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stellarik [79]
3 years ago
8

How many 5 digit codes can be formed from the digits 1,2,3,4,5,6 and 7 if digits cannot be repeated

Mathematics
1 answer:
Semmy [17]3 years ago
4 0

- You pick the first digit at random - you have 7 possible outcome. For the next digit, as you cannot have repetitions, you only have 6 possible outcomes. And for the second one 5, the third one 4 and the fourth one 3, and lastly the third one, which is 2.

- So you have 7×6×5×4×3×2 possible outcomes.

= 17640

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Answer:

It’s also worth pointing out that the presentation of the algebraic expressions in coin word problems are a bit different and not so straightforward compared to what we’re used to. For example, instead of saying “the number of nickels is 2 more than the number of dimes“, you’ll often see this expressed in coin word problems as “there are 2 more nickels than dimes“. Both algebraic expressions can be written in an equation as n=d+2n=d+2 but just expressed differently.

Step-by-step explanation:

Tamara has 35 coins in nickels and quarters. In all, she has $4.15. How many of each kind of coin does she have? Right off the bat, the problem gives us two important pieces of information. First, it tells us that there is a total number of 35 coins consisting of nickels and quarters. Secondly, the total value of the coins is $4.15. We need to translate these statements into algebraic equations to find how many nickels and how many quarters she has.

BTW hope this helps

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The given differential equation has characteristic equation

r^5 - 4r^4 + 4r^3 - r^2 + 4r - 4 = 0

Solve for the roots r.

r^3 (r^2 - 4r + 4) - (r^2 - 4r + 4) = 0

(r^3 - 1) (r^2 - 4r + 4) = 0

(r^3 - 1) (r - 2)^2 = 0

r^3 - 1 = 0 \text{ or } (r-2)^2=0

The first case has the three cubic roots of 1 as its roots,

r^3 = 1 = 1e^{i0} \implies r = 1^{1/3} e^{i(0+2\pi k)/3} \text{ for } k\in\{0,1,2\} \\\\ \implies r = 1e^{i0} = 1 \text{ or } r = 1e^{i2\pi/3} = -\dfrac{1+i\sqrt3}2 \text{ or } r = 1e^{i4\pi/3} = -\dfrac{1-i\sqrt3}2

while the other case has a repeated root of

(r-2)^2 = 0 \implies r = 2

Hence the characteristic solution to the ODE is

y_c = C_1 e^x + C_2 e^{-(1+i\sqrt3)/2\,x} + C_3 e^{-(1-i\sqrt3)/2\,x} + C_4e^{2x} + C_5xe^{2x}

Using Euler's identity

e^{ix} = \cos(x) + i \sin(x)

we can reduce the complex exponential terms to

e^{-(1\pm i\sqrt3)/2\,x} = e^{-x/2} \left(\cos\left(\dfrac{\sqrt3}2x\right) \pm i \sin\left(\dfrac{\sqrt3}2x\right)\right)

and thus simplify y_c to

y_c = C_1 e^x + C_2 e^{-x/2} \cos\left(\dfrac{\sqrt3}2x\right) + C_3 e^{-x/2} \sin\left(\dfrac{\sqrt3}2x\right) \\ ~~~~~~~~ + C_3 e^{-(1-i\sqrt3)/2\,x} + C_4e^{2x} + C_5xe^{2x}

For the non-homogeneous ODE, consider the constant particular solution

y_p = A

whose derivatives all vanish. Substituting this into the ODE gives

-4A = 69 \implies A = -\dfrac{69}4

and so the general solution to the ODE is

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