For the fourth term r = 3
Fourth term = 6C3(x)^(6 - 3) (4y)^3 = 20x^3 (64y^3) = 1,280x^3y^3
Answer:
9.6558916e+17
Scientific Notation: 3.45 x 10^5
E Notation: 3.45e5
= 9.26 × 1010
(scientific notation)
= 9.26e10
(scientific e notation)
= 92.6 × 109
(engineering notation)
(billion; prefix giga- (G))
its probably: 9.26 × 1010 as the question asks for scientific notation
Step-by-step explanation:
somewhere between those lines, good luck!
Answer:
![\large \boxed{\sf \ \ 2 \ \ }](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Csf%20%5C%20%5C%202%20%20%5C%20%5C%20%7D)
Step-by-step explanation:
Hello,
<u>The mean of five numbers is 8</u> so we can write
![\dfrac{x_1+x_2+x_3+x_4+x_5}{5}=8](https://tex.z-dn.net/?f=%5Cdfrac%7Bx_1%2Bx_2%2Bx_3%2Bx_4%2Bx_5%7D%7B5%7D%3D8)
<u>When another number is added the mean is 7</u>, let s note x the another number we can write
![\dfrac{x_1+x_2+x_3+x_4+x_5+x}{6}=7](https://tex.z-dn.net/?f=%5Cdfrac%7Bx_1%2Bx_2%2Bx_3%2Bx_4%2Bx_5%2Bx%7D%7B6%7D%3D7)
From the first equation we can say
![x_1+x_2+x_3+x_4+x_5=8*5=40](https://tex.z-dn.net/?f=x_1%2Bx_2%2Bx_3%2Bx_4%2Bx_5%3D8%2A5%3D40)
So the second equation becomes
![\dfrac{x_1+x_2+x_3+x_4+x_5+x}{6}=7\\\\ \dfrac{40+x}{6}=7\\\\40 + x = 6*7=42\\\\ x = 42-40 = 2\\](https://tex.z-dn.net/?f=%5Cdfrac%7Bx_1%2Bx_2%2Bx_3%2Bx_4%2Bx_5%2Bx%7D%7B6%7D%3D7%5C%5C%5C%5C%3C%3D%3E%20%5Cdfrac%7B40%2Bx%7D%7B6%7D%3D7%5C%5C%5C%5C%3C%3D%3E40%20%2B%20x%20%3D%206%2A7%3D42%5C%5C%5C%5C%3C%3D%3E%20x%20%3D%2042-40%20%3D%202%5C%5C)
The solution is then 2
Hope this helps
Hi!
The combination of 4 digit numbers using those numbers are: 3, 276; 3, 762; 3, 267; 3, 726; 3, 672 etc. This is pretty easy, now you try.
Hope I helped! Have a nice day!
Answer:
The first one luv
Step-by-step explanation: