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ivann1987 [24]
3 years ago
10

Two students, Jennifer and Jamal, factored the trinomial 6x2 − 3x − 9. Jamal factored it as 3(2x − 3)(x + 1) and Jennifer factor

ed it as (2x − 3)(3x + 3). Indicate which student factored the trinomial completely and which student did not, and explain why.
Mathematics
2 answers:
GaryK [48]3 years ago
7 0
Jamal factored it completely.

in Jennifer's answer, 3x+3 has a common factor of 3 which can be factored to become 3(x+1) which is what Jamal did.
vaieri [72.5K]3 years ago
3 0

no thats wrong Jamal factored it correctly

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Irina18 [472]
<h3><u>QUES</u><u>TION</u><u>:</u><u> </u></h3>

what is x/9 = 9/27 help plz

<h3><u>ANSWER</u><u> </u><u>AND</u><u> </u><u>SOLU</u><u>TION</u><u>:</u><u> </u></h3>

  • SOLVING FOR X

<u>Reduce</u><u> </u><u>first</u><u> </u><u>the</u><u> </u><u>fraction</u><u> </u><u>with</u><u> </u><u>9</u>

\large \sf  = \dfrac{x}{9}  =  \dfrac{ \cancel9}{ \cancel{27}}

<u>\large \sf =  \dfrac{x}{9}  =  \dfrac{1}{3}</u>

<u>Now</u><u> </u><u>simpli</u><u>fy</u><u> </u><u>using</u><u> </u><u>cross-multiply</u>

<u>\large \sf  = \dfrac{3x}{9}  = 1</u>

<u>\large \sf = 3x = 1 \times 9</u>

<u>\large \sf = 3x = 9</u>

<u>Lastly</u><u> </u><u>divide</u><u> </u><u>both</u><u> </u><u>sides</u>

<u>\large \sf = 3x \div 3 = 9 \div 3</u>

<u>\large \sf = x = 9  \div 3</u>

<u>\large \sf = x = 3</u>

<h3><u>FINAL</u><u> </u><u>ANSWER</u><u>:</u></h3>

\huge \orange{ \underline{ \boxed{ \sf{x = 3}}}}

HOPE THIS HELP YOU! HAVE A NICE DAY!

~kimtaetae92~

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ale4655 [162]

The vertex of the function f(x) exists (1, 5), the vertex of the function g(x) exists (-2, -3), and the vertex of the function f(x) exists maximum and the vertex of the function g(x) exists minimum.

<h3>How to determine the vertex for each function is a minimum or a maximum? </h3>

Given:

$\mathrm{f}(\mathrm{x})=-(\mathrm{x}-1)^{2}+5$ and

$\mathrm{g}(\mathrm{x})=(\mathrm{x}-2)^{2}-3$

The generalized equation of a parabola in the vertex form exists

$y=a(x-h)^{2}+k

Vertex of the function f(x) exists (1, 5).

Vertex of the function g(x) exists (-2, -3).

Now, if (a > 0) then the vertex of the function exists minimum, and if (a < 0) then the vertex of the function exists maximum.

The vertex of the function f(x) exists at a maximum and the vertex of the function g(x) exists at a minimum.

To learn more about the vertex of the function refer to:

brainly.com/question/11325676

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2 years ago
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alexgriva [62]
Recall that

|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x

There are three cases to consider:

(1) When x+2, we have |x+2|=-(x+2) and |x-2|=-(x-2), so

|x-2|+|x+2|=-(x-2)-(x+2)=-2x-4

(2) When x+2\ge0 and x-2, we get |x+2|=x+2 and |x-2|=-(x-2), so

|x-2|+|x+2|=-(x-2)+(x+2)=4

(3) When x-2\ge0, we have |x+2|=x+2 and |x-2|=x-2, so

|x-2|+|x+2|=(x-2)+(x+2)=2x

So

|x-2|+|x+2|=\begin{cases}-2x-4&\text{if }x
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Answer:

12 + 7 = 19

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2 years ago
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vovangra [49]
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