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3 years ago

Which arrangement for these numbers is from least to greatest?

Mathematics

1 answer:

laiz [17]3 years ago

8 0

Answer: D) -5/6 < -3/5 < -4/7 < -4/9

Step-by-step explanation:

Since putting fractions from least to greatest by using negatives the number or fraction has to be closed to 0. -5/6 is the farthest from 0 so it'll be the least -4/9 is closes to 0 so -4/9 is the most.

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Sofia is working two summer jobs, making $12 per hour babysitting and making $8 per hour clearing tables. In a given week, she c

Mamont248 [21]

The possible values for the number of whole hours clearing tables that she must work to meet her requirements is 2, 3 hours

Solution:

Amount earned in babysitting = $ 12 per hour

Amount earned in clearing tables = $ 8 per hour

In a given week, she can work a maximum of 17 total hours and must earn a minimum of $180

Sofia worked 14 hours babysitting

Therefore,

Amount earned at babysitting = 14 x 12 = 168

Thus, Sofia earned $ 168 at babysitting

Sofia must earn a minimum of $ 180

Remaining amount to be earned = 180 - 168 = 12

Thus, Sofia must earn $ 12 from clearing tables

Amount earned in clearing tables = $ 8 per hour

So, she must work for atleast 1.5 hours to get $ 12 from clearing tables

She can work a maximum of 17 total hours and Sofia worked 14 hours babysitting

Remaining is 17 - 14 = 3 hours

Thus possible values for the number of whole hours clearing tables that she must work to meet her requirements is 2 hours or 3 hours

3 0

4 years ago

Last year, your salary was $33,975. This year, your boss tells you your salary will be $34,960. What percent raise (change) did

Pavlova-9 [17]

Answer:

2.9%

Step-by-step explanation:

5 0

3 years ago

Draw 8 lines that are between 1 inch and 3 inches long measure each line to the nearest fourth inch and make a line plot

Paraphin [41]

The Supplemental Security Income (SSI) program, administered by the Social Security Administration (SSA), is the income source of last resort for thelow-income aged, blind, and disabled. As the nation's largest income-assistance program, it paid $38 billion in benefits in calendar year 2006 to roughly 7 million recipients per month. BecauseSSI is means tested, administering the program often requires month-to-month, recipient-by-recipient benefit recomputations. An increase in a recipient's income usually triggers a benefit recomputation. Or, an increase in the recipient's financial assets, which may render the recipient ineligible, would also prompt a recomputation. With this crush of ongoing recomputations, it is of little wonder that administrative simplification is a time-honored mantra for program administrators.

6 0

3 years ago

The app isn't letting me post the picture alone so this is just here so I can post the picture ;-;

labwork [276]

Answer:

(x, y) = (0, -2248)

Step-by-step explanation:

Substitution

Substitute for y:

300x -2248 = 350x -2248

0 = 50x . . . . . . . . add 2248-300x to both sides

x = 0

y = 300·0 -2248

The solution is (x, y) = (0, -2248).

Elimination

Subtract the first equation from the second.

(y) -(y) = (350x -2248) -(300x -2248)

0 = 50x

x = 0

y = 300·0 -2248

The solution is (x, y) = (0, -2248).

Additional comment

If you think for a bit about what the graph looks like, you realize both lines cross at their y-intercept, (0, -2248). That point of intersection is the solution to the system of equations.

7 0

3 years ago

PRECAL: Having trouble on this review, need some help.

ra1l [238]

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:

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

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

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

6 0

2 years ago