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Answers by Arthur D.

1+2+3+4+...+97+98+99+100 S(n)=(100/2)(2*1+[100-1]*1) S(n)=50(2+99) S(n)=50*101 S(n)=5050   3,6,9,12,...,96,99 there are 33 of these numbers that are divisible by 3 S(n)=(33/2)(2*3+[33-1]*3) S(n)=(33/2)(6+96...

Proof by mathematical induction: an odd number can by written as  2n-1 where n=1, 2, 3, ... prove that 1+3+5+7+...+2n-1=n2 show that this is true for k=1, assume this is true for some odd integer 2k-1, and finally show this is true for 2(k+1)-1 if...

Here's part a) using different reasoning. Row 5 starts with the 11th odd number. The previous rows have 4, 3, 2, and 1 odd numbers for a total of 10 odd numbers. The 99th row will begin with which odd number ? Add the number of odd numbers from each row. 1+2+3+4+...

V=(4/3)(∏)(r3) 62,000=(4/3)(∏)(r3) multiply both sides by 3/4 46,500=∏(r3) (46,500/∏)=r3 take the cube root of both sides (46,500/∏)^(1/3)=r factor 46,500 46,500=5*3*31*2*2*5*5=2*2*3*31*53 the cube root of 53 is 5 5[(2*2*3*31)/∏]^(1/3)=r 5[(372/∏)^(1/3)]=r...

3-5+7-9+11-13+15-17+19-21... find the sum to 80 terms break this up into two parts 3+7+11+15+19....  (there will be 40 numbers ) -5-9-13-17-21...      (there will be 40 numbers ) add the first numbers in each sequence...

(4/7)^0=1 (anything raised to the 0 power is equal to 1 except 0^0 which does not exist in mathematics)   (3/5)^3=(3/5)(3/5)(3/5)=27/125   2^(-5)=1/25=1/32

26=64 possible combinations of true-false answers the probability is 1/64 suppose you had 3 questions (R=right and W=wrong) RRR RRW WRR RWR WWW WWR RWW WRW the probability of getting all 3 right is 1/8 (23=8 and only 1 is RRR)

w=width w+10=length A=lw 119=(w+10)(w) 119=w^2+10w w^2+10w-119=0 119=7*17 you can really stop here because you have the answer: w=7 feet and l=17 feet however... (w+17)(w-7)=0 w-7=0 w=7 and w+10=17 basically what you are doing...