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Elden [556K]
3 years ago
12

Someone help meeee lollll

Mathematics
1 answer:
den301095 [7]3 years ago
6 0

Answer:

Step-by-step explanation:

The best way to do this is to first remove the parenthesis by distributing the negative into them. Then we will group the like terms together, moving their respective signs with them, then we will simplify.  For the first one, after distributing the negative into the parenthesis (remember that a negative in front of a set of parenthesis containing a polynomial changes the signs in front of each term within the polynomial; negatives become positives and positives become negatives):

3x^2-6x+11-10x^2+4x-6 and then grouping like terms next to each other:

3x^2-10x^2-6x+4x+11-6 (this step is not completely necessary; I just encourage new learners to do it so as to not miss any of the like terms)

The above then simplifies to

-7x^2-2x+5

The next one, after distributing the negative into the parenthesis:

-3x^2-5x-3+10x^2+7x-2 and then grouping like terms next to each other (remember to move their signs with them!):

-3x^2+10x^2-5x+7x-3-2 which simplifies to

7x^2+2x-5

Now for the last one:

12x^2+6x-5-5x^2-8x+12 and

12x^2-5x^2+6x-8x-5+12 which simplifies to

7x^2-2x+7

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What is the range of the function y = x 2?
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Answer:

\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

The graph is also attached below.

Step-by-step explanation:

Given the function

y=x^2

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\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)

\mathrm{If}\:a

\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v

a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(0,\:0\right)

f\left(x\right)\ge \:0

Thus,

\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}

The graph is also attached below.

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