Which of the following methods of timekeeping is the least precise?

Mathematics

2 answers:

katen-ka-za [31]2 years ago

8 0

I think C would be your answer

Verizon [17]2 years ago

5 0

Answer:

na D cause imagine always observing the seasons bro

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A firm has the following fixed proportion production function q = min(k, 2l). a. Find the firms long-run total, average, and mar

Yuri [45]

Answer:

I think its right

Step-by-step explanation:

a) In the long run, we have, 5k=10 and k=2l Thus, C=2l+3l= 5l=0.5q Thus, AC=5l/10l =0.5 MC=0.5 b) In the short run, k=10 Q=min(50,10l) If l<5, q=10l. C=10+3l= 10+0.3q Thus, AC=10/q + 0.3 If l>5, q=50. C= 10+3l AC= (10+3l)/50 If q>50, then MC...

3 0

2 years ago

Consider the line x+5y=6

Tems11 [23]

Answer:

$x + 5y = 6$ is perpendicular to $y = 5x + \frac{6}{5}$ and parallel to $y = -\frac{1}{5}x + 2\\$

Step-by-step explanation:

First, convert the equation to standard form so that y is isolated.

x + 5y = 6 --> x - 6 = -5y (divide both sides by -5) --> $y = -\frac{1}{5}x + \frac{6}{5}$

A perpendicular line will have a slope that is the opposite reciprocal of the original slope (meaning you flip the numerator and denominator then make it negative).

$-\frac{1}{5}$ is perpendicular to $-(\frac{-5}{1} )$ which simplifies to 5.

A parallel line will have the same slope, but the y-intercept will be different. It can be pretty much any number as long as the original slope is used in the new equation.

$y = -\frac{1}{5}x + \frac{6}{5}\\$ is parallel to $y = -\frac{1}{5}x + 2\\$ just like $y = -\frac{1}{5}x - \frac{100}{23}$ .