Answer:
Using either method, we obtain: 
Step-by-step explanation:
a) By evaluating the integral:
![\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cint%5Climits%5Et_0%20%7B%5Csqrt%5B8%5D%7Bu%5E3%7D%20%7D%20%5C%2C%20du)
The integral itself can be evaluated by writing the root and exponent of the variable u as: ![\sqrt[8]{u^3} =u^{\frac{3}{8}](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7Bu%5E3%7D%20%3Du%5E%7B%5Cfrac%7B3%7D%7B8%7D)
Then, an antiderivative of this is: 
which evaluated between the limits of integration gives:
and now the derivative of this expression with respect to "t" is:
b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:
"If f is continuous on [a,b] then
is continuous on [a,b], differentiable on (a,b) and 
Since this this function 
is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:
The value of m is 35 degrees, because the right angle is 90 degrees, therefore, 90-55=35.
The value of t is 55 degrees because is perpendicular to the other 55 degree angle.
Answer:
15 seconds
Step-by-step explanation:
If you make a table of values for the dog and the squirrel using d = rt, then the rates are easy: the dog's rate is 150 and the squirrel's is 100. The t is what we are looking for, so that's our unknown, and the distance is a bit tricky, but let's look at what we know: the dog is 200 feet behind the squirrel, so when the dog catches up to the squirrel, he has run some distance d plus the 200 feet to catch up. Since we don't know what d is, we will just call it d! Now it seems as though we have 2 unknowns which is a problem. However, if we solve both equations (the one for the dog and the one for the squirrel) for t, we can set them equal to each other. Here's the dog's equation:
d = rt
d+200 = 150t
And the squirrel's:
d = 100t
If we solve both for t and set them equal to each other we have:
Now we can cross multiply to solve for d:
150d = 100d + 20,000 and
50d = 20,000
d = 400
But we're not looking for the distance the squirrel traveled before the dog caught it, we are looking for how long it took. So sub that d value back into one of the equations we have solved for t and do the math:
That's 1/4 of a minute which is 15 seconds.
[Rewritten and Fixed]
Joint relative frequency is the ratio of the frequency in a certain category and the total number of data points in that category.
Answer:
Step-by-step explanation:
1.
Simplify the expression by combining like terms. Remember, like terms have the same variable part, to simplify these terms, one performs operations between the coefficients. Please note that a variable with an exponent is not the same as a variable without the exponent. A term with no variable part is referred to as a constant, constants are like terms.
2.
Use a very similar method to solve this problem as used in the first. Please note that all of the rules mentioned in the first problem also apply to this problem; for that matter, the rules mentioned in the first problem generally apply to any pre-algebra problem.
3.
Use the same rules as applied in the first problem. Also, keep the distributive property in mind. In simple terms, the distributive property states the following (
). Also note, a term raised to an exponent is equal to the term times itself the number of times the exponent indicates. In the event that the term raised to an exponent is a constant, one can simplify it. Apply these properties here,
4.
The same method used to solve problem (3) can be applied to this problem.
