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Aloiza [94]
3 years ago
13

Subtract (6a^2+3a) - (4a^2 + 2a)

Mathematics
2 answers:
kicyunya [14]3 years ago
8 0

Answer:

2a^2+a

Step-by-step explanation:

This is an algebra. An algebra is an expression with no equality sign, and it will be expressed till the simplest form.

(6a^2+3a) - (4a^2 + 2a)

Remove the brackets

6a^2+3a-4a^2-2a

6a^2-4a^2+3a-2a

2a^2+a

Alexus [3.1K]3 years ago
7 0

Answer:

2a² + a   (   or a(2a+1)   )

Step-by-step explanation:

(6a²+3a) - (4a² + 2a)  (expand parentheses)

= 6a²+3a - 4a² - 2a   (group like terms)

= 6a²- 4a²+3a- 2a

= 2a² + a

= a(2a+1)

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The function f(x) is to be graphed on a coordinate plane.
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Answer:

f(x) is going to be on the y axis or intercept and what number it is on the line is in quotes f(x)="6"+2(1.05)^x it would be at 6 on the y axis.

5 0
2 years ago
PLEASE PLEASE HELP ME I REALLY NEED HELP
Gekata [30.6K]

Answer:

(a) yes it does

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7 0
3 years ago
Unit rate $20 for 3 pies sold
rjkz [21]

Answer:

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6 0
3 years ago
Could someone help me real quick.
Keith_Richards [23]
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3 0
3 years ago
Find a12 of the sequence 1/4, 7/12, 11/12, 5/4
UNO [17]

1/4 = 3/12, and 5/4 = 15/12, so it looks like there's a common difference between terms of 4/12 = 1/3. The the n-th term in the sequence is given recursively by

\begin{cases}a_1=\frac14\\a_n=a_{n-1}+\frac13&\text{for }n>1\end{cases}

By substitution, we get

a_n=a_{n-1}+\dfrac13\implies a_n=\left(a_{n-2}+\dfrac13\right)+\dfrac13

a_n=a_{n-2}+\dfrac23

and doing this again and again until we stop with an expression containing a_1, we find that

a_n=a_1+\dfrac{n-1}3

a_n=\dfrac{4n-1}{12}

Then the 12th term in the sequence is

a_{12}=\dfrac{4\cdot12-1}{12}=\boxed{\dfrac{47}{12}}

7 0
3 years ago
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