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viva [34]
3 years ago
15

The combination for a lock uses three of the digits 1, 2, 3, and 4. How many three-digit combinations can the lock have? Explain

your answer.
Mathematics
1 answer:
Elenna [48]3 years ago
3 0

Answer:

24

Step-by-step explanation:

First you start by writing down all the combinations that start with each number. Then you just write the left over numbers and you'll end up with 24 because the is six of each starting number. Ex: 123, 132, 134, 143, 124, 142...

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U xy, for u = 2, x = 9, and y = 6
Yuki888 [10]
First thing you gotta do is to sub in the numbers into the equation/expression
<span>uxy = (2)(9)(6)
</span>      = 108

* brackets means multiplying

Final answer is 108
8 0
3 years ago
Bus A and Bus B leave the bus depot at 8 am.
LenaWriter [7]

Answer: 10:55

Step-by-step explanation:

Taking statement at face value and the simplest scenario that commencing from 08:00am the buses take a route from depot that returns bus A to depot at 25min intervals while Bus B returns at 35min intervals.

The time the buses will be back at the depot simultaneously will be when:

N(a) * 25mins = N(b) * 35mins

Therefore, when N(b) * 35 is divisible by 25 where N(a) and N(b) are integers.

Multiples of 25 (Bus A) = 25, 50, 75, 100, 125, 150, 175, 200 etc

Multiples of 35 (Bus B) = 35, 70, 105, 140, 175, 210, 245 etc

This shows that after 7 circuits by BUS A and 5 circuits by Bus B, there will be an equal number which is 175 minutes.

So both buses are next at Depot together after 175minutes (2hr 55min) on the clock that is

at 08:00 + 2:55 = 10:55

6 0
3 years ago
Is 5x-7y=3 standard form?
barxatty [35]
Yes it is. Ax+By=C is standard form. :)
7 0
3 years ago
Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator
SpyIntel [72]
\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)

When x=3, you're left with

147=49a_1\implies a_1=\dfrac{147}{49}=3

When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

Adding the two equations together gives -10=2a_5, or a_5=-5. Subtracting them gives 2\sqrt2=2\sqrt2a_4, a_4=1.

Now, you have

5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)
5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)
2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that a_2x^4=2x^4 and a_3(-3)(-2)=6a_3=-6. These alone tell you that you must have a_2=2 and a_3=-1.

So the partial fraction decomposition is

\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}
7 0
3 years ago
What is the value of this expression when a = 3 and b = negative 2? (StartFraction 3 a Superscript negative 2 Baseline b Supersc
erastova [34]

Answer:

625

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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