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Flauer [41]
3 years ago
11

Help please will give brainliest

Mathematics
1 answer:
Mnenie [13.5K]3 years ago
6 0

Answer: (-2,-11)

Step-by-step explanation:

We looking for values that will satisfy the equation upon inspection, when x is -2, the equation gives y=-11 .

I.e y=4x-3 ,when x=2

Y= 4(-2)-3

Y= -8-3

Y= -11

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How to evaluate the limit
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\displaystyle\lim_{x\to2}\frac{x^2-x+6}{x+2}

Both the numerator and denominator are continuous at x=2, which means the quotient rule for limits applies:

\dfrac{\displaystyle\lim_{x\to2}(x^2-x+6)}{\displaystyle\lim_{x\to2}(x+2)}=\dfrac{2^2-2+6}{2+2}=\dfrac84=2

Perhaps you meant to write that x\to-2 instead? In that case, you would have

\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5
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Factor -8x^3-2x^2-12x-3
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shtirl [24]
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Help me! Need answers to questions ii and iii please. Thanks x
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Ii.First ,expand: 5(x-2)=32
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