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3 years ago

7

40% of the children in a sports club play badminton. 25% of the children who play badminton also play squash. There are 11 child

ren in the club who play both badminton and squash.
How many children are there in the sports club altogether?

1 answer:

Keith_Richards [23]3 years ago

6 0

Answer:

There are 110 children total in the sports club

Step-by-step explanation:

To get this answer, its actually easier than it seems. You might need a calculator however.

First you start with the 11 children who play both badminton and squash, then, you divide that by 0.25 (25%) to get 44.

Next you take 44 and divide it by 0.4 (40%) to get 110.

And there you go! If you want to make sure you got it right simply start with 110 and multiply it by 0.4 then multiply the number/decimal you get by 0.25. You should get 11 to confirm your answer :)

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Answer:

Donuts(b) = 12*b

Step-by-step explanation:

Notice that the question is:

What's an expression to model the number of donuts in b boxes?

We know that we are talking of boxes of a dozen donuts.

We know that in one dozen, we have 12 donuts.

Then if we have b boxes, and in each box we have 12 donuts, we will have a total of b times 12 donuts, the expression is:

Donuts(b) = 12*b

There is a lot of information here that we did not use (like the cost of cardboard and donuts), and that may be there just to distract us, the important thing to do here is:

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3 years ago

Evaluate the integral of the quantity x divided by the quantity x to the fourth plus sixteen, dx . (2 points) one eighth times t

Anika [276]

Answer:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

Step-by-step explanation:

Given

$\int\limits {\frac{x}{x^4 + 16}} \, dx$

Required

Solve

Let

$u = \frac{x^2}{4}$

Differentiate

$du = 2 * \frac{x^{2-1}}{4}\ dx$

$du = 2 * \frac{x}{4}\ dx$

$du = \frac{x}{2}\ dx$

Make dx the subject

$dx = \frac{2}{x}\ du$

The given integral becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{x}{x^4 + 16}} \, * \frac{2}{x}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{1}{x^4 + 16}} \, * \frac{2}{1}\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

Recall that: $u = \frac{x^2}{4}$

Make $x^2$ the subject

$x^2= 4u$

Square both sides

$x^4= (4u)^2$

$x^4= 16u^2$

Substitute $16u^2$ for $x^4$ in $\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{x^4 + 16}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16u^2 + 16}} \,\ du$

Simplify

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \int\limits {\frac{2}{16}* \frac{1}{8u^2 + 8}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{2}{16}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

In standard integration

$\int\limits {\frac{1}{u^2 + 1}} \,\ du = arctan(u)$

So, the expression becomes:

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}\int\limits {\frac{1}{u^2 + 1}} \,\ du$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(u)$

Recall that: $u = \frac{x^2}{4}$

$\int\limits {\frac{x}{x^4 + 16}} \, dx = \frac{1}{8}*arctan(\frac{x^2}{4}) + c$

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2 years ago

Explain why the "+ C" is not needed in the antiderivative when evaluating a definite integral.

Aliun [14]

The reason the "+ C" is not needed in the antiderivative when evaluating a definite integral is; The C's cancel each other out as desired.

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Let us say we want to estimate the definite integral;

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Now, for any C, f(x) + C is an antiderivative of f′(x).

From fundamental theorem of Calculus, we can say that;

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where Ф(x) is any antiderivative of f'(x). Thus, Ф(x) = f(x) + C would not work because the C's will cancel each other.

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Answer:

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Step-by-step explanation:

a) As P(0) is true

we will assume that

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this simply means that ; P(n) is true for all 'n' in the set

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we will assume

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