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mestny [16]
3 years ago
9

Factor completely 3x ^2 -27

Mathematics
2 answers:
pentagon [3]3 years ago
6 0
Both terms are multiply of 3, so you can factor 3 out;

3(x² - 9)

Now, we can see that x² - 9 is difference of two perfect squares

Recall that a² - b² = (a+b)(a-b)

So we have x²-9 = (x+3)(x-3)

So we can factor more here; 3(x²-9) = 3(x+3)(x-3)

Final answer: 3(x+3)(x-3)
Tom [10]3 years ago
3 0
First, factor out a 3.
3(x² - 9)

In any quadratic ax² + bx + c, we can split the bx term up into two new terms which we want to equal the product of a and c.
In this case, we have x² + 0x - 9. (the 0x is a placeholder)
We want two numbers that add to 0 and multiply to get -9.
Obviously, these numbers are 3 and -3.

Now we have 3(x² + 3x - 3x - 9).
Let's factor.
3(x(x+3)-3(x+3))
<u>3(x-3)(x+3)</u>

There are multiple shortcuts which you could make here, FYI:
Instead of splitting the middle, if your a value is 1, you can go straight to that step (x+number)(x+other number).
Whenever you have a difference of squares, like a²-b², that factors to (a+b)(a-b).
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Solve the proportion x+1/3 = x/2<br><br> 5<br> 1<br> 2<br> 10
s344n2d4d5 [400]

Value of x is 2.

Step-by-step explanation:

  • Step 1: Given that x+1/3 = x/2

Cross-multiply to find the value of x.

⇒ 2(x + 1) = 3x

⇒ 2x + 2 = 3x

⇒ x = 2

4 0
3 years ago
Please help me with this question, I'm stuck ;( .
sveticcg [70]

i think C but i dont know forsure

7 0
3 years ago
Pls help its due today
Simora [160]

Answer:

the answer is a, unchecked growth

5 0
3 years ago
Read 2 more answers
Can anyone help me with this question please :( .
mel-nik [20]

Answer:

I think it's non linear because the 1 to 1 wouldn't work

3 0
2 years ago
The solution set for the inequality x + 4 ≤ 1/2 (x - 3) includes -11 as an element.
Sindrei [870]

Answer:

<h2>x ≤ -11, Yes. The solution set include -11.</h2>

Step-by-step explanation:

x+4\leq\dfrac{1}{2}(x-3)\qquad\text{multiply both sides by 2}\\\\2x+8\leq1(x-3)\\\\2x+8\leq x-3\qquad\text{subtract 8 from both sides}\\\\2x\leq x-11\qquad\text{subtract x from both sides}\\\\\boxed{x\leq-11}

6 0
2 years ago
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