Is the value of 4 in 46,395 36 times the value of 4 in 14,906

Mathematics

1 answer:

expeople1 [14]3 years ago

4 0

Yesthe value of 4 in 46,395 36 times the value of 4 in 14,906

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Write the first five terms of the sequence defined by the recursive formula

Arada [10]

Answer:

The second option is the correct answer

The sequence is : 0, -1, -6, -31, -156

Step-by-step explanation:

It is given that,

The recursive formula , an = 5a(n-1) - 1 and a1 = 0

a2 = 5a1 -1 = 5x0 - 1 = -1

a3 = 5a2 - 1 = (5x -1 ) - 1 = - 5 - 1 = -6

a4 = 5a3 - 1 =(5x -6) - 1 = -30 - 1= -31

a5 = 5a4 - 1 = (5 x -31 ) - 1 = -155 -1 = -156

Therefore the resulting sequence is

0, -1, -6, -31 ,-156

6 0

3 years ago

Read 2 more answers

Three years back , a father was 24 years older than his son .at present,the father is five times as old as the son.how old will

azamat

Answer:

The son is 5 I think

Step-by-step explanation:

24+3=27

27 divided by 5 = 5.4

round 5.4 to 5 since it is under .5

and you get 5

3 0

3 years ago

Express this to single logarithm

zzz [600]

Answer: $\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right)$

We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.

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Explanation:

It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.

I'm going to use these three log rules, which apply to any base.

log(A) + log(B) = log(A*B)
log(A) - log(B) = log(A/B)
B*log(A) = log(A^B)

From there, we can then say the following:

$\frac{1}{2}\log_{2}\left(m\right)-3\log_{2}\left(n\right)+2\log_{2}\left(q\right)\\\\\log_{2}\left(m^{1/2}\right)-\log_{2}\left(n^3\right)+\log_{2}\left(q^2\right) \ \text{ .... use log rule 3}\\\\\log_{2}\left(\sqrt{m}\right)+\log_{2}\left(q^2\right)-\log_{2}\left(n^3\right)\\\\\log_{2}\left(\sqrt{m}*q^2\right)-\log_{2}\left(n^3\right) \ \text{ .... use log rule 1}\\\\\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right) \ \text{ .... use log rule 2}$