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2 years ago

Have a great day lads.Don't forget to smile :)

Mathematics

2 answers:

natulia [17]2 years ago

8 0

║∵ω ∵║

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zysi [14]2 years ago

4 0

Answer:

Aww thank you you too stay safe! :)

Step-by-step explanation:

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The square below has a side length of 14 inches. What is the area of the shaded region?

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14 square units

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2 years ago

Match the measures with the correct type of measurement. a.72m b. 125mm3 c. 50cm2 Volume, Surface Area, Perimeter

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a. Perimeter

b. Volume

c.Suface Area

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Solve 5(85) using mental math with the help of distributive properties

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3 years ago

Read 2 more answers

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

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3 years ago

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Answer:

This is your answer ☺️☺️☺️

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2 years ago