If a is 3 / 4 of B and B is 5 / 6 of C, how many parts of C is a, and how many less is a than C?

If a is 3 / 4 of B and B is 5 / 6 of C, how many parts of C is a, and how many less is a than C?

If a is 3 / 4 of B and B is 5 / 6 of C,
A=B×3/4；
B=C×5/6;
∴A÷3/4=C×5/6;
A/C=5/6×3/4=5/8;
Then a is 5 / 8 of C;
Less = 1-5 / 8 = 3 / 8;
A is 3 / 8 less than C;
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5 out of 6 = a fraction = a fraction

5 out of 6 = 10 out of 12 = 15 out of 18

2 out of 60-5 = a few. 1 out of 60-3 = a few

60-2/5=300/5-2/5=298/5
60-1/3=180/3-1/3=179/3

The solution equation: ① x-0.84x = 0.8 ② 5 × 3.1-2x = 4.5

①X-0.84X=0.8，  （1-0.84）X=0.8，        0.16X=0.8，  0.16X÷0.16=0.8÷0.16，            X=5；②5×3.1-2X=4.5，    15.5-2X=4.5， 15.5-2X+2X=4.5+2X，     4.5+2X=15.5，  4.5+2X-4.5=15.5-4.5，         2X=11，      2X÷2=11÷2，          X=5.5．

X-0.84 = 0.8 how to solve the equation

x-0.84=0.8
x=0.8+0.84=1.64

8 + X + X multiplied by 3 = 84 (solving equation)

8 + X + x times 3 = 84
X+3X=84-8
4X=76
X=76÷4
X=19

81 64 49 36 （ ） （ ）

25 16

36+49+64+81.+400

Second order arithmetic sequence means that the difference of each item is arithmetic sequence. Its general formula is an = a1 + (a2-a1) (n-1) + (a3-2a2 + A1) (n-1) (n-2) / 2. Tolerance: D = a1-2a2 + a3
Sum formula Sn = Na1 + (a2-a1) (n-1) n / 2 + [n (n + 1) (2n + 1) / 6 + n (1-3n) / 2] (a3-2a2 + A1) / 2]
Substituting 400 into the general formula, n = 15, so there are 15 terms
Substituting into the summation formula: S15 = 2315

Calculation: 36 + 49 + 64 + 81 + +400?
Relatives-

1^2+2^2+3^2+...n^2=n(n+1)(2n+1)/636+49+64+81+… +400=6²+7²+… 20²=1²+2²+...+5²+6²+… 20²-（1²+2²+...+5²）=（20×21×41）÷6-（5×6×11）÷6=2815...

1.4.9.16. (). (). 49.64. Fill in the number according to the rule

1.4.9.16.( 25).(36 ).49.64
It's the square of one to eight
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