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ludmilkaskok [199]
3 years ago
12

Work out an estimate for the value of 41.2 x 19.8 ——————- 0.49

Mathematics
1 answer:
frez [133]3 years ago
3 0

Answer:

820

Step-by-step explanation:

41.2 * 19.8 = 41 *20 = 820

41.2

In this tenth value is 2 which is less than 5. So ignore it and write the whole number alone. 41

19.8

In this tenth value is 8 awhich greater than 5. So add 1 to whole part and write.

19+1 = 20

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50% is 30, it’s half of 60. 25% is the same as dividing by 4, 60/4 is 15.
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A table measures x+4 inches long and 5x+12 inches wide. What is the perimeter of the table?
tangare [24]

Answer:

P = 12x +32

Step-by-step explanation:

The perimeter is found by using the formula

P =2(l+w)

We know the length and the width

P = 2(x+4 + 5x+12)

Combine like terms

P = 2(6x+16)

Distribute

P = 12x +32

5 0
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Need help!!!!!!!!!!!!!!!!!!!!!!!!!!
Pani-rosa [81]

9514 1404 393

Answer:

  a) GH or HG

  b) KL

  c) YZ or ZY

  d) EF

Step-by-step explanation:

A segment or line can be named by naming two points on it in any order. The points naming a line segment are its end points.

A ray is named by first naming its end point, then any other point on the ray.

a) GH or HG

b) KL

c) YZ or ZY

d) EF

4 0
3 years ago
PLEASE HELP ILL GIVE 10 POINTS X2
Molodets [167]
I’m still thinking about the equation part give me a second

3 0
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
4 0
3 years ago
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