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Korvikt [17]
3 years ago
14

Which inequality represents all numbers x on a number line that are farther from −8 than from −4?

Mathematics
1 answer:
pentagon [3]3 years ago
7 0

Answer:

x - 8>-4-x

Step-by-step explanation:

Looking at x - 8>-4-x

Collect the like terms;

x+x > -4 + 8

2x < 4

x > 4/2

x < 2

Since the values of x are greater than 2,this shows that they are positive values and will be farther from -8 than -4

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ipn [44]
There are 2 way to solve this.

one using Pythagoras theorem and 2nd using trigonometry

so lets solve it by both

using Pythagoras theorem we know

base^2 + perpendicular^2 = hypotanes^2

6^2 + x^2 = 12^2

36 + x^2 = 144

x^2 = 144- 36 = 108

x = √(108) = √( 2×2×3×3×3)

= (2×3) √ (3) = 6 √3

so answer is option 2

bow lets use trigonometry

we know
sin theta = perpendicular / hypotanes
sin 60 = x /12
x = 12 × sin 60
we kNow sin 60 = √3/ 2
so
x = 12×√3 /2 = 6√3
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3 years ago
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:

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

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