Solution equation: 19-20% x = 7, 4 / 9: 1 / 6 = x: 15

Solution equation: 19-20% x = 7, 4 / 9: 1 / 6 = x: 15

1、19-20％x=7
Change the percentage into a decimal, and the original equation is as follows:
19-1/5 x=7
If the item is transferred, the following results can be obtained:
19-7=1/5 x
That is to say:
1/5 x=19-7
1/5 x=12
x=12÷1/5=12×5=60
2、4/9 :1/6=x：15
1 / 6 x = 4 / 9 × 15 (inner multiplied by inner equals outer multiplied by outer)
1/6 x=20/3
x=20/3 ÷1/6=20/3 ×6=40

At present, m students can complete the classroom cleaning task in a minute. What is the time required for such (M + n) students to complete the classroom cleaning task (assuming that everyone's efficiency is the same)?

Let the time required be X
am=(m+n)x
x=am/(m+n)
The time required is: AM / (M + n)

x^2-3x

A:
x^2-3x

Finding derivative f (x) = λ e ^ (- λ x)

　　f'(x) = λ[e^(-λx)]*(-λ)
　　= -(λ^2)*[e^(-λx)].

Sequence A1 = 1 / 3, an + 1 = an + an & # 178 / N & # 178; prove an > 1 / 2 + 1 / 4N
Change to an > 1 / 2-1 / 4N
sorry

The problem is obviously wrong, let alone the following, when n = 1, the inequality does not hold

The absolute value of solving inequality 2x + 1 is greater than x + 2

Given that K belongs to R, X1 and X2 are the two zeros of the function g (x) = x2 -- 2kx -- K2 + 2, find the minimum value of X1 + x2

Because g (x) has two zeros,
So the discriminant 4K ^ 2 - 4 (- K ^ 2 + 2) > = 0
That is, K ^ 2 > = 1
From Veda's theorem, it is concluded that
x1+x2=2k,x1*x2=-k^2+2
therefore
x1^2+x2^2=(x1+x2)^2-2x1*x2
=4k^2-2(-k^2+2)
=6k^2-4
>=6-4=2
Thus, the minimum value of X1 ^ 2 + x2 ^ 2 is 2

Given the quadratic function y = x ^ 2 = 6x-9, when x = what value, y decreases with the increase of X
Y = x ^ 2 + 6x-9 sorry, wrong number

a> 0, opening up
So y on the left side of the axis of symmetry decreases as x increases
Axis of symmetry - B / 2A = - 3
So when x ≤ - 3, y decreases with the increase of X

Let a be the integer part of the root 24-1 and B be the decimal part of the root 24, and find the value of b-a

5

-The third power of 6x - the second power of 16x + 32x is solved by the cross multiplication method of grade one,

-The third power of 6x - the second power of 16x + 32x
Original solution = - 2x (3x + 8x-16)
=-2x(3x-4)(x+4)
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