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3 years ago

(factor out) 3x(to the power of 2) -39x+36

1 answer:

7 0

3x² - 39x + 36
3(x²) - 3(13x) + 3(12)
3(x² - 13x + 12)
3(x² - 12x - x + 12)
3(x(x) - x(12) - 1(x) + 1(12))
3(x(x - 12) - 1(x - 12))
3(x - 1)(x - 12)

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Helppp plsssss!!show your work.<br>Thanks

gregori [183]

You are basically just adding, so combine like terms.

3x^2 + 5x^2 = 8x^2

6x - 2x = 4x

4 - 8 = -4

So the answer is D.

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3 years ago

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NARA [144]

Answer:

The rearrangement can be 45, 987 , 310

Step-by-step explanation:

Here, we want to rearrange the number such that 9 is worth 10 times as what it is worth presently

The value of 9 presently is 90,000

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So we can have the new arrangement as;

45, 987, 310

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3 years ago

What is the sign of 39 11/13 + -41 1/13

Radda [10]

Answer:

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Step-by-step explanation:

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3 years ago

NOT GRADED. <br>FIND THE MEASURE OF THE INDICATED ANGLE IN EACH TRIANGLE.

Setler [38]

Answer:

The angle in the box is always 50 and that other acute angle is 46

Hope this helps

Step-by-step explanation:

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3 years ago

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For what value of constant c is the function k(x) continuous at x = 0 if k =

nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x) is continuous, we take the limit of the parameter as follows:

$\mathbf{ \lim_{x \to 0^-} k(x) = \lim_{x \to 0^+} k(x) = 0 }$

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= c }$

Provided that:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }$

Using l'Hospital's rule:

$\mathbf{\implies \lim_{x \to 0} \ \ \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}= \lim_{x \to 0} sec \ x \ tan \ x = 0}$

Therefore:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}=0 }$

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

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2 years ago