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OverLord2011 [107]
3 years ago
12

Will give brainliest to correct answer.

Mathematics
1 answer:
ryzh [129]3 years ago
4 0
We can solve this problem by seeing at which part both of the parts of the graphs of the function are discontinued. Both of the parts of the graph of the function are discontinued at -2, so we will have to find a function that has a value that is undefined for x = -2. We can do this using the denominator of the fraction that's in each of the functions. The function where x = -2 will cause it to be undefined is the third one, so the answer to this question is C., f (x) = \frac{1}{x+2} +5. 
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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
4 0
3 years ago
_ 1)
Alina [70]

Answer: A)

Step-by-step explanation: If You just look at the numbers you can single out which one is correct

3 0
3 years ago
1. Type mmnn using exponents.
kompoz [17]

Step-by-step explanation:

\bold{1.}\\\\\ mmnn=m^2n^2=(mn)^2\\\\\bold{2.}\\\\3x=5+7\\3x=12\qquad\text{divide both sides by 3}\\x=4\\\\\bold{3.}\\\\\dfrac{x}{3}=7\qquad\text{multiply both sides by 3}\\x=21\\\\\bold{4.}\\\\0.5x=12\qquad\text{multiply both sides by 2}\\1x=24\\x=24\\\\\bold{5.}\\\\4x-6=2x+4\qquad\text{add 6 to both sides}\\4x=2x+10\qquad\text{subtract 2x from both sides}\\2x=10\qquad\text{divide both sides by 2}\\x=5

\bold{6.}\\\\3x^2\ for\ x=6\ /\text{put x = 6 to the expression}/\\\\3(6)^2=3(36)=108\\\\\bold{7.}\\\\3y^2+2y\ for\ y=3\ /\text{put y = 3 to the expression}/\\\\3(3)^2+2(3)=3(9)+6=27+6=33

7 0
3 years ago
Read 2 more answers
Write the equation in the standard linear equation form (Ax+By=C) if possible. Is the equation linear? y/3+x=5
mojhsa [17]

Answer:

3x + y = 15

Step-by-step explanation:

y/3 + x = 5

You can first move the x and y variable around so it fits the standard linear equation format.

x + y/3 = 5

In order to make it look more like the standard format, you can multiply both sides of the equation by 3 to get rid of the 3 in the denominator of the y variable.  It would look like this:

3 ( x +y/3) = 3 ( 5 )

When you multiply this out, you get this:

3x + y = 15

8 0
3 years ago
A piece of fabric is yard long. A piece of ribbon is yard long. How many
atroni [7]

Answer:

the piece of fabric and ribbon are the same length. in that case, its 0

Step-by-step explanation:

3 0
3 years ago
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