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:
(4,1)
Step-by-step explanation:
when points are reflected on the y-axis you switch your x coordinate from positive to negative or the other way around the y coordinate stays the same.
(2,1) reflected is (-2,1)
(7,4) reflected is (-7,4)
(-8,-4) reflected is (8,-4)
so (-4,1) reflected is (4,1)
Answer:
Choice A
Step-by-step explanation:
That's because IJ and UV are both in between two angles which are congruent to each other
Harold: y = 3x + 18
Bill: y = 2x+24
Hope this helps with your mid-term too guys !
if a whole number, put the number above a one like so : if the number was 8, write it as 8/1
multiply the number by the reciprocal of the fraction
example: 8 divided by 1/3
8/1 times 3/1 =24/1
=24
