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The geometric mean of two numbers, ( a ) and ( b ), is calculated using the formula ( \sqrt{a \times b} ). For 9 and 30, this is ( \sqrt{9 \times 30} = \sqrt{270} ). Simplifying ( \sqrt{270} ), we get ( 3\sqrt{30} ), which is approximately 16.43. Thus, the geometric mean of 9 and 30 is ( 3\sqrt{30} ) or approximately 16.43.
To determine the relationship between a, b, and c given the equations ( a + b = 2 ), ( b + c = 5 ), and ( c + a = 9 ), we can solve these equations simultaneously. From the first equation, we can express ( a ) as ( a = 2 - b ). Substituting ( a ) into the third equation gives ( c + (2 - b) = 9 ), which simplifies to ( c - b = 7 ). Combining this with the second equation ( b + c = 5 ) allows us to solve for ( b ) and subsequently find ( a ) and ( c ). Ultimately, we find that ( a = -1 ), ( b = 3 ), and ( c = 2 ).
9
A geometric sequence is : a•r^n while a quadratic sequence is a• n^2 + b•n + c So the answer is no, unless we are talking about an infinite sequence of zeros which strictly speaking is both a geometric and a quadratic sequence.
The geometric mean of 9 and 81 is 27.0
1/9 + 1/b = c 1/b = c - 1/9 1 = cb - b/9 9 = 9cb - b 9 + b = 9cb (9 + b)/9c = b *parenthesis here means its all over 9c.
The geometric mean of two numbers, ( a ) and ( b ), is calculated using the formula ( \sqrt{a \times b} ). For 9 and 30, this is ( \sqrt{9 \times 30} = \sqrt{270} ). Simplifying ( \sqrt{270} ), we get ( 3\sqrt{30} ), which is approximately 16.43. Thus, the geometric mean of 9 and 30 is ( 3\sqrt{30} ) or approximately 16.43.
To determine the relationship between a, b, and c given the equations ( a + b = 2 ), ( b + c = 5 ), and ( c + a = 9 ), we can solve these equations simultaneously. From the first equation, we can express ( a ) as ( a = 2 - b ). Substituting ( a ) into the third equation gives ( c + (2 - b) = 9 ), which simplifies to ( c - b = 7 ). Combining this with the second equation ( b + c = 5 ) allows us to solve for ( b ) and subsequently find ( a ) and ( c ). Ultimately, we find that ( a = -1 ), ( b = 3 ), and ( c = 2 ).
well, i think if you use this you can find out. A = 1-9 ,B = 0-9 , C = 0-9 , D = 0-9 , E = 0-9 for 2digit numbers = A A for 3 digit numbers = A B A for 4 digit numbers = A B B A and so on till you get to for 8 digit numbers = A B C D D C B A for 9 digit numbers = A B C D E D C B A and last for 10 digit number = A B C D E E D C B A this should work...
9
This is the associative law for addition: If a, b, c are any numbers then: a + (b + c) = (a + b) + c
You haven't provided any choices for the "which of the following" part of your question. Such questions are best avoided here. However, assuming a, b and c are all natural numbers, all of the following are true for a<b AND b+c=10: a=1, b=2, c=8 a=1, b=3, c=7 a=1, b=4, c=6 a=1, b=5, c=5 a=1, b=6, c=4 a=1, b=7, c=3 a=1, b=8, c=2 a=1, b=9, c=1 a=2, b=3, c=7 a=2, b=4, c=6 a=2, b=5, c=5 a=2, b=6, c=4 a=2, b=7, c=3 a=2, b=8, c=2 a=2, b=9, c=1 a=3, b=4, c=6 a=3, b=5, c=5 a=3, b=6, c=4 a=3, b=7, c=3 a=3, b=8, c=2 a=3, b=9, c=1 a=4, b=5, c=5 a=4, b=6, c=4 a=4, b=7, c=3 a=4, b=8, c=2 a=4, b=9, c=1 a=5, b=6, c=4 a=5, b=7, c=3 a=5, b=8, c=2 a=5, b=9, c=1 a=6, b=7, c=3 a=6, b=8, c=2 a=6, b=9, c=1 a=7, b=8, c=2 a=7, b=9, c=1 a=8, b=9, c=1
A geometric sequence is : a•r^n while a quadratic sequence is a• n^2 + b•n + c So the answer is no, unless we are talking about an infinite sequence of zeros which strictly speaking is both a geometric and a quadratic sequence.
The geometric mean of 9 and 81 is 27.0
The geometric mean of 9 and 25 is: 15.0
5-2x3=9
Yes. By the associative property, a + b + c = a + c + b.