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svlad2 [7]
3 years ago
13

In order to pay for his expenses, Chuck needs to earn $2400 at least every

Mathematics
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

5400

Step-by-step explanation:

2400+3000

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56 / 4 = 14
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The graphs below have the same shape. What is the equation of the red<br> graph?
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What is the factorization of the following trinomial? 15x2 - 44x - 36
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15x^2-44x-36

=(15x^2+10x)+(-54x-36)

factor out  5x from 15x^2+10x  = 5x(3x+2)

Factor out -18 from -54x-36 =  -18(3x+2)

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PRECAL:<br> Having trouble on this review, need some help.
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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}

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\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:

\displaystyle \lim_{x\to6}\frac{2x^2-12x}{x^2-4x-12}=\lim_{x\to6}\frac{2x(x-6)}{(x+2)(x-6)} = \lim_{x\to6}\frac{2x}{x+2} = \frac{2\times6}{6+2}=\boxed{\frac32}

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

where the last equality holds because x is approaching +∞, so we can assume x ≠ -1. Then the limit is

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

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