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Irina18 [472]
3 years ago
12

6 less than the number y

Mathematics
1 answer:
Nonamiya [84]3 years ago
3 0

Answer:

6 less a number y can be represented by the expression 6 - y

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I think it would be each girl scored 2 points, and each boy only scored one. :)

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What is the formula for the following arithmetic sequence?<br><br> 12, 6, 0, -6, ...
Lady_Fox [76]

Answer:

12+35x3.14

Step-by-step explanation:

7 0
3 years ago
Jack has two more than half as many cups as
Rufina [12.5K]

Step-by-step explanation:

(x \div 2 + 2) +  \frac{(x \div 2 + 2)}{2}  + x

its hard to determine without the known value of x, but the above would be the equation needed to solve it if you knew what x was

4 0
3 years ago
A photocopy machine can print 735 copies in 7 minutes. What is the rate at which the machine prints the copies per minutes
qaws [65]

Answer:

105

Step-by-step explanation:

Divide 735 by 7

5 0
3 years ago
Read 2 more answers
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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