+C$

$w=e^{-2\ln(\cos u)-u+C}$

$w=Ce^{-u}\sec^2u$

then for $w(x)$ ,

$w=Ce^{-\ln x}\sec^2(-\ln x)$

$w=C\dfrac{\sec^2(\ln x)}x$

Solve for $v(x)$ by integrating both sides.

$v=\displaystyle C_1\int\frac{\sec^2(\ln x)}x\,\mathrm dx$

Substitute $u=\ln x$ again and solve for $v(u)$ :

$v=\displaystyle C_1\int\sec^2u\,\mathrm du$

$v=C_1\tan u+C_2$

then for $v(x)$ ,

$v=C_1\tan(\ln x)+C_2$

So the second solution would be

$y_2=x^2\cos(\ln x)(C_1\tan(\ln x)+C_2)$

$y_2=C_1x^2\sin(\ln x)+C_2x^2\cos(\ln x)$

$y_1(x)$ already accounts for the second term of the solution above, so we end up with

$\boxed{y_2=x^2\sin(\ln x)}$

as the second independent solution.

6 0

4 years ago

Segment tq is 26 units long what is the length of qv 8 units 26 31 unit 32 units

Vlada [557]

Answer: 31 units I just did the assignment

8 0

4 years ago

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Jacob earned $370 walking dogs over the summer. He put 30% of what he earned into his savings account.

zaharov [31]

Step-by-step explanation:

Jacob earned $370 walking dogs over the summer. He put 30% of what he earned into his savings account

$\mapsto \: 30 \times \frac{370}{100}$

5 0

3 years ago