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vagabundo [1.1K]
3 years ago
13

There are 330 walnuts in 6 bags. If each bag has the same number of walnuts, how many walnuts are in 2 bags?

Mathematics
2 answers:
Lerok [7]3 years ago
7 0

Answer:

110 walnuts

Step-by-step explanation:

6 bags -> 330 walnuts

Divide 6:

1 bag -> 330\div 6=55

Multiply 2:

2 bags -> 55\times 2 =110

user100 [1]3 years ago
7 0

Answer:

in six bags there are 330 walnuts

in 1 bag there are 330/6 walnuts

in 2 bags there are 330/6*2 =110 walnuts

Step-by-step explanation:

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kiruha [24]

Answer:it is 14 hun.

Step-by-step explanation:

4 0
3 years ago
What two numbers multipled get -48 and when added get 8?
Katarina [22]
X * y = -48
x + y = 8

y = -48/x

x -48/x = 8
x^2 - 48 = 8x
x^2 - 8x -48 = 0
x = 12
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5 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
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3 years ago
In which of these numbers does the digit 4 represent 4/100
Tpy6a [65]
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The first place after the decimal represents the tenths place, the second place represents the 100ths place and so on.
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Which expression calculates the speed in meters per second of an object that travels a distance of 100 m every 20 s?
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A. 100/20

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