The quotient of 3 times of a number divided by 8 is 5, and the remainder is 2

The quotient of 3 times of a number divided by 8 is 5, and the remainder is 2

8*5+2=3X
40+2=3X
42=3X
X=14

2x-5=3x+10

2x-5=3x+10
2x-3x=5+10
-x=15
x=-15

66×11+22×67=（）
Simple calculation

A:
66x11+22x67 66x11+22x67
=66x22/2+66x22+22 =66x11+66x11+66x11+11+11
=66x22x1.5+22 =11(66+66+66+1+1)
=22(66x1.5+1) =11x200
=22x100 =2200
=2200

If n is any integer, the value of (n + 11) 2-n2 can always be divided by K, then K is equal to ()
A. Multiple of 11b. 22c. 11 or 12D. 11

∵ (n + 11) 2-n2, = (n + 11 + n) (n + 11-n), = 11 (2n + 11), ∵ (n + 11) 2-n2 can always be divided by 11

How to solve the equation 55-8x = 3x

-3X-8X=55
-11X=-55
X=-55/-11
X=5

(2 × 4 × 8 × 16 × 32) × (25 × 6.25 × 1.25 × 0.25) is calculated by a simple method

(2 × 4 × 8 × 16 × 32) × (25 × 25 / 4 × 5 / 4 × 1 / 4) let's divide it first. It's about 8 × 8 × 8 × 25 × 25 × 5 = 25 × 8 × 25 × 8 × 5 × 8 = 1600000

Given that the sum of the first n terms of the sequence {an} is Sn, and an = Sn · sn-1 (n is greater than or equal to 2, Sn is not 0), A1 = 2 / 9, find the general formula of {an}

an=Sn·Sn-11=an/(Sn·Sn-1)=1/Sn-1 - 1/Sn1=1/Sn-2 - 1/Sn-1…………………… 1 = 1 / S1 - 1 / S2 n-1 formula has n-1 = 1 / S1 - 1 / Sn, S1 = A1, so 1 / Sn = 1 / A1 - (n-1) = 11 / 2-n1 / sn-1 = 11 / 2 - (n-1) = 13 / 2-N, so an = sn-sn-1 = 1 / (11 / 2-N) - 1 / (...)

The remainder of 66.6 (2000 6) divided by 7 is
emergency

666666 / 7 = 95238. Six sixes are cycled once. After 333 cycles, there are two sixes left. So the remainder is the remainder of 66 divided by 7, that is, 3

Simple calculation (17 / 2 × 9 / 50) × 30 / 17 × 125

(17/2×9/50)×30/17×125
=17/2×30/17×9/50×125
=15×9/2×5
=15×22.5
=337.5

If the hyperbolic equation is x Λ 2 / 2 + y Λ 2 / 9 = 2, what is its asymptote equation
There is a question in the exercise book. Its asymptote equation is x / 4 + Y / 3 = 0, so let the hyperbolic equation be x ∧ 2 / 16-y ∧ 2 / 9 = K (k is not equal to 0). Why should the right side be K? Instead of 1? Is the asymptote equation just on the left side of the hyperbolic equation?

If the hyperbolic equation is x ∧ 2 / 2-y ∧ 2 / 9 = 2, the solution of the asymptote equation is as follows: (1) let the right side of the equation be equal to 0, X ∧ 2 / 2-y ∧ 2 / 9 = 0 (2) shift term x ∧ 2 / 2 = y ∧ 2 / 9 (3) root Y / 3 = ± X / √ 2, the asymptote equation y = ± 3 √ 2x / 2, because the hyperbola can not be determined by the hyperbolic asymptote