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Mazyrski [523]
4 years ago
9

What's the property to (x/y)^3=x^3/y^3

Mathematics
1 answer:
Sphinxa [80]4 years ago
7 0

Answer:

Power of a quotient property

Step-by-step explanation:

It is power of a quotient property, when you have different bases, whose quotient(division) is raised to the same power, it is equal to dividing, each base raised to same power.

Here, x and y are different bases and 3 is the common power to both of them.

In general for any power, say m :

(\frac{x}{y} )^{m} = \frac{x^{m} }{y^{m}}

Because, (\frac{x}{y} )^{m} = \frac{x}{y} \times \frac{x}{y} \times\frac{x}{y}.... m times

And, (\frac{x}{y} )^{3} = \frac{x}{y} \times \frac{x}{y} \times\frac{x}{y}

i.e (\frac{x}{y} )^{3} = \frac{x^{3} }{y^{3}}

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What is the solution to y=12
Nimfa-mama [501]

Answer:

y=12

Step-by-step explanation:

This equation cannot be simplified further.

Hope this helped! :)

8 0
3 years ago
An analog signal with a bandwidth of 36MHz is sampled at a frequency (fs) of 36,000,000 samples per second during the ADC (analo
Elza [17]

Answer:

The signal would have experienced aliasing.

Step-by-step explanation:

Given that:

the bandwidth of the signal f_m = 36MHz

= 36 × 10⁶ Hz

The sampling frequency f_s = 36 × 10⁶ Hz

Suppose the sampling frequency is equivalent to the bandwidth of the signal, then aliasing will occur.

Therefore, according to the Nyquist criteria;

Nyquist criteria posit that if the sampling frequency is more above twice the maximum frequency to be sampled, a repeating waveform can be accurately reconstructed.

∴

By Nyquist criteria, for perfect reconstruction of an original signal, i.e. the received signal without aliasing effect;

Then,

f_s \geq 2f_m

∴

The signal would have experienced aliasing.

4 0
3 years ago
Which pair shows equivalent expressions?
AVprozaik [17]
The 3rd one because the rest you would have to distribute them and for the 3rd one it’s the same
3 0
3 years ago
Eloise made a list of some multiples of 8/5. Write 5 fractions that could be in Eloise’s list.
Sonbull [250]
We can get the 5 fractions by multiplying 8/5 by 2, 3, 4, 5, and 6.
8/5*2=16/5
8/5*3=24/5
8/5*4=32/5
8/5*5=40/5 or 8.
8/5*6=48/5.
Hope this helps!
-Raiden
5 0
4 years ago
eight cards are drawn from a standard deck of 52 cards. how many hands off with cards contain exactly three queens and three jac
d1i1m1o1n [39]

Answer:

15,136 hands off with cards contain exactly three queens and three jacks.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

Standard deck:

4 queens and 4 jacks.

The other 52 - 8 = 44 cards are neither queens nor jacks.

Wow many hands off with cards contain exactly three queens and three jacks?

3 queens from a set of 4.

3 jacks from a set of 4.

2 other cards(not queens neither jacks) from the other 44. So

C_{4,3}C_{4,3}C_{44,2} = \frac{4!}{1!3!} \times \frac{4!}{1!3!} \frac{44!}{2!42!} = 4*4*22*43 = 15136

15,136 hands off with cards contain exactly three queens and three jacks.

4 0
3 years ago
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