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otez555 [7]
3 years ago
11

-2 < 3x+4 < 31 find x

Mathematics
1 answer:
Eva8 [605]3 years ago
8 0

First rewrite into 2 inequalities,

-2\lt3x+4

3x+4\lt31

Solve each of the inequalities,

-2\lt3x+4\implies3x\gt-6\implies x\gt-2

3x+4\lt31\implies 3x\lt27\implies x\lt9

So our x is between -2 and 9 which can be written as,

-2\lt x\lt9

or as an interval,

x\in(-2,9)

Hope this helps :)

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What is the The range of y = square root X-5 – 1
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Answer:

[-1, infinity)

Step-by-step explanation:

I must assume that you meant y = √(x - 5) - 1.

The range of the square root function is [0, infinity.

If 1 is subtracted from the square root function, the range of the resulting function is [-1, infinity).

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if each of the 4 shelves holds 14 movies, how many movies does betty have on her shelving unit? show your work
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The mass of a carbon atom is 1.994 × 10-23 grams. The mass of a hydrogen atom is 1.67 × 10-24 grams. What is the difference of t
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 1.994 x 10^-23
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1. As you can tell from the function definition and plot, there's a discontinuity at x = -2. But in the limit from either side of x = -2, f(x) is approaching the value at the empty circle:

\displaystyle \lim_{x\to-2}f(x) = \lim_{x\to-2}(x-2) = -2-2 = \boxed{-4}

Basically, since x is approaching -2, we are talking about values of x such x ≠ 2. Then we can compute the limit by taking the expression from the definition of f(x) using that x ≠ 2.

2. f(x) is continuous at x = -1, so the limit can be computed directly again:

\displaystyle \lim_{x\to-1} f(x) = \lim_{x\to-1}(x-2) = -1-2=\boxed{-3}

3. Using the same reasoning as in (1), the limit would be the value of f(x) at the empty circle in the graph. So

\displaystyle \lim_{x\to-2}f(x) = \boxed{-1}

4. Your answer is correct; the limit doesn't exist because there is a jump discontinuity. f(x) approaches two different values depending on which direction x is approaching 2.

5. It's a bit difficult to see, but it looks like x is approaching 2 from above/from the right, in which case

\displaystyle \lim_{x\to2^+}f(x) = \boxed{0}

When x approaches 2 from above, we assume x > 2. And according to the plot, we have f(x) = 0 whenever x > 2.

6. It should be rather clear from the plot that

\displaystyle \lim_{x\to0}f(x) = \lim_{x\to0}(\sin(x)+3) = \sin(0) + 3 = \boxed{3}

because sin(x) + 3 is continuous at x = 0. On the other hand, the limit at infinity doesn't exist because sin(x) oscillates between -1 and 1 forever, never landing on a single finite value.

For 7-8, divide through each term by the largest power of x in the expression:

7. Divide through by x². Every remaining rational term will converge to 0.

\displaystyle \lim_{x\to\infty}\frac{x^2+x-12}{2x^2-5x-3} = \lim_{x\to\infty}\frac{1+\frac1x-\frac{12}{x^2}}{2-\frac5x-\frac3{x^2}}=\boxed{\frac12}

8. Divide through by x² again:

\displaystyle \lim_{x\to-\infty}\frac{x+3}{x^2+x-12} = \lim_{x\to-\infty}\frac{\frac1x+\frac3{x^2}}{1+\frac1x-\frac{12}{x^2}} = \frac01 = \boxed{0}

9. Factorize the numerator and denominator. Then bearing in mind that "x is approaching 6" means x ≠ 6, we can cancel a factor of x - 6:

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10. Factorize the numerator and simplify:

\dfrac{-2x^2+2}{x+1} = -2 \times \dfrac{x^2-1}{x+1} = -2 \times \dfrac{(x+1)(x-1)}{x+1} = -2(x-1) = -2x+2

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\displaystyle \lim_{x\to\infty} \frac{-2x^2+2}{x+1} = \lim_{x\to\infty} (-2x+2) = \boxed{-\infty}

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2 years ago
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igomit [66]

Answer:

Production cost is $20 per item.

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Fixed cost is $100 and 20 items cost $500 to produce.

C=100+x*production cost

500=100+20*production cost

400=20*production cost

Production cost = $20.

So, C(x)=20x+100, where C is total cost and x is the number of items produced.

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