Answer: 3:12 pm (C.)
Explanation:
The 4 train from battery park to e 161st is 35 minutes. However, it takes a few minutes to walk from the 161st stop to the stadium, so you would need to budget in extra time (so 3:22 would be too late even though it gets in before 4:00 pm). The best train would be the 3:12 because it would get in at 3:47, giving you plenty of time to walk from the station to the stadium.
Answer:
Answer to the following question is as follows;
Explanation:
The primary argument of a writer in an article is the assertion that regulates the entire article, that really is, it is the writer's opinion on a subject matter that will evaluate all of the knowledge described in the paper, in which the person who wrote this article will display evidence that demonstrates how his point of view is completely accurate and therefore should be treated seriously.
Answer:
double a;
double b;
double distance = Math.sqrt( (Math.pow(a, 2) + Math.pow(b, 2));
System.out.println("The distance is: " + distance);
Explanation:
The code snippet is written in Java.
First, variable a is declared as a type of double and variable b is also declared as a type of double.
The given formula for the distance is:
√(a² + b²)
So we use the inbuilt function of Java to calculate the power and the square root.
Math.sqrt is to find the square root and it returns a value of type double.
Math.pow is use to calculate the value of a² and b².
a² = Math.pow(a, 2)
b² = Math.pow(b, 2)
Then Math.sqrt is applied to the sum of Math.pow(a, 2) and Math.pow(b, 2), the value is assigned to distance of type double.
The value of distance is now displayed to the user:
System.out.println("The distance is: " + distance);
The easiest way to find such limits, where there is a numerator and a denominator is to apply <span><span>Hospital's Rule.
1st find the derivative of the numerator and the derivative of the denominator, if it still gives an indeterminate value, find the second derivative of N and D
3) lim sin(2x)/x when x →0
Derivative sin2x → 2cos2x
Derivative x→ 1
2cos2x/1 when x→0 , 2cos2x → 2
and lim sin(2x)/x when x →0 is 2
4) lim(sinx)/(2x²-x)
→cosx/(2x-1) when x →0 cosx/(2x-1) = -1
and lim(sinx)/(2x²-x) when x →0 is -1
and so on and so forth. Try to continue following the same principle
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