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3 years ago

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Consider rectangle abcd (not shown) with ab=30 and bc=16. if m and n are the midpoints of sides ab¯¯¯¯¯ and bc¯¯¯¯¯ respectively

, find mn.

1 answer:

max2010maxim [7]3 years ago

5 0

M = 30/2 = 15
N = 16/2 =8

Hope this helps.

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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.

A)

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:

$A= 1320 u^{2}$

c)

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

3 years ago

ABCD is a rectangle. The measure of angle BCE is 4x-23 and the angle measure of angle DCE is 5+5x. Find the measure of angle BEC

Temka [501]

<h2>Explanation:</h2><h2></h2>

The diagram for this problem is shown below. Point E is the intersection of the two diagonals and we know that:

$\angle BCE=\beta \\ \\ \angle DCE=\theta$

Every internal angle of any rectangle measures 90 degrees, so:

$\beta +\alpha=90 \\ \\ (4x-23)+(5+5x)=90 \\ \\ \\ Isolating \ x: \\ \\ (4x+5x)+(5-23)=90 \\ \\ 9x-18=90 \\ \\ 9x=108 \\ \\ x=\frac{108}{9}=12$

So:

$\beta=4x-23=4(12)-23=25^{\circ}$

So the measure of angle BEC can be found as follows:

We know that triangle ΔCEB is an isosceles triangle because the diagonals of any rectangle measure the same. So

$\angle EBC=\beta$

The sum of internal angles of any triangle add up to 180 degrees, so:

$\beta + \beta+\angle BEC=180^{\circ} \\ \\ 2\beta+\angle BEC=180^{\circ} \\ \\ \angle BEC=180^{\circ}-2\beta \\ \\ \angle BEC=180^{\circ}-2(25^{\circ}) \\ \\ \angle BEC=180^{\circ}-50^{\circ} \\ \\ \boxed{\angle BEC=130^{\circ}}}$

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4 years ago

dominic rides hus bike at a rate of 4 minutes per mile how many miles will it take him to ride 12 miles

earnstyle [38]

It would be 48 miles

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3 years ago

Read 2 more answers

The line with x-intercept of 10 and y-<br> intercept of -2.

Yuri [45]

Answer:

y = $\frac{1}{5}$ x - 2

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ = x- intercept (10, 0) and (x₂, y₂ ) = y- intercept (0, - 2)

m = $\frac{-2-0}{0-10}$ = $\frac{-2}{-10}$ = $\frac{1}{5}$

The y- intercept c = - 2

y = $\frac{1}{5}$ x - 2 ← equation of line

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3 years ago

Simplify the following expressions

qaws [65]

Answer:

1. -15c - 4

2. 3x - 5

3. 14q

4. -195 + 39m

5. 55b + 130

Step-by-step explanation:

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3 years ago