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Verdich [7]
3 years ago
15

PLEASE HELP! MATH

Mathematics
2 answers:
nikdorinn [45]3 years ago
5 0
6x^4+9x^2+12x/3x = 2x^3+3x+4

So the answer is D. 2x3 + 3x + 4
Usimov [2.4K]3 years ago
3 0

Answer:

D) 2x^{3} +3x + 4.

Step-by-step explanation:

Given : quantity 6 times x to the 4th power plus 9 times x to the 2nd power plus 12 times x all over 3 times x.

To find : Simplify completely the quantity.

Solution : We have given that a statement .

According to question:

6 times x to the 4th power = 6x^{4}.

9 times x to the 2nd power  =  9x^{2}.

12 times x = 12x.

6x^{4} +  9x^{2} +  12x over 3x.

\frac{6x^{4}}{3x} +\frac{9x^{2}}{3x} +\frac{12x}{3x}.

On dividing we get ,

2x^{3} +3x + 4.

Therefore, D) 2x^{3} +3x + 4.

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7 0
3 years ago
-x + 2y = 3<br> 2x – 3y = -6
s2008m [1.1K]

Answer:

x = -3

y = 0

Step-by-step explanation:

<u>Given</u><u> </u><u>equations</u><u> </u><u>:</u><u>-</u><u> </u>

<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u>

<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u>

<u>From</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u><u> </u><u> </u>

<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u>

<u>-x</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>-</u><u> </u><u>2</u><u>y</u><u> </u>

<u>x</u><u> </u><u>=</u><u> </u><u>2</u><u>y</u><u> </u><u>-</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u>

<u>From</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u><u> </u>

<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u>

<u>2</u><u>x</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>+</u><u> </u><u>3</u><u>y</u><u> </u>

<u>x =  \frac{ - 6 + 3y}{2}</u>

<u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>

<u>Equating</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u><u> </u><u>and</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>

<u>x</u><u> </u><u>=</u><u> </u><u>x</u><u> </u>

<u>2y - 3 =  \frac{ - 6 + 3y}{2}</u>

4y - 6 = -6 + 3y

4y - 3y = -6 + 6

y = 0

Putting value of y in ( iii )

x = 2y - 3

x = 2 ( 0 ) - 3

x = -3

4 0
3 years ago
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