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VLD [36.1K]
3 years ago
10

The width of a rectangle is 10 inches less than its length. If the perimeter of the rectangle is 36 inches. Write and solve an e

quation to find its width in inches.
Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
8 0

Answer:

Step-by-step explanation:

Let us suppose that length of Triangle is equal to x.

that is

Length = x

Than

width = x - 10

perimeter of Rectangle is

2(length) + 2(width) = 36

2x + 2(x-10) = 36

2x +2x - 20 = 36

4x -20 = 36

4x = 36 +20

4x = 56

x = 56/4

x = 14

so Length of Rectangle is 14 inches

and

width = x - 10

width = 14 - 10

width = 4

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2 years ago
4c=-40 i don't know this and i need help<br> ill do what i can in return
melamori03 [73]

Answer:

4c=-40

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2 years ago
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Use the substitution method to solve the system of equations. Choose the correct ordered pair.
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Here you have 2 linear equations and are to solve this system.  Both equations have already been solved for y, so you can set one of them = to the other one:

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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\}.

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

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Answer:

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2 years ago
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