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3 years ago

(19+8)x11+4 Order of operations

Mathematics

2 answers:

Allushta [10]3 years ago

6 0

Answer:

$301$

Step-by-step explanation:

<h2><u>Order of Operations is another term for PEMDAS:</u></h2><h2><u /></h2><h2><u>Parentheses </u></h2><h2><u>Exponents</u></h2><h2><u>Multiplication and Division </u></h2><h2><u>Addition and Subtraction</u></h2>

<u />

<u>So we first do the problem in the parenthesis:</u>

$(19+ 8) ==> 27$

<u></u>

<u>Next do the multiplication:</u>

$27 * 11 ==> 297$

<u>Then lastly, do the addition:</u>

$297 + 4 ==> 301$

likoan [24]3 years ago

4 0

Answer:

19+8=27

27*11=297

297+4=301

Step-by-step explanation:

BODMAS

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2 years ago

How do you work this out?

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Because ABCD is a rectangle, the length of CD is 12 cm.

We need to determine the length of DE. If we can do that, then the sum of the lengths of CD and DE represents the unknown: the length of CE.

To find the length of CE, we have to "solve" the upper triangle.

Here's an outline of what to do:

1. Show that BC=AD and find the length.
2. Note that angle CAD is 60 degrees. Why?
3. Note that angle EAD is 30 degrees. Why?
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5. Add ED and DC, that is, ED + 12 cm. This is your answer.

Please ask questions if need be.

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3 years ago

An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

USPshnik [31]

The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$

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