If 3p − 2 ≥ 1, what is the least possible value of 3p + 2 ? A) 5 B) 3 C) 2 D) 1

Eva, the owner of eva's second time around wedding dresses, currently has five dresses to be altered, shown in the order in whic

cupoosta [38]

To answer this question, an assumption must be made, that Eva spends 8 hours a day working. If this is the case, then Eva will complete jobs w, x, and v on day one, for a total of six hours. Since the next job (y) requires 4 hours, she will spend two hours working that day, leaving 2 more hours to go on that job. The next day she will spend 2 hours finishing job y, completing it, and finish the longest job z (hours) that day. This means she had 4 jobs on day one, and 2 jobs on day 2 for and average of 3 jobs per day.
This answer assumes an 8 hour work day, and that Eva can start a job she cannot finish that day.

5 0

3 years ago

Which of the following BEST describes a pyramid?

Jet001 [13]

Answer:

C

Step-by-step explanation:

Pyramids must have bases and a point off the base

3 0

3 years ago

For what value of c is the function defined below continuous on (-\infty,\infty)?

kozerog [31]

$f(x)= \left \{ {{x^2-c^2,x \ \textless \ 4} \atop {cx+20},x \geq 4} \right
$

It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
A function, f, is continuous at x = 4 if
 $\lim_{x \rightarrow 4} \ f(x) = f(4)$

In notation we write respectively
 $\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)$

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just 4c + 20.

On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
$\lim_{x \rightarrow 4-} f(x) = \lim_{x \rightarrow 4-} (x^2 - c^2) = 16 - c^2$

Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c²

$c^2+4c+4=0
\\(c+2)^2=0
\\c=-2$

That is to say, if c = -2, f(x) is continuous at x = 4.

Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers $(-\infty, +\infty)$