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3 years ago

If x,y and z are the first three terms of an geometric sequence, show that x^2,y^2 and z^2 form another geometric sequence

1 answer:

8 0

x y z are in geometric
x/y = y/z
OR y^2=xz

x^2, y^2, z^2

x^2, xz, z^2
If they are in geometric sequence then
xz/x^2 = z^2/xz
z/x = z/x
The ratios are common
so x^2,y^2 & z^2 is a GP.

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if you are dealing from a standard deck of 52 cards, how many ways could you choose two eighths one after the other

tekilochka [14]

Hearts spades
hearts clubs
hearts diamonds
spades clubs
spades diamonds
clubs diamonds

You could choose two eights one after the other 6 ways.

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3 years ago

I need serious help with this! I'll give 20 points!

GrogVix [38]

Answer:

glizzy

Step-by-step explanation:

7 0

3 years ago

PLEASE HELP ive been stuck on this for a long time

Tomtit [17]

Answer:

Below

Step-by-step explanation:

You can use Pythagorean theorem to find the hypotenuse of a right triangle.

a2 + b2 = c2

The square root of 15 = 3.8729.... (a bunch of other numbers)

4^2 + 3.8729^2 = c2

16 + 15 = 31

Now that you have this, you need to find the square root

The square root of 31 = 5.56775436

Rounding this to the tenth it would be: 5.57

Final answer: C = 5.57

hope this helps! :)

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3 years ago

A fraction can be changed to a decimal by dividing the

77julia77 [94]

A fraction can be changed to a decimal by dividing the Numerator by the Denominator. The Numerator on the fraction is always the top number, and the Denominator is always the bottom number.

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3 years ago

Read 2 more answers

Assume a jar has five red marbles and three black marbles. Draw out two marbles with and without replacement. Find the requested

Doss [256]

Answer:

For probabilities with replacement

$P(2\ Red) = \frac{25}{64}$

$P(2\ Black) = \frac{9}{64}$

$P(1\ Red\ and\ 1\ Black) = \frac{15}{32}$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{64}$

For probabilities without replacement

$P(2\ Red) = \frac{5}{14}$

$P(2\ Black) = \frac{3}{28}$

$P(1\ Red\ and\ 1\ Black) = \frac{15}{28}$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{56}$

Step-by-step explanation:

Given

$Marbles = 8$

$Red = 5$

$Black = 3$

For probabilities with replacement

(a) P(2 Red)

This is calculated as:

$P(2\ Red) = P(Red)\ and\ P(Red)$

$P(2\ Red) = P(Red)\ *\ P(Red)$

So, we have:

$P(2\ Red) = \frac{n(Red)}{Total} \ *\ \frac{n(Red)}{Total}\\$

$P(2\ Red) = \frac{5}{8} * \frac{5}{8}$

$P(2\ Red) = \frac{25}{64}$

(b) P(2 Black)

This is calculated as:

$P(2\ Black) = P(Black)\ and\ P(Black)$

$P(2\ Black) = P(Black)\ *\ P(Black)$

So, we have:

$P(2\ Black) = \frac{n(Black)}{Total}\ *\ \frac{n(Black)}{Total}$

$P(2\ Black) = \frac{3}{8}\ *\ \frac{3}{8}$

$P(2\ Black) = \frac{9}{64}$

(c) P(1 Red and 1 Black)

This is calculated as:

$P(1\ Red\ and\ 1\ Black) = [P(Red)\ and\ P(Black)]\ or\ [P(Black)\ and\ P(Red)]$

$P(1\ Red\ and\ 1\ Black) = [P(Red)\ *\ P(Black)]\ +\ [P(Black)\ *\ P(Red)]$

$P(1\ Red\ and\ 1\ Black) = 2[P(Red)\ *\ P(Black)]$

So, we have:

$P(1\ Red\ and\ 1\ Black) = 2*[\frac{5}{8} *\frac{3}{8}]$

$P(1\ Red\ and\ 1\ Black) = 2*\frac{15}{64}$

$P(1\ Red\ and\ 1\ Black) = \frac{15}{32}$

(d) P(1st Red and 2nd Black)

This is calculated as:

$P(1st\ Red\ and\ 2nd\ Black) = [P(Red)\ and\ P(Black)]$

$P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{n(Red)}{Total} *\ \frac{n(Black)}{Total}$

So, we have:

$P(1st\ Red\ and\ 2nd\ Black) = \frac{5}{8} *\frac{3}{8}$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{64}$

For probabilities without replacement

(a) P(2 Red)

This is calculated as:

$P(2\ Red) = P(Red)\ and\ P(Red)$

$P(2\ Red) = P(Red)\ *\ P(Red)$

So, we have:

$P(2\ Red) = \frac{n(Red)}{Total} \ *\ \frac{n(Red)-1}{Total-1}$

We subtracted 1 because the number of red balls (and the total) decreased by 1 after the first red ball is picked.

$P(2\ Red) = \frac{5}{8} * \frac{4}{7}$

$P(2\ Red) = \frac{5}{2} * \frac{1}{7}$

$P(2\ Red) = \frac{5}{14}$

(b) P(2 Black)

This is calculated as:

$P(2\ Black) = P(Black)\ and\ P(Black)$

$P(2\ Black) = P(Black)\ *\ P(Black)$

So, we have:

$P(2\ Black) = \frac{n(Black)}{Total}\ *\ \frac{n(Black)-1}{Total-1}$

We subtracted 1 because the number of black balls (and the total) decreased by 1 after the first black ball is picked.

$P(2\ Black) = \frac{3}{8}\ *\ \frac{2}{7}$

$P(2\ Black) = \frac{3}{4}\ *\ \frac{1}{7}$

$P(2\ Black) = \frac{3}{28}$

(c) P(1 Red and 1 Black)

This is calculated as:

$P(1\ Red\ and\ 1\ Black) = [P(Red)\ and\ P(Black)]\ or\ [P(Black)\ and\ P(Red)]$

$P(1\ Red\ and\ 1\ Black) = [P(Red)\ *\ P(Black)]\ +\ [P(Black)\ *\ P(Red)]$

$P(1\ Red\ and\ 1\ Black) = [\frac{n(Red)}{Total}\ *\ \frac{n(Black)}{Total-1}]\ +\ [\frac{n(Black)}{Total}\ *\ \frac{n(Red)}{Total-1}]$

So, we have:

$P(1\ Red\ and\ 1\ Black) = [\frac{5}{8} *\frac{3}{7}] + [\frac{3}{8} *\frac{5}{7}]$

$P(1\ Red\ and\ 1\ Black) = [\frac{15}{56} ] + [\frac{15}{56}]$

$P(1\ Red\ and\ 1\ Black) = \frac{30}{56}$

$P(1\ Red\ and\ 1\ Black) = \frac{15}{28}$

(d) P(1st Red and 2nd Black)

This is calculated as:

$P(1st\ Red\ and\ 2nd\ Black) = [P(Red)\ and\ P(Black)]$

$P(1st\ Red\ and\ 2nd\ Black) = P(Red)\ *\ P(Black)$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{n(Red)}{Total} *\ \frac{n(Black)}{Total-1}$

So, we have:

$P(1st\ Red\ and\ 2nd\ Black) = \frac{5}{8} *\frac{3}{7}$

$P(1st\ Red\ and\ 2nd\ Black) = \frac{15}{56}$

7 0

3 years ago