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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You could solve this using a proportion or just by common sense.
Proportion: 8/12 = x/126
Cross multiply: 12x = 1008
1008/12 = 84
Or the common sense route is just dividing 12 by 8 then multiplying the answer by 126. Either way, you get the same answer. Hope this helps! :)
~Ash
Answer:
f(y-2)+f(4-y)=4
Step-by-step explanation:
Assume (let) x=y-2
So: y=x+2
f(y-2)+f(4-y)=f(x)+f(-x+2)=f(x)+f(2-x)
The value of that expression is 4 from the given.
Answer:
Step-by-step explanation:
h =2
area of triangle = 16x9x0.5= 72 cm^2
area of parallelogram = 72 x 5 = 360 cm^2
area of parallelogram = base x height(h)
360 = 30h
divide both sides by 30
h = 12
Translate into math language:
7 - 3x = 23→ -3x = 23-7 → - 3x = 16 and x = - 16/3
