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3 years ago

9+(6w+3w) how do I do this it is to hardThe expression (9+6w)+3w

Mathematics

1 answer:

Mama L [17]3 years ago

6 0

Answer: 9w+9

Step-by-step explanation:

First combine like terms: 6w+3w

You can't combine 9 and 9w so that would be your answer.

Then if it calls for it, use the commutitive property to move 9w before 9.

Hope this helps!

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for sale, a store decrease it's price on all items by 25%. An item that cost $ 120 before the sale now costs $120 - 0.25 ( $ 120

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3 years ago

It’s for my geometry class and I don’t know how to solve it. Can someone please help me?

dusya [7]

Answer:

Below in bold.

Step-by-step explanation:

Because the 2 sides marked with the double lines are equal:

5y-1 = 2y+11

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3 years ago

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oksano4ka [1.4K]

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8 0

3 years ago

Read 2 more answers

What is the best method to solve the system of equations? 2x + 3y = 8 4x - 6y = 2?

Nana76 [90]

Answer: x = 9/4, y = 7/6

║<span>2x + 3y = 8
</span>║<span>4x - 6y = 2

</span>║4x + 6y = 16
║4x - 6y = 2

Add both equations together:

8x = 18
x = 9/4

Sub x = 9/4 into 2x + 3y = 8
2(9/4) + 3y = 8
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5 0

3 years ago

Evaluate the sum of the following finite geometric series.

rjkz [21]

Answer:

$\large\boxed{\dfrac{156}{125}\approx1.2}$

Step-by-step explanation:

<h3>Method 1:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}$

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}$

<h3>Method 2:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}$

$a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1$

$\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}$

5 0

3 years ago