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krok68 [10]
3 years ago
14

Plz help me I beg u plz plz!!

Mathematics
2 answers:
Goshia [24]3 years ago
5 0
The graph increases then stays constant
Serga [27]3 years ago
4 0
The last option. The graph increases then remains constant. :)
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at a school there are 10 students taking only chemistry, 9 students taking only physics 5 students taking both, and 16 students
sveticcg [70]

Answer:

If there are 10 students taking only chemistry, 9 students taking only physics, and 5 students only taking both chemisty and 16 students are taking neither; I would add 10+9+5+16=40 (total students) and divide 10/40 (25% chemistry) 9/40 (22.5% physics) 5/40 (12.5% both) 16/40 (40% neither)

Step-by-step explanation:

1. Determine a single event with a single outcome.

2. Identify the total number of outcomes that can occur.

3. Divide the number of events by the number of possible outcomes.

8 0
3 years ago
Read 2 more answers
The sum of two number is 27. One number is 3more than the other number. Write and solve a system of equations to find the two nu
Sladkaya [172]

Answer:

One number is 15 and 3 less than that is 12 so that is your other number

3 0
3 years ago
Math the number to the value of its 5
Ugo [173]
700,353 = 50
133,533 = 500
596,967 = 500,000
632,295 = 5

5 0
3 years ago
Find the location of the absolute maximum and absolute minimum of the function on the interval [ -4,0).
N76 [4]
Maximum is 0 and minimum is -4. Due to the brackets, that means that this is where the value terminates. The parentheses next to the zero means equal to
3 0
3 years ago
Evaluate the following limit, if it exists : limx→0 (12xe^x−12x) / (cos(5x)−1)
Amanda [17]

Answer:

\lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=-\frac{24}{25}

Step-by-step explanation:

Notice that \lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=\frac{12(0)e^{0}-12(0)}{cos(5(0))-1}=\frac{0}{0}, which is in indeterminate form, so we must use L'Hôpital's rule which states that \lim_{x \to c} \frac{f(x)}{g(x)}=\lim_{x \to c} \frac{f'(x)}{g'(x)}. Basically, we keep differentiating the numerator and denominator until we can plug the limit in without having any discontinuities:

\frac{12xe^x-12x}{cos(5x)-1}\\\\\frac{12xe^x+12e^x-12}{-5sin(5x)}\\ \\\frac{12xe^x+12e^x+12e^x}{-25cos(5x)}

Now, plug in the limit and evaluate:

\frac{12(0)e^{0}+12e^{0}+12e^{0}}{-25cos(5(0))}\\ \\\frac{12+12}{-25cos(0)}\\ \\\frac{24}{-25}\\ \\-\frac{24}{25}

Thus, \lim_{x \to 0} \frac{12xe^x-12x}{cos(5x)-1}=-\frac{24}{25}

3 0
2 years ago
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