The sum of all the even integers between 99 and 301 is 20200
To find the sum of even integers between 99 and 301, we will use the arithmetic progressions(AP). The even numbers can be considered as an AP with common difference 2.
In this case, the first even integer will be 100 and the last even integer will be 300.
nth term of the AP = first term + (n-1) x common difference
⇒ 300 = 100 + (n-1) x 2
Therefore, n = (200 + 2 )/2 = 101
That is, there are 101 even integers between 99 and 301.
Sum of the 'n' terms in an AP = n/2 ( first term + last term)
= 101/2 (300+100)
= 20200
Thus sum of all the even integers between 99 and 301 = 20200
Learn more about arithmetic progressions at brainly.com/question/24592110
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Answer:
The answer is 6
Step-by-step explanation:
The graph moved 6 spots to the left therefore your answer would be positive.
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Answer:
1a) Length = 7x + 3 & Width = 4x - 2
1b) Area = 
1c) Area = 2774 sq. m
2. 
Step-by-step explanation:
1a)
The length given as words is "3 more than 7 times x"
The width given as words is "4 times x minus 2"
The expression for length would be 7x + 3
The expression for width would be 4x - 2
1b)
The area is length * width
Since we already know the algebraic expressions for length and width from part (a) above, we use the formula:
Area = (7x+3)(4x-2) = 28x^2 -14x + 12x - 6 = 28x^2 -2x -6
Area = 
1c)
Given x = 10, we put this into the area expression we found in (b) above.Let's see:

Area = 2774 sq. m
2.
We can group the first two terms and next two terms and write up:

That's the factored form.