n3 + 2n2 - 15n = 0 : extract n as a factor of all the left hand terms then, n(n2 + 2n - 15) = 0 : the expression n2 + 2n - 15 can be factored as (n + 5)(n - 3) then, n(n + 5)(n - 3) = 0 : This holds true when either n = 0, or n + 5 = 0 or n - 3 = 0 When n + 5 = 0 then n = -5 : When n - 3 = 0 then n = 3. The solutions are n = 0 or n = -5 or n = 3.
n plus 15.n plus 15.n plus 15.n plus 15.
Because 0 is nothing. Therefore, you can add infinity 0s=0 Also, 0*N=0, N-0=0, and 0+N=N, but you can't do N/0, because 0 is nothing. ( N= any number)
its 5035the summarian notation tells you that(sum of all #from 0 to a number'N'(sum of all #from 0 to N) = (n)+(n-1)+(n-2)+(n-3)+...+(2) +(1) or(sum of all #from 0 to N) = (1)+ (2) + (3) + (4)+...+(n-1)+(n)the two different sums are aligned by columns. now add the two colunms accordingly and you'll get2x(sum of all #from 0 to N)=(n+1)+(n+1)+...+(n+1) (n+1)is added n timesso2x(sum of all #from 0 to N) =n(n+1)(sum of all #from 0 to N) =n(n+1)/2so (sum of 5 to 100) = (sum of 0 to 100)- (sum of 0 to 5)=100(101)/2 - 5(6)/2 = 5050 - 15 = 5035
n=number of term (ie 4 has n=1, 8 has n=2) t=value or term number 'n' (n=3 has t=15) Easy way: 4 --> 8 =+4 8 --> 15 = +7 15 --> 25 =+10 25 --> 38 = +13 Note: increases are also increasing by 3 Note: next increase should be +16 38+16=54 4 --> 8 = +4 4 --> 15 =+11 4 --> 25 =+21 4 --> 38 =+34 +4 = 4+0 +11= (4+0)[previous increase]+(4+0+3) [initial increase plus 3] +21= ((4+0)+(4+0+3))[previous increase]+((4+0+3)+3)[initial increase plus 3 again] +34= ((4+0)+(4+0+3))+((4+0+3)+3)[previous increase]+4+0+3+3+3 +4= 4*1+3*0 +11 = 4*2+3*1 +21 = 4*3+3*3 +34 = 4*4+3*6 Note: 0,1,3,6 are the sums of an arithmetic series that increases by 1 and starts at 0 (ie. 0=0, 1=0+1, 3=0+1+2) S=(n/2)(2n(initial)+(n-1)d) where n = term number where n(initial) = n where t=1 where d = increase in series S=(n/2)(2*0+(n-1)*1) S=(n/2)(n-1) Therefore additions: 4n+3*(n/2)(n-1) So, for the first increase (ie 4 --> 8). The increase = 4*1+3*0.5*0=4 1st increase = 4*1+3*0.5*0=4 2nd increase = 4*2+3*1*1=11 3rd increase = 4*3+3*1.5*2= 21 Note if you add 4(the first term of the initial series you get the right answers but shifted one term: ie: with: 4+4n+3*(n/2)(n-1) = 4(n+1)+3*(n/2)(n-1) t1=8 t2=15 t3=25 t4=38 Merely shift the n values to correct (ie. n --> n-1, n-1 --> n-2 etc.): t = 4n+3((n-1)/2)(n-2) t6= 4*6+3*2.5*4=24+30=54
15 N 15 E is the African continent.
15 N 15 E is Kanem, Chad. The continent is Africa.
That would place you in the country of Chad - close to its border with Niger, on the continent of Africa.
n3 + 2n2 - 15n = 0 : extract n as a factor of all the left hand terms then, n(n2 + 2n - 15) = 0 : the expression n2 + 2n - 15 can be factored as (n + 5)(n - 3) then, n(n + 5)(n - 3) = 0 : This holds true when either n = 0, or n + 5 = 0 or n - 3 = 0 When n + 5 = 0 then n = -5 : When n - 3 = 0 then n = 3. The solutions are n = 0 or n = -5 or n = 3.
OKAY..I'm a 6th grader in 2010,and I think I know the correct answer...Here it goes;The continent located at 15 N and 75 E is in Asia. Thank you(:
45 N 15 E is Croatia, and the continent is Europe.
North America, Africa, and Asia are all located at 15˚N.
15° 0′ 0″ N, 75° 0′ 0″ W
15° 25′ 0″ N
15° 0′ 0″ N, 32° 0′ 0″ E
South America
7° 15′ 0″ N, 80° 54′ 0″ E7.25, 80.9