It is known that the quadratic trinomial x2-2 (M + 1) x + 9 is a complete square, then M=______ ．

It is known that the quadratic trinomial x2-2 (M + 1) x + 9 is a complete square, then M=______ ．

∵ x2-2 (M + 1) x + 9 is the complete square form, ∵ - 2 (M + 1) = ± 6, the solution is m = 2 or - 4, so the answer is: 2 or - 4

All velocity formula of physics in grade one of senior high school

The formula: (1) VT = V0 + at (2) s = v0t + at2 (3) vt2-v02 = 2As (4) s = (V0 + VT) t / 2
(l) The displacement difference of an object moving in a straight line with uniform velocity change is a constant, that is, Δ s = s Ⅱ - s Ⅰ = at2 = constant
(2) The average speed of an object moving in a straight line with uniform speed change is equal to the instantaneous speed in the middle of the time, that is, V (T / 2) = v = (V0 + VT) t / 2. The above two inferences are often used in students' experiments such as "measuring the acceleration of uniform speed change in a straight line", which should be mastered
(3) The instantaneous velocity of an object moving in a straight line with uniform velocity is
(4) Uniform acceleration linear motion with zero initial velocity (let t be equal time interval)
① At the end of it, at the end of 2T, at the end of 3T The ratio of instantaneous velocity is VL ∶ V2 ∶ V3 ∶Vn＝1∶2∶3∶…… ∶n；
② In the first, second, third The displacement ratio is SL ∶ S2 ∶ S3 ∶ Sn=1^2∶2^2∶3^2∶…… ∶n^2；
③ In the first t, in the second t, in the third t The ratio of displacement is Si: s Ⅱ: s Ⅲ: Si ∶SN=l∶3∶5∶…… ∶（2n－1）；
④ If you can't type the root number, go to the library of my space

. quick
To write the original formula is equal to. 1, (- 2) ^ 4 ^ (2 and 2 / 3) ^ 2 + 5 and 1 / 2 * (- 1 / 6) - 0.25
2. 7 / 6 * (1 / 6-1 / 3) - 3 / 14 △ 3 / 5
3. [(- 5) ^ - 4 ^ - (- 3) ^] - (7 / 8-5 / 11) * (- 7) ^ 4
4. - 1 ± (- 1 / 4) ± (- 4)
5、-|-3^2+(-3)^2|+|-3^3-3^3|
6. - 5 ÷ (- 5) ^ 2 * 125 / 8
7. - 1 and 1 / 2 * (1 / 3-1 / 4) / / 2 and 1 / 2
8. 0 - (- 2) ^ 3-13 ÷ (- 1 / 2) ^ 2
9、-1^4-(0-1)*(-2)^2÷[3-（-1）]
10. - 2 1 / 3 * (- 1 2 / 7) + (- 5 1 / 3) / (- 1 7 / 9)
11. [3 / 4 + (- 1 / 2) ^ 3 - (- 7 / 8)] / (- 7 / 8)
The original form=
Make the process clear

1. (- 2) ^ 4 ÷ (2 and 2 / 3) ^ 2 + 5 and 1 / 2 * (- 1 / 6) - 0.25
=16÷64/9+11/2×(-1/6)-1/4
=9/4-11/12-1/4
=2-11/12
=13/12;
2. 7 / 6 * (1 / 6-1 / 3) - 3 / 14 △ 3 / 5
=(7/6)×(-1/6)-(3/14)×(5/3)
=-7/36-5/14
=-49/252-90/252
=-139/252;
3. [(- 5) ^ - 4 ^ - (- 3) ^] - (7 / 8-5 / 11) * (- 7) ^ 4
Is there something wrong with this topic
4. - 1 ± (- 1 / 4) ± (- 4)
=-1×(-4)÷(-4)
=-1;
5、-|-3^2+(-3)^2|+|-3^3-3^3|
=3²-3²+2×3³
=2×27
=54;
6. - 5 ÷ (- 5) ^ 2 * 125 / 8
=-5/25×125/8
=-25/8;
7. - 1 and 1 / 2 * (1 / 3-1 / 4) / / 2 and 1 / 2
=(-3/2)×(1/12)×(2/5)
=-1/20;
8. 0 - (- 2) ^ 3-13 ÷ (- 1 / 2) ^ 2
=0+8-13÷(1/4)
=8-13×4
=8-52
=-44;
9、-1^4-(0-1)*(-2)^2÷[3-（-1）]
=-1+4÷(4)
=-1+1
=0;
10. - 2 1 / 3 * (- 1 2 / 7) + (- 5 1 / 3) / (- 1 7 / 9)
=(-7/3)×(-9/7)+(-16/3)×(-9/16)
=3+3
=6;
11. [3 / 4 + (- 1 / 2) ^ 3 - (- 7 / 8)] / (- 7 / 8)
=(3/4-1/8+7/8)×(-8/7)
=3/2×(-8/7)
=-12/7;
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
If you don't understand this question, you can ask,

