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2 years ago

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An equation is shown below:

1 answer:

Oksana_A [137]2 years ago

8 0

I hope this helped I've only done part A

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Find the approximate value of the circumference of a circle with the given radius. Use = 3.14. Round your results to one more de

Schach [20]

Circumference=2pir=<span>2∗3.14∗5.1</span><span>=30.028=30.03.</span>

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2 years ago

1. Solve exponential equations: ax=b, where a>0, a≠1, b>0

Brums [2.3K]

Answer:

$x=log_{a}b$

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2 years ago

How do you simplify 2 15/10?

sdas [7]

First you take 15/10 and see that you can take 5 out of the top and bottom, so it would become 3/2. Now you have 2 3/2. 2 goes into 3 once with 1 remainder so 3/2= 1 1/2
So then you take the original 2 and add it to 1 1/2. The answer is now 3 1/2

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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

4 0

3 years ago

Pam has taken 4 tests. She scored an 82, 94, 89 and 92. What would she need to score on her 5th test to have a mean of 90?

zavuch27 [327]

Answer:

93

Step-by-step explanation:

5*90 = 450

82+94+89+92 = 357

so

450-357 = 93

hope this helped :)

5 0

3 years ago