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2 years ago

Are there any limits to the value of Sine, Cosine and Tangent? if so, what are they and why?

1 answer:

8 0

With Sine, Cosine, and Tangent there are no limits with the numbers you can use and they can be found on a calculator. When you press one of the 3 it will show up like:

Sine(
Cos(
Tan(

Than just add your number and press enter

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Mr. Sanchez bought 2 magazines for $9.95 each and 1 book for $14.95. If the sales tax is 6%, what is the total cost of Mr. Sanch

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add up the purchases

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3 years ago

PLEASE HELP ASAP In this task, you will practice finding the area under a nonlinear function by using rectangles. You will use g

mrs_skeptik [129]

Answer:

a) $1280 u^{2}$

b) $1320 u^{2}$

c) $\frac{4000}{3} u^{2}$

Step-by-step explanation:

In order to solve this problem we must start by sketching the graph of the function. This will help us visualize the problem better. (See attached picture)

You can sketch the graph of the function by plotting as many points as you can from x=0 to x=20 or by finding the vertex form of the quadratic equation by completing the square. You can also do so by using a graphing device, you decide which method suits better for you.

So we are interested in finding the area under the curve, so we divide it into 5 rectangles taking a right hand approximation. This is, the right upper corner of each rectangle will touch the graph. (see attached picture).

In order to figure the width of each rectangle we can use the following formula:

$\Delta x=\frac{b-a}{n}$

in this case a=0, b=20 and n=5 so we get:

$\Delta x=\frac{20-0}{5}=\frac{20}{5}=4$

so each rectangle must have a width of 4 units.

We can now calculate the hight of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:

h1=64

h2=96

h3=96

h4= 64

h5=0

so now we can use the following formula to find the area under the graph. Basically what the formula does is add the areas of the rectangles:

$A=\sum^{n}_{i=1} f(x_{i}) \Delta x$

which can be rewritten as:

$A=\Delta x \sum^{n}_{i=1} f(x_{i})$

So we go ahead and solve it:

$A=(4)(64+96+96+64+0)$

so:

$A= 1280 u^{2}$

B) The same procedure is used to solve part B, just that this time we divide the area in 10 rectangles.

In order to figure the width of each rectangle we can use the following formula:

$\Delta x=\frac{b-a}{n}$

in this case a=0, b=20 and n=10 so we get:

$\Delta x=\frac{20-0}{10}=\frac{20}{10}=2$

so each rectangle must have a width of 2 units.

We can now calculate the hight of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:

h1=36

h2=64

h3=84

h4= 96

h5=100

h6=96

h7=84

h8=64

h9=36

h10=0

so now we can use the following formula to find the area under the graph. Basically what the formula does is add the areas of the rectangles:

$A=\sum^{n}_{i=1} f(x_{i}) \Delta x$

which can be rewritten as:

$A=\Delta x \sum^{n}_{i=1} f(x_{i})$

So we go ahead and solve it:

$A=(2)(36+64+84+96+100+96+84+64+36+0)$

so:

$A= 1320 u^{2}$

In order to find part c, we calculate the area by using limits, the limit will look like this:

$\lim_{n \to \infty} \sum^{n}_{i=1} f(x^{*}_{i}) \Delta x$

so we start by finding the change of x so we get:

$\Delta x =\frac{b-a}{n}$

$\Delta x =\frac{20-0}{n}$

$\Delta x =\frac{20}{n}$

next we find $x^{*}_{i}$

$x^{*}_{i}=a+\Delta x i$

so:

$x^{*}_{i}=0+\frac{20}{n} i=\frac{20}{n} i$

and we find $f(x^{*}_{i})$

$f(x^{*}_{i})=f(\frac{20}{n} i)=-(\frac{20}{n} i)^{2}+20(\frac{20}{n} i)$

cand we do some algebra to simplify it.

$f(x^{*}_{i})=-\frac{400}{n^{2}}i^{2}+\frac{400}{n}i$

we do some factorization:

$f(x^{*}_{i})=-\frac{400}{n}(\frac{i^{2}}{n}-i)$

and plug it into our formula:

$\lim_{n \to \infty} \sum^{n}_{i=1}-\frac{400}{n}(\frac{i^{2}}{n}-i) (\frac{20}{n})$

And simplify:

$\lim_{n \to \infty} \sum^{n}_{i=1}-\frac{8000}{n^{2}}(\frac{i^{2}}{n}-i)$

$\lim_{n \to \infty} -\frac{8000}{n^{2}} \sum^{n}_{i=1}(\frac{i^{2}}{n}-i)$

And now we use summation formulas:

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (\frac{n(n+1)(2n+1)}{6n}-\frac{n(n+1)}{2})$

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (\frac{2n^{2}+3n+1}{6}-\frac{n^{2}}{2}-\frac{n}{2})$

and simplify:

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (-\frac{n^{2}}{6}+\frac{1}{6})$

$\lim_{n \to \infty} \frac{4000}{3}+\frac{4000}{3n^{2}}$

and solve the limit

$\frac{4000}{3}u^{2}$

4 0

2 years ago