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2 years ago

How do you solve: 2(b - 5) + 7b-4

Mathematics

2 answers:

Blizzard [7]2 years ago

8 0

Answer:

9b-14

Step-by-step explanation:

2(b-5)+7b-4

First you want to do what is in the parenthesis:

2 times b 2 times 5

The two is on the outside of the parenthesis which means you multiply it times the variable and the number.

Simply: 2b-10+7-+4

Now, you want to break the equation down. Both 7 and 2 have 'b' so you can add the two together.

2b+7b=9b

-10 and -4 can be put together. 4 is negative so you would subtract the two numbers.

-10-4= -14

Next, you want to take the two numbers 9b and -14 and put them into equation form.

9b-14

<u>I hope this helped! </u>

Alinara [238K]2 years ago

4 0

2(b-5)+7b-4
Solve 2(b-5) to get 2b-10
Combine your like terms 2b-10+7b-4
Solve 2b+7b to get 9b
Solve -10+(-4) to get -14
So your answer is
9b-14

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I am in the same row as 83 you say me when you start at 10 and count by tens.

yKpoI14uk [10]

Answer:90

Step-by-step explanation:

10,20,30,40,50,60,70,80,90,100

Counting in tens the same row as 83 is 90 .

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3 years ago

Add the group of numbers below. Make sure<br> your answer is fully reduced.<br> 1/3+1/5+2/5

RideAnS [48]

Answer:

14/15

Step-by-step explanation:

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2 years ago

Read 2 more answers

What is the solution to the equation w+6=5w-10

docker41 [41]

Answer:

w = 4

Step-by-step explanation:

you would do this by re-arranging the equation

6 = 5w-10-w (move w to the other side)

6 = 4w-10 (because 5w-w = 4w)

6+10=4w (move 10 to the other side)

16 = 4w (add 6 and 10)

4 = w (divide both sides by 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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Chelsea played her tuba from 4:25 pm until 5:07

melisa1 [442]

Answer:

42 minutes

Step-by-step explanation:

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3 years ago