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Levart [38]
3 years ago
6

What is the sum of all even numbers between 1 and 1000

Mathematics
1 answer:
weeeeeb [17]3 years ago
4 0
There are 500 even numbers from 1 to 1000 starting with 2 and ending with 998. The sequence of even numbers is an arithmetic sequence with a common difference of 2. Using the equation for the arithmetic series,

                                       S = (a1 + an)(n/2)
and substituting, 
                                     
                                        S = (2 + 998)(500/2)
we get, 

                                        S = 250,000

Therefore, the sum of all even numbers between 1 and 1000 is 250,000.

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Write a division problem that has a 3-digit dividend and a divisor between 10 and 20. Show how to solve it by drawing a quick pi
stich3 [128]

Answer:

In the division problem,

The number which is divided is called dividend,

While, the number which divides, is called divisor,

Let dividend = 150 ( a three digit number ),

Divisor = 15 ( a number between 10 and 12 ),

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150\div 15

In the below figure,

Take a grid that shows 150,

Divide this grid such that each box represents 15,

Thus, there are such 10 boxes,

That is,

150\div 15 = 10

7 0
3 years ago
Read 2 more answers
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Natali5045456 [20]
1/3(12x - 6)

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4 0
3 years ago
Write in simplest form.
Jobisdone [24]
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3 0
3 years ago
Use a matrix to solve the system:
Romashka-Z-Leto [24]

Answer:

(2.83 , 1 , 4)

Step-by-step explanation:

2x+2y-z=4\\4x-2y-2z=2\\3x+3y-4z=-4\\

Rewrite these equations in matrix form

\left[\begin{array}{ccc}2&2&-1\\4&-2&-2\\3&3&-4\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\

we can write it like this,

AX=B\\X=A^{-1}B

so to solve it we need to take the inverse of the 3 x 3 matrix A then multiply it by B.

We get the inverse of matrix A,

A^{-1}=\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right]  \\

now multiply the matrix with B

X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}7/15&1/6&-1/5\\1/3&-1/6&0\\3/5&0&-2/5\end{array}\right]\left[\begin{array}{ccc}4\\2\\-4\end{array}\right] \\\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}2.83\\1\\4\end{array}\right] \\

4 0
3 years ago
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xxTIMURxx [149]

Answer:

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4 0
3 years ago
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