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3 years ago

PLEASE HELP with this question. it is in the picture file below!!!!

Mathematics

1 answer:

ryzh [129]3 years ago

3 0

The line g(x) is 3 units above the line f(x). k = 3.

_____

The y-intercept of g(x) is 5; the y-intercept of f(x) is 2. 5-2 = 3, the distance between the lines where they cross the y-axis.

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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

Spencer and his sister watched a cartoon movie that was 1 hour and 20 minutes long. After the movie, they played a card game for

liberstina [14]

Answer:

10:45 - they started watching the film at this time

Step-by-step explanation:

4 0

2 years ago

Una máquina demora 2 minutos y 30 segundos en fabricar una silla. Si en un día la máquina funciona sin parar 8 horas ¿Cuántas si

vaieri [72.5K]

Una máquina demora 2 minutos y 30 segundos en fabricar una silla. Si en un día la máquina funciona sin parar 8 horas ¿Cuántas sillas fabricará?

5 x 12
24 x 8
192

5 0

2 years ago

In ΔMNO, the measure of ∠O=90°, the measure of ∠N=70°, and OM = 3.3 feet. Find the length of MN to the nearest tenth of a foot.

QveST [7]

Answer:3.5

Step-by-step explanation:

8 0

3 years ago

Simplify. 5√ 10 <br> 1) 2√ 2) 2√2 3) 10√2 4) 5√

Nataly_w [17]

For this case we have that the expression in its exact form is the same, that is:

$5 \sqrt {10}$