AC = 2 BD
Distance BD = √[(2.5+1)^2 +(3-3)^2]
Distance BD = √(3.5)^2
Distance BD = 3.5
AC = 2 * 3.5 = 7
Answer
d. 7
(3n+2)/(n-4) - (n-6)/(n+4)
common denominator (n-4)(n+4)
{(n+4)(3n+2)-(n-4)(n-6)}/{(n-4)(n+4)}
Use the foil method:
{(3n²+14n+8)-(n²-10n+24)}/{(n-4)(n+4)}
distribute negative sign:
{(3n²+14n+8-n²+10n-24)}/{(n-4)(n+4)}
subtract:
(2n²+24n-16)/{(n-4)(n+4)}
take out 2:
2{n²+12n-8}/{(n-4)(n+4)}
The biggest 3 digits whole number is 999.
Hence, the biggest sum one can do with two of those is 999+999=1998
Here is a photo of how to solve it
Answer:
- 3, - 11, - 35, - 107
Step-by-step explanation:
Using the recursive formula with f(1) = - 3 , then
f(2) = - 2 + 3f(1) = - 2 + 3(- 3) = - 2 - 9 = - 11
f(3) = - 2 + 3f(2) = - 2 + 3(- 11) = - 2 - 33 = - 35
f(4) = - 2 + 3f(3) = - 2 + 3(- 35) = - 2 - 105 = - 107
Thus the first four terms are
- 3, - 11, - 35, - 107