Who can help me to list out the formula of physical acceleration in Volume 1 of grade 1 of senior high school

S=vt
S=v.t+(1/2)at^2
5. ^ 2-vt ^ 2 = 2As (. And T are subscripts)

Practical problems fast fast
1. Li Ming and Zhang Hong need to input one minute of the manuscript. Li Ming inputs 46 words per minute, while Zhang Hong inputs 44 words per minute. They work together and finish the manuscript in 2 / 1 hour. How many words are there in the manuscript?

Solution: Li Ming input 46 words per minute, Zhang Hong input 44 words per minute, then the two cooperate to input (44 + 46) words per minute
Half an hour = 30 minutes
(44 + 46) 30 = 2700
A: there are 2700 words in the manuscript
Hope to help you^_^

The velocity formula of senior one physics
What's the speed formula of grade one
It's the complete formula of VO + at VO + at * t / 2 V ^ 2-VO ^ 2 / 2a, as long as the formula is a little fuzzy

Vt = V0 + at, is the final velocity equal to the initial velocity plus acceleration times time;
S = v0t + 1 / 2A (T ^ 2), is the displacement equal to the initial velocity times time plus half times acceleration times time square;
Vt ^ 2-v0 ^ 2 = 2As, is the square of the final velocity minus the square of the initial velocity, which is equal to twice the acceleration times the displacement

How to write the three practical problems on page 19 of the first volume of primary school mathematics class training
1. The turnover of a supermarket in February is 1.2 million yuan, which is one fifth higher than that in February. How many million yuan is the turnover in March?
2. The school used 640 tons of water last month, which is a quarter less than last month. How many tons of water is used this month?
3. The number of days of precipitation in February 2007 accounts for two seventh of the whole month. How many days of precipitation are there in this month?

1.120 times one fifth
2.640 divided by one fourth
3.28 except two out of seven
That's it. I'll figure out the answer myself

Why the speed of the middle position is higher than that of the middle moment when an object moves in a straight line with uniform speed change
Explain the uniform acceleration and deceleration respectively.

The most obvious problem of this problem is to draw a V-T graph, which is a straight line; to draw a V-X graph, because of uniform acceleration, V increases faster and faster with X (think about it carefully), which is a concave curve. And because the two graphs have the same starting speed and the same end speed, of course, the middle point of the former is higher

Help me see how to write it
1. On a 1:10000 scale map, the distance between a and B is 4cm. If on a 1:8000 scale map, how many cm is the distance between a and B?
2. A batch of maps are bound by a printing factory. If 3000 volumes are bound every day, they will be finished in 16 days. If 4 days in advance, how many volumes should be bound every day?
3. The ratio of tons of grain stored in warehouse A and warehouse B is 9:11. After 15 tons are transported from warehouse B, the ratio of tons of grain stored in warehouse A and warehouse B is 3:2. How many tons of grain does a have?

1. On a 1:10000 scale map, the distance between a and B is 4cm. If on a 1:8000 scale map, how many cm is the distance between a and B?
4x10000 △ 8000 = 5 (CM)
2. A batch of maps are bound by a printing factory. If 3000 volumes are bound every day, they will be finished in 16 days. If 4 days in advance, how many volumes should be bound every day?
(3000x16) / (16-4) = 4000 (volume)
3. The ratio of tons of grain stored in warehouse A and warehouse B is 9:11. After 15 tons are transported from warehouse B, the ratio of tons of grain stored in warehouse A and warehouse B is 3:2. How many tons of grain does a have?
A: B = 9:11
A: B = 3:2 = 9:6
Each: 15 (11-6) = 3 (tons)
A: 3x9 = 27 (tons)

The relationship between the instantaneous velocity in the middle of a period of time and the instantaneous velocity in the middle of a period of time

From 2A · s / 2 = median V ^ 2 to initial V ^ 2
The median V ^ 2 = initial V ^ 2 + as = initial V ^ 2 + a (initial VT + at ^ 2 / 2)
That is, median V ^ 2 = initial V ^ 2 + initial VAT + A ^ 2T ^ 2 / 2
From middle time v = initial V + at / 2
We can get the middle time v ^ 2 = initial V ^ 2 + initial VAT + A ^ 2T ^ 2 / 4
It can be proved that the velocity in the middle position is larger
Inference conclusion: the speed of the middle position is greater than the speed of the middle time, no matter whether it is adding or subtracting (it is very simple to draw the line segment process diagram with method 1)