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Ulleksa [173]
3 years ago
5

Y

Mathematics
2 answers:
algol133 years ago
8 0

Answer: The pre-image moves horizontally left.

The pre-image moved horizontally 3 units.

The pre-image moved vertically up.

The pre-image moved vertically 2 units.

Step-by-step explanation:

Ray Of Light [21]3 years ago
6 0

Answer:

The pre-image moves horizontally left.

The pre-image moved horizontally 3 units.

The pre-image moved vertically up.

The pre-image moved vertically 2 units.

Step-by-step explanation:

got it right on edgenuity

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Which graph shows a function whose inverse is also a function?
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Step-by-step explanation:

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Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive
Nostrana [21]

Answer:

The first four terms of the series are

(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})

\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n} = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

\sum_{k=0}^\infty a(r)^k

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is \sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}

Given series,

\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}

=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......

The first four terms of the series are

(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})

Let

S_n=\sum_{n=0}^\infty \frac{9}{7^n}    and     t_n=\sum_{n=0}^\infty \frac{3}{5^n}

Now for S_n,

S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......

    =\sum_{n=0}^\infty9(\frac 17)^n

It is a geometric series.

The common ratio of S_n is \frac17

The sum of the series

S_n=\sum_{n=0}^\infty \frac{9}{7^n}

    =\frac{9}{1-\frac17}

    =\frac{9}{\frac67}

    =\frac{9\times 7}{6}

    =10.5

Now for t_n

t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......

    =\sum_{n=0}^\infty3(\frac 15)^n

It is a geometric series.

The common ratio of t_n is \frac15

The sum of the series

t_n=\sum_{n=0}^\infty \frac{3}{5^n}

    =\frac{3}{1-\frac15}

    =\frac{3}{\frac45}

    =\frac{3\times 5}{4}

    =3.75

The sum of the series is \sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}

                                        = S_n+t_n

                                       =10.5+3.75

                                       =14.25

4 0
3 years ago
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