If the maximum value of y = - X & # 178; + 2x + k is 1, then K=

If the maximum value of y = - X & # 178; + 2x + k is 1, then K=

y=-x²+2x+k
=-x²+2x-1+k+1
=k+1-(x-1)²
≥ K + 1 -------- then (x-1) &# 178; = 0
Because the maximum value of y = - X & # 178; + 2x + k is 1
So K + 1 = 1
So k = 0

If the maximum value of the parabola y = - x2 + 4x + k is 3, then K=______ ．

The maximum value of ∵ parabola y = - x2 + 4x + k is 3, ∵ 4K − 16 − 4 = 3, ∵ k = - 1

Who knows how to write English words in January February March April May June July August September October November December?

january
february
march
apirl
may
june
july
august
september
october
november
december

The mass of the scale plate and spring of a spring platform scale is not taken into account. A static object P with mass m = 12kg is placed in the plate, and the stiffness coefficient K of the spring is 800N / m. now, a vertical upward tension f is applied to p to make p make a uniform acceleration linear motion from the rest. It is known that f is variable in the first 0.2S
1 / 2A * t * t = 0.15, a = 7.5 M / (S2), so Fmin = 12 * 7.5 = 90N,
After 0.2S, n decreases to 0, f increases to the maximum, Fmax = 12 * (7.5 + 10) = 210n, why Fmin is Ma, Fmax is m (G + a)

The maximum is 210n, the minimum is 90N. Because the object starts to be stationary, the spring compresses x = g / k = 120 / 800 = 0.15 (m); in the first 0.2 seconds, the spring is in a compressed state, and F + n = g, the spring becomes looser and looser, n becomes smaller, f becomes larger, until the object completely breaks away from the scale, f becomes a constant force

There are 100 oral arithmetic questions in grade four,

640÷80=
15×5=
23×3=
12×2×5=
480÷80=
16×5=
27×3=
90÷15=
48÷4=
640÷16=
39÷3=
24×20=
32×3=
48÷16=
12×8=
27×3=
56÷14=
24÷8=
14×2=
83－45=
560÷80=
96÷24=
40÷20=
40×30=
37＋26=
76－39=
605＋59=
30×23=
12×8=
27＋32=
48＋27=
4500×20=
73＋15 =
120×600 =
200×360=
6800×400=
280＋270=
4×2500=
6000÷40=
5×1280=
310－70=
400×14=
470＋180=
1000÷25=
160×600=
20×420=
290×300=
8100÷300=
7600÷200=
7600÷400=
680＋270=
980÷14=
4200÷30=
6×1300=
1300×50=
200×48=
930－660=
530＋280=
9200÷400=
840÷21=
180×500=
8000÷500 =
1900÷20=
200×160=
8700÷300=
300×330=
3×1400=
7000÷14=
600÷12=
9600÷80=
140×300=
8800÷40=
9600÷800=
750－290=
5×490=
760×20=
7500÷500=
370×200=
650÷13=
8600－4200=
240×4=
640÷80=
15×10=
12×11=
160×30=
220×40=
104×5=
4500÷50=
120×2=
90÷30=
270÷30=
270×30=
84÷21=
76÷9=
66÷7=
100－54=
123＋15=
360÷4=
55÷5=
32×6= 7000÷70＝
200÷40＝
180÷30=
240÷40=
35×2=
140×7=
13×6=
280×3=
350×2=
50×11=
250×6=
7200+900=
410-201=

An elevator transports 1.21t goods from the ground floor to the top of the fourth floor at a constant speed. If the height of each floor is 3.5m, the power of the elevator to the goods is 1260
How many seconds does the elevator rise? (g = 10N / kg)

p*t=mgh
1260*t=1.21*1000*10*3.5*4
t=134.4s

Given that the tolerance of the arithmetic sequence {an} is 2, the number of items is even, the sum of all odd items is L5, and the sum of all even items is 25, then the number of items of this sequence is ()
A. 10B. 20C. 30D. 40

Let the number of terms of this sequence be 2K, then the sum of odd terms = a1 + a3 + +A2k-1 = 15, sum of even terms = A2 + A4 + +a2k=25，∴（a2-a1）+（a4-a3）+… +(a2k-a2k-1) = 25-15 = 10, the tolerance of ∵ arithmetic sequence {an} is 2, ∵ 2K = 10, ∵ the number of terms of this sequence is 10

What is the displacement of a car in 20 seconds when a car runs at a constant speed of 54 km / s for 10 seconds and then accelerates at a constant speed of 0.5 m / s for 10 seconds?
km/h

s=325m
v=16.25m/s
The first parameter is suspected to be 54km / h

Xiao Ming and Xiao Gang are good friends. They both like collecting stamps,
Xiao Ming finds that two fifths of his stamps are exactly 80. He is going to exchange three fifths of the stamps with Xiao Gang. How many stamps is Xiao Ming going to exchange? It takes Master Wang four hours and master Li six hours to process a batch of parts. If they work together, how many hours can they complete the task?

Let's find out how many stamps Xiao Ming has. Two fifths of his stamps are 80, and one fifths of his stamps are 40,
Five out of five is his total, which is five times as many as 40, 200 pieces
Then take 3 / 10 of 200 pieces to exchange, 1 / 10 is 20, and 3 / 10 is 60. The answer comes out
It takes 4 hours for Master Wang to process, and 6 hours for Master Li to process. You can assume a number, a common multiple of them, 4x6 = 24, and assume that there are 24 pieces in total
Master Wang takes four hours, so how many pieces can he make per hour? Then 24 divided by 4 = 6, and he can make six pieces per hour
Master Li takes 6 hours, just like above, taking 24 divided by 6 = 4, he has 4 pieces per hour
How much can they do in an hour together? Add up 4 + 6 to 10 pieces per hour
That's a total of 24 pieces, 10 pieces per hour, that's 24 divided by 10 = 2.4, that's 2.4 hours

It is known that the solution of the equation 3x = 2-4a about X is negative, and the value range of a is obtained

x=(2-4a)/3;
∴(2-4a)/3＜0;
4a＞2;
∴a＞1/2;
If you don't understand this question, you can ask,