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Judy8x9= 72r
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What is the product of seven and the total of r and 8?
7 × (r + 8) = 7(r + 8) = 7r + 56
Q 5th term of a GP is 2 then product of its 9 terms is?
I'll try to answer the question, "If the 5th term of a geometric progression is 2, then the product of its FIRST 9 terms is --?" Given the first term is A and the ratio is r, then the progression starts out... A, Ar, Ar^2, Ar^3, Ar^4, ... So the 5th term is Ar^4, which equals 2. The series continues... Ar^5, Ar^6, Ar^7, Ar^8, ... Ar^8 is the 9th term. The product P of all 9 terms is therefore: P = A * Ar * Ar^2 *...*Ar^8 Collect all the A's P = (A^9)*(1 * r * r^2 ...* r^8) P = A^9 * r^(0+1+2+...+8) There's a formula for the sum of the first n integers (n/2)(n+1), or if you don't know just add it up. 1+2+...+8 = 36 Therefore P = A^9 * r^36 Since 36 is a multiple of 9, you can simplify: P = (Ar^4)^9 Still with me? Remember that Ar^4=2 (a given fact). So finally P = 2^9 = 512. Cute problem.
How many times can 8 go into 77?
9 R 5
How many times does 15 go into 129?
15 goes into 129 8.6 times or 8 with 9 left over.
How many times does 9 go into 62?
9 goes into 62 only six times with a remainder of 8.
8 minus the product of 9 and r?
8 - 9*r
What is the product of seven and the total of r and 8?
7 × (r + 8) = 7(r + 8) = 7r + 56
how to factor the binomial eighty one r squared minus sixty four?
(9r - 8)(9r + 8)
What is -476 divided by 90?
-566
Q 5th term of a GP is 2 then product of its 9 terms is?
I'll try to answer the question, "If the 5th term of a geometric progression is 2, then the product of its FIRST 9 terms is --?" Given the first term is A and the ratio is r, then the progression starts out... A, Ar, Ar^2, Ar^3, Ar^4, ... So the 5th term is Ar^4, which equals 2. The series continues... Ar^5, Ar^6, Ar^7, Ar^8, ... Ar^8 is the 9th term. The product P of all 9 terms is therefore: P = A * Ar * Ar^2 *...*Ar^8 Collect all the A's P = (A^9)*(1 * r * r^2 ...* r^8) P = A^9 * r^(0+1+2+...+8) There's a formula for the sum of the first n integers (n/2)(n+1), or if you don't know just add it up. 1+2+...+8 = 36 Therefore P = A^9 * r^36 Since 36 is a multiple of 9, you can simplify: P = (Ar^4)^9 Still with me? Remember that Ar^4=2 (a given fact). So finally P = 2^9 = 512. Cute problem.
Find the possible values of r in in the inequality 5 r minus 3?
iF THE QUESTION IS WRITTEN LIKE THIS: WHAT IS THE VALUE IN r IN THE INEQUALITY 5>r=3. THEN THE BEST POSSIBLE ANSWER WOULD BE...D) R<8
How many 9 in 71?
7 R 8
What is 9 divided by 4878?
0.0018
What is 75 and divide 8?
9.375
What are the release dates for Out 'N' About with R-J- Fritz - 1998 9-8?
Out 'N' About with R-J- Fritz - 1998 9-8 was released on: USA: 29 April 2006
Is r equals 9 a solution of 128 equals 16r?
No, it is not. Substitute and check. You have 128 + 16r. So now, if r=9, then the equation would be 128 = 16x9. But, 16x9 is 145 and is not equal to 128. Thus, r=9 is not the solution. The solution is r+ 128/16 which is equal to 8. Thus, the actual solution is r = 8.
How many combinations of 8 numbers are possible from 10 numbers?
In general, the number of combinations of n things taken r at a time isnCr = n!/[(n - r)!r!]Thus, we have:10C8= 10!/[(10 - 8)!8!]= 10!/(2!!8!)= (10 x 9 x 8!)/(8! x 2 x 1)= (10 x 9)/2= 5 x 9= 45
