1981+25
Since she was born in the year 1981, and her son was born 25 years later so you add 25
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
#SPJ4
I'm going to assume you mean:
13 * (3/4) + x = 7 * (1/4)
Let's go ahead and simplify the right side.
7 * (1/4) = 1.75
So now we have:
13 * (3/4) + x = 1.75
Now let's take care of the right side. Multiplication before addition.
13 * (3/4) = 9.75
9.75 + x = 1.75
To isolate x, we subtract 9.75 from both sides.
9.75 - 9.75 + x = 1.75 - 9.75
x = -8
Hopefully I spaced this well. \('-')/
Answer:
282.74
Step-by-step explanation:
Answer is square with the side of 212 inches. It's area is 212×212 which is bigger than rectangular area.
