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2 years ago

5x-3y=-12 in slope-intercept form

2 answers:

6 0

5x-3y= -12
-3y= -12 - 5x Move 5x to the right side
3y= 12 + 5x Change signs on both sides
y= 4 + 5/3x Divide both sides by 3
y= 5/3x + 4 Rewrite

y= 5/3x +4 Answer

If you don’t mind, please give me brainliest!

madreJ [45]2 years ago

4 0

Answer:

Subtract 5x from both sides of the equation

-3y = 12 - 5x

Then, divide each term by -3 and simplify.

y = -4 + 5x

------

After that, you'll do this

y= 5x

----- - 4

Lastly, you put that into slope-intercept form.

y = 5

---- x - 4

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3 years ago

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

Help me on problem 4 !!!!I WILL GIVE BRAINLIEST!!!!!

Aleksandr-060686 [28]

Answer:

b=12.20

Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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