Answer:
X^2 -5x -36
Step-by-step explanation:
formula of an area of a rectangle: ab
a-length
b-width
(x+4)*(x-9) = x^2 -5x -36
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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(7 + i) - (6 - 2i) = 7 + i - 6 + 2i = 1 + 3i
Answer:
Mr.Robinson; 27/40 i think.
Step-by-step explanation: To begin solving you have to make both of the denominators the same by multiplying 2/5 by 4 and 1/4 by 5 giving you 8/20 and 5/20. From there i'm pretty sure you just add the two fractions to get 13/40 leaving 27/40 to be washed.
