Can someone help me please
1 answer:
Answer:
6a. r^5
6b. 18(4+√2)
Step-by-step explanation:
6a. The first term of the sequence is √r, and the common ratio is √r. Hence the 10th term will be ...
a1·r^(n-1) . . . for a1=√r and n=10.
√r·(√r)^(10-1) = (√r)^10 = r^5
_____
6b. The sum of n terms is given by ...
Sn = a1·(R^n -1)/(R -1)
For a1=√8 and the common ratio R = √8, the sum of 4 terms is ...
Or, you could add up the 4 terms:
You might be interested in
Answer:
{(1,4) (2,8) (3,12) (4,16)}
Set up the given triangle on x-y coordinates with right angle at (0,0). So the two vertices are at (5,0) and (0,2![sqrt{x} n]{3}](https://tex.z-dn.net/?f=sqrt%7Bx%7D%20n%5D%7B3%7D%20)
)
let (a,0) and (0,b) be two vertices of the<span> equilateral triangle. So the third vertex must be at </span>
for a pt (x,y) on line sx+ty=1, the minimum of
equals to
smallest value happens at
so area is
hence m=75, n=67, p=3
m+n+p = 75+67+3 = 145
let (a,0) and (0,b) be two vertices of the<span> equilateral triangle. So the third vertex must be at </span>
for a pt (x,y) on line sx+ty=1, the minimum of
equals to
smallest value happens at
so area is
hence m=75, n=67, p=3
m+n+p = 75+67+3 = 145