Let the numbers be m & m+1 (consecutive) Hence their squares are m^2 & ( m + 1)^2 => m^2 & m^2 + 2m + 2 Hence Their sum is m^2 + m^2 + 2m + 1 = 85 2m^2 + 2m - 84 = 0 m^2 + m - 42 =0 Factor (m + 7)(m - 6) = 0 Hence the numbers are 6 & 7 .
The equation for n layers is S(n) = n(n+1)(2n+1)/6It is simplest to prove it by induction.When n = 1,S(1) = 1*(1+1)(2*1+1)/6 = 1*2*3/6 = 1.Thus the formula is true for n = 1.Suppose it is true for n = m. That is, for a pyramid of m levels,S(m) = m*(m+1)*(2m+1)/6Then the (m+1)th level has (m+1)*(m+1) oranges and soS(m+1) = S(m) + (m+1)*(m+1)= m*(m+1)*(2m+1)/6 + (m+1)*(m+1)= (m+1)/6*[m*(2m+1) + 6(m+1)]= (m+1)/6*[2m^2 + m + 6m + 6]= (m+1)/6*[2m^2 + 7m + 6]= (m+1)/6*(m+2)*(2m+3)= (m+1)*(m+2)*(2m+3)/6= [(m+1)]*[(m+1)+1)]*[2*(m+1)+1]/6Thus, if the formula is true for n = m, then it is true for n = m+1.Therefore, since it is true for n =1 it is true for all positive integers.
(m-2)/(m+2) * m/(m-1) = [(m-2)*m]/[(m+2)*(m-1)] = (m2 - 2m)/m2 + m - 2)
m^2 6n - 9n2 - 2m = (m - 3n)(m + 3n - 2)
Let the integers be 'm' & 'm+`1'. Hence m + m+ 1 = 57 2m + 1 = 57 2m = 56 m = 28 M+ 1 = 28 + 1 = 29
m-2+1-2m+1 When simplified: -m
m=-0.2
1/m + 1/2m = 6 LCD is 2m (2 + 1) / 2m = 6 3/2m = 6 3/6 = 2m = 1/2 m = 1/4
1 m^(2) + 1 m^(2) = 2m ^(2)
m=1/2 -2m=16m-9 -18m=-9 m=1/2
Let the numbers be m & m+1 (consecutive) Hence their squares are m^2 & ( m + 1)^2 => m^2 & m^2 + 2m + 2 Hence Their sum is m^2 + m^2 + 2m + 1 = 85 2m^2 + 2m - 84 = 0 m^2 + m - 42 =0 Factor (m + 7)(m - 6) = 0 Hence the numbers are 6 & 7 .
-2m + 7 = m - 2 +7 +2 = m +2m 9 = 3m 3m = 9 m = 9/3 = 3 to test: -2(3) + 7 = (3) - 2 -6 + 7 = 3 - 2 1 = 1
0 = 2m^2 - 2m + 36 0 = 2(m^2 - m + 18) 0 = m^2 - m + 18
The equation for n layers is S(n) = n(n+1)(2n+1)/6It is simplest to prove it by induction.When n = 1,S(1) = 1*(1+1)(2*1+1)/6 = 1*2*3/6 = 1.Thus the formula is true for n = 1.Suppose it is true for n = m. That is, for a pyramid of m levels,S(m) = m*(m+1)*(2m+1)/6Then the (m+1)th level has (m+1)*(m+1) oranges and soS(m+1) = S(m) + (m+1)*(m+1)= m*(m+1)*(2m+1)/6 + (m+1)*(m+1)= (m+1)/6*[m*(2m+1) + 6(m+1)]= (m+1)/6*[2m^2 + m + 6m + 6]= (m+1)/6*[2m^2 + 7m + 6]= (m+1)/6*(m+2)*(2m+3)= (m+1)*(m+2)*(2m+3)/6= [(m+1)]*[(m+1)+1)]*[2*(m+1)+1]/6Thus, if the formula is true for n = m, then it is true for n = m+1.Therefore, since it is true for n =1 it is true for all positive integers.
Kinetic energy = 1/2 (mv)2 if m increase by 2m, then Kinetic increases by 2
-1/2m = -9 To solve this: Multiply by 2m: -1 = -9*2m Divide by -9: -1/-9 = 2m Divide by 2: (-1/-9)/2 = m Work it out: -1/-9 = 0.111... (the 1 is recurring) 0.111... / 2 = 0.0555... (the five is recurring) So x = 0.055... or 5/90
There is NO single value (THE vallue) for 2m when m takes three different values.