I honestly don’t know but I really need this so I can get my answer as well , good luck though .
Let n = the smaller of the two numbers, and since the other number is 5 more than twice the smaller number n, then ...
Let 2n + 5 = the second and larger number.
Since the sum of the two unknown numbers is 26, then we can write the following equation to be solved for n as follows:
n + (2n + 5) = 26
n + 2n + 5 = 26
Collecting like-terms on the left, we get:
3n + 5 = 26
3n + 5 - 5 = 26 - 5
3n + 0 = 21
3n = 21
(3n)/3 = 21/3
(3/3)n = 21/3
(1)n = 7
n = 7
Therefore, ...
2n + 5 = 2(7) + 5
= 14 + 5
= 19
CHECK:
n + (2n + 5) = 26
7 + (19) = 26
7 + 19 = 26
26 = 26
Therefore, the two desired numbers whose sum is 26 are indeed 7 and 19.
Hope this helps :)
Ok so,,,, first you find the area of a trapezoid which is b1 plus b2 times h over 2 so it is 9 plus 7 times 6 divided by two then you get 48 and you subtract by 20 times 24 bc its the area of rectangle so 480 minus 48 which is 432 cm squared
Y=2/(x+1)
x=2/(y+1)
y-1=2/x
y=(2/x)+1
Answer is -48x^11+40x^10+40x^9
