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2 years ago

3(2x + 5) 2 2(x + 6)

Mathematics

1 answer:

puteri [66]2 years ago

6 0

Answer:

i recommend you watch videos on youtuube about equations and take notes.

Step-by-step explanation:

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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

The coordinates of the vertices of quadrilateral ABCD are A(-5, 1), B(-2,5), C(5, 3),

sladkih [1.3K]

Step-by-step explanation:

To calculate a slope, we need to apply: $m=\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Applying formula to each slope:

$m_{AB}=\frac{5-1}{-2-(-5)} = \frac{4}{3} \\m_{BC}=\frac{3-5}{5-(-2)} =\frac{-2}{7}\\m_{CD}=\frac{-1-3}{2-5}=\frac{-4}{-3} =\frac{4}{3}$

So, as you can see, there are equal pair of slopes, meaning that they are parallels, which<em> demonstrate that it's actually a quadrilateral figure.</em>

6 0

3 years ago

Read 2 more answers

A rectangle with a length of x + 5 has a perimeter of 4x + 14.

geniusboy [140]

Answer:

hope this helps you

8 0

2 years ago

Wastewater is filling barrels at the rate of 7 quarts per hour. The recycling facility picks up 25 full barrels on each trip, an

In-s [12.5K]

7*24 = 168 quarts of water per day goes into barrel.

25*48 = 1200 quarts of waste water picked up

1200÷168 =7.142857143

8 0

2 years ago

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PLEASE HELP ME WITH BOTH OF THESE PROBLEMS THIS IS DUE SOON PLEASE HELP ME

iogann1982 [59]

Answer:

Step-by-step explanation:

P = 2width + 2length

8x = 2x + 2 length; 8x-2x = 2 length; 6x=2 length

length = 3x

sides= 1/4 of 3x = 3x/4

perimeter of square = 4* side = 4*3x/4 = 3x

x= 4

P rectangle =8x= 8*4= 32

P square = 3x = 3*4 = 12

32-12 = 20 cm greater is the P rectangle than P of square

A rectangle = 2x*x = 2*4*4 = 32

A square =(3x/4)² = (3*4/4)² = 9

32-9 = 23 cm² greater is the A rectangle than A of square

4Comic+1 Nonfiction

4comic = 8y so 1 comic = 2y

1 nonfiction = (3+7y)+ 2y

1 nonfiction = 7y+2y+3 = 9y+3

total= 8y+9y+3 = 17y+3

if y=4

total = 17*4 +3 = $71

7 0

3 years ago

Read 2 more answers