Let the square of P: X - 8x - 20 > 0, Q: [x - (1 + a)] > 0 be known. If q is not a sufficient and unnecessary condition for P, the value range of positive real number a is obtained

Let the square of P: X - 8x - 20 > 0, Q: [x - (1 + a)] > 0 be known. If q is not a sufficient and unnecessary condition for P, the value range of positive real number a is obtained

Non P: x ^ 2-8x-20

Calculation: (1) 23-17 - (- 7) + (- 16) (2) 23 + (- 15) - 1 + 13 (3) (- 26.54) + (- 6.4) - 18.54 + 6.4 (4) (- 478) -(- 512) + (- 414) - 318 (5) 0 + 1 - [(- 1) -(- 37) -(+ 5) -(- 47)] + | - 4|

(1) Original formula = 23-17 + 7-16, = 23 + 7-17-16, = - 3. (2) original formula = (23 + 13-1) + - 15, = - 15. (3) original formula = (- 26.54) - 18.54 + [(- 6.4) + 6.4], = (- 26.54) - 18.54, = - 45.08. (4) original formula = (- 478) + 512 + (- 414) - 318, = (- 478-414-318) + 512, = - 1214 + 512 = - 634. (5) original formula = 1 - [(- 1) + 37-5 + 47] + 4, = 1 - [(- 1 + 37 + 47) - 5] + 4, = 10

Find the equation of the line L whose slope is 3 / 4 and the circumference of the triangle is 12

Take the special points (0, b) and (- B / K, 0) on the two coordinate axes as two points on the straight line. According to the meaning of the problem, it is a right triangle, k = 3 / 4, and according to the Pythagorean theorem, if the length of the three sides is 3 + 4 + 5 = 12 (perimeter), then the linear equation is y = 3 / 4K + 3

How much is 20-5 and one sixth - 4 and two fifths - 1 and five sixths - three fifths

=25-6 31-5 22-6 11-5 2
=25 - (31 / 6 + 11 / 6) - (22 / 5 + 2 / 5)
=25-7-4.8
=13.2

Which option is the formula of y = 4x & # 178; - 4x + 2 in the form of y = a (X-H) &# 178; + k?
【A】y=（2x-1）²+1 【B】y=（2x-1）²+2 【C】y=（x-½）²+1 【D】y=4（x-½）²+2
I first mention the common factor, and the result is y = 4 (x + ½) ² + 1. Why is there no option =. What's wrong
I got the wrong answer. I turn y = 4 (x - ½) ² + 1

Hello
It doesn't have to be 1
(4x^2-4x+1)+1
=(2x-1)^2+1
So a is right

How much is 100 minus 65 floors and 5 minus 5

24

a. B, C, D are real numbers that are not equal to zero, and a is not equal to B, C is not equal to D, find
a. B, C, D are real numbers that are not equal to zero, and a is not equal to B, C is not equal to D, AD + BC is not equal to 0,
Let M1 = a + B / (a-b), M2 = C + D / (C-D), m3 = (AC BD) / (AD + BC),
Then M1 + M2 + m3 = M1 × M2 × m3
Why···

Is the title wrong?
Should be M1 = (a + b) / (a-b), M2 = (c + D) / (C-D)?
Then M1 + M2 = 2 (AC BD) / ((a-b) (C-D))
(m1+m2)/m3=2(ad+bc)/((a-b)(c-d))
m1*m2=(ac+ad+bc+bd)/((a-b)(c-d))
m1*m2-1=2(ad+bc)/((a-b)(c-d))
That is, (M1 + m2) / m3 = M1 * M2-1
So M1 + M2 + m3 = M1 × M2 × m3

Simple calculation of 163 × 25 × 25 × 237

163×25×25×237
=25×(163+237)
=25×400
=10000

Merge similar items, remove brackets 7a + 3 (a + 3b)
7a+3(a+3b)
3x-(4y-2y+1)

7a+3(a+3b)
=7a+3a+9b
=10a+9b
3x-(4y-2y+1)
=3x-4y+2y-1
=3x-2y-1

There is a bottle of alcohol. The first time you pour out 2 / 3 of it, it is more than 80g. Then you pour back 140g. The second time you pour out 3 / 4 of the alcohol in the bottle. At this time, there is 90g of alcohol left in the bottle. How many grams are there in the original bottle?
(no equations!)

90 / (1-3 / 4) = 360 G
360-140 = 220 G·
(500 + 80) / (1-2 / 3) = 900g
Hope to help you!