The vertices of a triangle are A(0,3) B(-2,-4) and C(1,5) find the new vertices

Mathematics

1 answer:

worty [1.4K]3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

Using the translation rule (x, y ) → (x - 2, y + 4 )

Subtract 2 from the original x- coordinate and add 4 to the original y- coordinate, thus

A(0, 3 ) → A'(0 - 2, 3 + 4 ) → A'(- 2, 7 )

B(- 2, - 4 ) → B'(- 2 - 2, - 4 + 4 ) → B'(- 4, 0 )

C(1, 5 ) → C'(1 - 2, 5 + 4 ) → C'(- 1, 9 )

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Which fraction represents the decimal 0. ModifyingAbove 1 2 with Bar? One-twelfth StartFraction 3 Over 25 EndFraction StartFract

mestny [16]

The considered decimal number is expressed by the fraction given by: Option C: 4/33

<h3>How to convert from decimal to fraction?</h3>

For conversion from decimal to fraction, we write it in the form a/b such that the result of the fraction comes as the given decimal. Usually, to get the decimal of the form a.bcd, we count how many digits are there after the decimal point (if its finite), then we write 10 raised to that many power as denominator, and the considered number without any decimal point as numerator.

For example:

$0.99 = \dfrac{99}{10^2} = \dfrac{99}{100}$

(10 raised to power k = 1 followed by k zeroes)

<h3>What is the sum of an infinite geometric series?</h3>

The sum of an infinite geometric series having the initial term "a" and the multiplication per term is done by "r" such that r < 1

is given by

$S_{\infty} = \dfrac{a}{1-r}$

The series would look like $a+ ar+ ar^2+\cdots$

For the considered case, the given decimal number is

$0.\overline{12} = 0.121212\cdots$

We can check all the options available one by one, so as to know which one ends up giving this pattern, or we can use another technique as shown below.

$0.12121... = (0.1 + 0.001 + 0.00001 + ... ) + (0.02 + 0.0002 + 0.000002 + ... )$

Each of the two series obtained are geometric series with the multiplication factor as 1/100

Using the sum formula for evaluation of an infinite geometric series, and the fact that initial term of first series is 0.1 and that of second series is 0.02, we get:
$0.121212... = \dfrac{0.1}{1-1/100} + \dfrac{0.02}{1 - 1/100} = \dfrac{1}{99/100}(\dfrac{1}{10} + \dfrac{2}{100})\\0.121212... = \dfrac{100}{99}\times(\dfrac{10}{100} + \dfrac{2}{100})\\\\\\0.121212 ... = \dfrac{12}{99} = \dfrac{4}{33}$