3 - n=13;s=0;p=0;c=1;printc," - ",n;cycle: s=(n%10);p=p+s^3;n=int(n/10);if(n!=0, goto cycle,0);c++;printc," - ",p;n=p;p=0;if(c<=100, goto cycle, 0);printc," - "
OUTPUT =:
1 - 13 2 - 28 3 - 520 4 - 133 5 - 55 6 - 250 7 - 133 8 - 55 9 - 250 10 - 133.....................So, the 100th term = 133
I just found a way easier way, plug in 9 for n and then that would be the sum of the first nine terms of the sequence, so then you subtract that from when n=10, and then you get the tenth term: 10(10+1)(10+2)-9(9+1)(9+2)=330
ah ok thank you!
No, not in this case, because both sqrt are the same: 2 + i and 2 + i.
Roots of complex numbers are multi valued.
A complex number has two square roots, three cube roots, four fourth roots and so on.
You will already have come across this with the square roots of positive numbers. The square root of 4 for example is plus or minus 2.
For your example, the second root is -2 - i .
Let the distance from the restaurant to his final destination =D The distance from his house to the restaurant =290 - D
[290 - D] / 40 =1/2 + D/50, solve for D D = 150 miles
290 - 150 = 140 miles - distance from his house to the restaurant. 140 / 40 =3.5 hours - it took him to drive to the restaurant. 10:00 am + 3.5 =1:30 pm - when he stopped for lunch till he finished at 2:00 pm. 150 /50 =3 hours that he drove from 2:00pm till 5:00 pm, when he reached his final destination.
yeah, but i was thinking what if there are more solutions that i don't know about?
Ah, that's 90 POINTS on a circle, a line segment is two points touching each other, which means it's 90/2 = 45
Who says 2 + i is wrong? That is the sqrt of 3 + 4i, because it factors into: (2 +i)^2 and sqrt(2 +i)^2 =2 + i
1. the answer that i agree with is here:
https://web2.0calc.com/questions/help-me-please_35236
2. \(\binom{10}{2} = 45\) line segments
1. Jeff the fly can reach 10^2 + 9^2 = 181 points.
2. There are 10 ways to choose the first point, then 9 ways to choose the second points, so there are 10*9 = 90 line segments.
Sorry everyone! I worked both of these out and got them correct, sorry for taking your time. No need to help me, have a good day!
Sorry, just figured out number 2, so no need to work it out! Thanks.
Melody, is there a reason you used natural log? I used log10 and got the same answer.
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∑[1 / (sqrt(n) + sqrt(n+1)), n, 1, 99] = 9
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 balls
a - There are 5 EVEN numbers. The probability =5/11 x 5/11 x 5/11=125 / 1331
b- Probability =5/11 x 4/10 x 3/9 =60 / 990 =2 / 33
∑[{4n^3 - n^2 - n + 1}{n^6 - n^5 + n^4 - n^3 + n^2 - n}, n, 2, infinity] =converges to 1
Let the total numbers of hamsters =H The number of white hamsters =5/6H H - 5/6H =1/6H - the remaining hamsters 1/6H x 8/9 =4/27H are black hamsters H - [5/6H + 4/27H] =1/54H - yellow hamsters H =54 hamsters 5/6 x 54 = 45 are white hamsters [54 - 45] * 8/9 =8 are black hamsters 9 - 8 = 1 is a yellow hamster.
Yellow hamsters =1 / 54
Could also approach this as follows:
A 40% dye solution is to mixed with a 53% dye solution to get 260L of a 50% solution. How many liters of the 40% and 50% solutions will be needed.
Find the number of positive integer solutions to x + 2y + 3z = 100.
Hello Guest!
x: 2 4 6 8 10
y: 40 39 38 37 36
z: 6 6 6 6 6
!
QR/2 = cos(45°) * 6sqrt(2) = 6
Height of triangle h = sin(45°) * 6sqrt(2) = 6
The area of triangle PQR = 6² = 36 u²
The sum of the first \(n\) terms of a certain sequence is \(n(n + 1)(n + 2)\). Find the tenth term of the sequence.
\(\begin{array}{|c|l|l|l|l|} \hline n & s_n=n(n+1)(n+2) & \\ \hline 1 & s_1=1*2*3=6 & s_1 = a_1 & 6= a_1 & a_1 = 6 \\ \hline 2 & s_2=2*3*4=24 & s_2 = s_1+a_2 & 24=6+a_2 & a_2 = 18 \\ \hline 3 & s_3=3*4*5=60 & s_3 = s_2+a_3 & 60=24+a_3 & a_3 = 36 \\ \hline 4 & s_4=4*5*6=120 & s_4 = s_3+a_4 & 120=60+a_4 & a_4 = 60 \\ \hline 5 & s_5=5*6*7=210 & s_5 = s_4+a_5 & 210=120+a_5 & a_5 = 90 \\ \hline 6 & s_6=6*7*8=336 & s_6 = s_5+a_6 & 336=210+a_6 & a_6 = 126 \\ \hline 7 & s_7=7*8*9=504 & s_7 = s_6+a_7 & 504=336+a_7 & a_7 = 168 \\ \hline 8 & s_8=8*9*10=720 & s_8 = s_7+a_8 & 720=504+a_8 & a_8 = 216 \\ \hline 9 & s_9=9*10*11=990 & s_9 = s_8+a_9 & 990=720+a_9 & a_9 = 270 \\ \hline 10 & s_{10}=10*11*12=1320 & s_{10} = s_9+a_{10} & 1320=990+a_{10} & \mathbf{a_{10} = 330} \\ \hline \end{array}\)
