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stepladder [879]
3 years ago
7

A certain four​-cylinder combination lock has 50 numbers on it. to open​ it, you turn to a number on the first​ cylinder, then t

o a second number on the second​ cylinder, and then to a third number on the third cylinder and so on until a four​-number lock combination has been affected. repetitions are​ allowed, and any of the 50 numbers can be used at each step to form the combination.​ (a) how many different lock combinations are​ there
Mathematics
2 answers:
Kaylis [27]3 years ago
8 0
There are 230,300 combinations.

This is given by

_nC_r = _{50}C_4=\frac{50!}{46!4!} = 230,300
kumpel [21]3 years ago
7 0

A certain four​-cylinder combination lock has 50 numbers on it.

It is a four cylinder combination lock

Each cylinder has 50 numbers

So we have to choose 4 numbers for 4 cylinders

_____ , ______ , ______, _____

you turn to a number on the first​ cylinder

We have 50 numbers so we have 50 combinations

__50___ , ______ , ______, _____

Repetitions are allowed so we do the same for remaining three

50 , 50 , 50 , 50

50 * 50 * 50* 50 = 6250000

There are 6250000 different lock combinations .

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Answer:

a. \frac {2} {s-1} converges to s> 1.

b. \frac{3}{e^3 \left(s-5 \right)} converges to s> 5.

c. - \frac {2}{s + 3} converges to s> - 3.

d. \frac {s}{s^2 + 25} converges to s> 0.

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f. \frac {12}{s^2 + 4} converges to s> 0.

g. -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4} converges to s> 0.

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Step-by-step explanation:

a. L \left\{2e^t \right\} = 2L \left\{e^t \right\} = 2 \cdot \frac {1} {s-1} = \frac {2} {s-1} converges to s> 1.

b. L \left\{3e^{5t-3} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = \frac{3}{e^3 \left(s-5 \right)} converges to s> 5.

c. L \left\{-2e^{-3t} \right\} = -2L \left\{e^{-3t} \right\} = - \frac {2}{s + 3} converges to s> - 3.

d. L \left\{\cos\left (5t \right)\right\} = \frac {s}{s^2 + 25} converges to s> 0.

e. L \left\{10 \sin\left(t\right)\right\} = 10L\left\{\sin\left(t\right)\right\} = \frac {10} {s^2 + 1} converges even s> 0.

f. L \left\{6\sin \left(2t \right) \right\} = 6L\left\{\sin\left (2t\right)\right\} = \frac {12}{s^2 + 4} converges to s> 0.

g. L \left\{-5\cos\left(2t + 1\right) \right\} = -5L\left\{\cos\left(2t + 1 \right)\right\} = -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4} converges to s> 0.

h. L\left\{\sin \left(t\right)\cos \left(t\right)\right\} = L\left\{\sin\left(2t\right)\frac{1}{2}\right\} =\frac{1}{2}\cdot \frac{2}{s^2+4} = \frac {1} {s ^ 2 + 4} converges to s> 0.

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3 years ago
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Let n be old painters time. Then the new painters are 2n. So:
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3 0
2 years ago
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