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3 years ago

Whats the common ratio of the sequence 7.04 Q14

Mathematics

2 answers:

loris [4]3 years ago

8 0

Answer:

1/4.

Step-by-step explanation:

You divide any term after the first by the previous one.

So the common ratio for this sequence is

-2/5 / - 8/5

= -2/5 * -5/8

= 2/8

= 1/4 (answer)

julsineya [31]3 years ago

4 0

Answer:

$\frac{1}{4}$

Step-by-step explanation:

The given sequence is a geometric sequence.

We can find the common ratio using any two consecutive terms of the sequence.

$r=\frac{\frac{-1}{10} }{\frac{x-2}{5} }=\frac{-1}{10}\times \frac{-5}{2}=\frac{1}{4}$

The common ratio is $\frac{1}{4}$

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Answer:

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Step-by-step explanation:

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3 years ago

Find the value of the following expression: (2^8 ⋅ 5^−5 ⋅ 19^0)^−2 ⋅ 5 to the power of negative 2 over 2 to the power of 3, whol

koban [17]

Answer:

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

Step-by-step explanation:

$\left(2^8\cdot5^{-5}\cdot19^0\right)^{-2}\cdot\left(\dfrac{5^{-2}}{2^3}\right)^4\cdot228\\\\\text{use}\ a^{-n}=\dfrac{1}{a^n}\ \text{and}\ a^0=1\ \text{and}\ (a^n)^m=a^{nm}\\\\=\left(2^8\cdot\dfrac{1}{5^5}\cdot1\right)^{-2}\cdot\left(\dfrac{\frac{1}{5^2}}{2^3}\right)^4\cdot228=\left(\dfrac{2^8}{5^5}\right)^{-2}\cdot\left(\dfrac{1}{2^35^2}\right)^4\cdot228$

$=\dfrac{(2^8)^{-2}}{(5^5)^{-2}}\cdot\dfrac{1^4}{(2^3)^4(5^2)^4}\cdot228=\dfrac{2^{-16}}{5^{-10}}\cdot\dfrac{1}{2^{12}5^8}\cdot228\\\\\text{use}\ a^n=\dfrac{1}{a^{-n}}\to\dfrac{1}{a^n}=a^{-n}\\\\=2^{-16}\cdot5^{10}\cdot2^{-12}\cdot5^{-8}\cdot228\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{-16+(-12)}\cdot5^{10+(-8)}\cdot228=2^{-28}\cdot5^2\cdot228\\\\=2^{-28}\cdot5^2\cdot4\cdot57=2^{-28}\cdot5^2\cdot2^2\cdot57=2^{-28+2}\cdot5^2\cdot57\\\\=2^{-26}\cdot5^2\cdot57=\dfrac{5^2\cdot57}{2^{26}}$

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

4 0

3 years ago

the probability density for a particle in a box is an oscillatory function even for very large energies. Explain how the classic

77julia77 [94]

Answer:

This is achieved for the specific case when high quantum number with low resolution is present.

Step-by-step explanation:

In Quantum Mechanics, the probability density defines the region in which the likelihood of finding the particle is most.

Now for the particle in the box, the probability density is also dependent on resolution as well so for large quantum number with small resolution, the oscillations will be densely packed and thus indicating in the formation of a constant probability density throughout similar to that of classical approach.

8 0

3 years ago

Whats the common ratio of the sequence 7.04 Q14​

Whats the common ratio of the sequence 7.04 Q14