Chapter 7 section 3 those who know will come back soon!

Chapter 7 section 3 those who know will come back soon!

What? Please make it clear

Review questions in the seventh chapter of mathematics exercise book

1、 (2) A-2 four (1) 2x + 10Y of x-3y (2) 5t-3h of - 2T + 6h

Seventh grade mathematics exercise book (1) Chapter 2 review questions group C
An investigation team started from the camp at 8 a.m. with a car of 50 km / h. The car first drove 40 km eastward, worked for 30 minutes, then drove westward for 2.5 o'clock, worked for one hour, and then drove westward all the time?

40-2.5 × 50 = 40-125 = - 85 (km)
A: the team is 85 meters to the west of the camp

Let m be a real number, f (x) = 2x ^ 2 + (x-m) | x-m |, H (x) = f (x) / x, X is not equal to 0, x x = 0 (1) if f (1) > = 4, find the value range of M (2) when m
When m > 0, we prove that h (x) is a monotone increasing function on [M, + ∞)

（1）f(x)=2x^2+(x-m)|x-m|
① When x > m
f(x)=2x^2+(x-m)^2
f(1)=2+(m-1)^2≥4,（1＞m）
The solution is m ＜ 1 - √ 2, m ＞ 1 + √ 2, m ＜ 1
So we get m ＜ 1 - √ 2
② When x < m
f(x)=2x^2-(x-m)^2
f(1)=2-(m-1)^2≥4,（1＜m）
No solution to give up
③ When x = M = 1
F (1) = 2 is not consistent with F (1) > = 4
Give up
To sum up, m ＜ 1 - √ 2
（2）f(x)=2x^2+(x-m)|x-m|
① When x > m > 0
f(x)=2x^2+(x-m)^2
h(x)=3x+(m^2/x)-2m
It can be seen that this is an antithesis function
It decreases in (0, √ 3m / 3) and increases in (√ 3m / 3, + ∞)
Because m ＞ 0, so √ 3m / 3 ＜ M
So it is a monotone increasing function on [M, + ∞)
② When x < m
f(x)=2x^2-(x-m)^2
h(x)=x-(m^2/x)+2m
Y = x is constant increasing, y = - (m ^ 2 / x) is increasing in (- ∞, 0), (0, + ∞)
So h (x) = x - (m ^ 2 / x) + 2m increases in (- ∞, 0), (0, + ∞)
M > 0, so it is monotone increasing function on [M, + ∞)
③ When x = m
f(x)=2x^2
H (x) = 2x is an increasing function
So it is a monotone increasing function on [M, + ∞)
In conclusion, H (x) is a monotone increasing function in [M, + ∞]

A cuboid water tank has a bottom area of 10 square meters and 2 meters. The tank can hold () cubic meters of water

This tank holds 12 cubic meters of water

Mn is the diameter of circle O, AB, CD is the chord, Mn is perpendicular to AB, CD / / ab

Connect Ao, Bo, Co, do
The isosceles triangle ABO, from the isosceles triangle three lines in one know Mn through the center o
And Mn is perpendicular to AB, AB is parallel to CD, so Mn is perpendicular to CD
Isosceles triangle CDO, from isosceles triangle three lines in one know Mn is the vertical bisector of CD

The center of the known ellipse is at the origin, the focus is on the x-axis, the eccentricity is √ 3 / 2, and the line L passing through the point m (- 1,0)
It is known that the center of the ellipse is at the origin, the focus is on the x-axis, the eccentricity is √ 3 / 2, and the line L passing through the point m (- 1,0) intersects the ellipse at P and Q
(1) If the slope of the straight line L is 1 and the vector PM = - 3 / 5 and the vector QM, the elliptic standard equation is obtained
(2) If the right vertex of the ellipse in (1) is a and the inclination angle of the straight line L is α, the vector AP × vector AQ will get the maximum value when we ask what the value of α is

Let the elliptic equation be x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0), because the eccentricity is (root sign 3) / 2, so (a ^ 2-B ^ 2) / A ^ 2 = 3 / 4, a ^ 2 = 4B ^ 2. (1) the equation of L is y = x + 1, substituting x ^ 2 / (4B ^ 2) + y ^ 2 / b ^ 2 = 1, simplifying to 5x ^ 2 + 8x + 4-4b ^ 2 = 0, let P (x1, Y1), q (X2, Y2), then X1 + x2 = - 8 / 5, x1x2 = (4-4b ^ 2

As shown in the figure, fold one side ad of rectangle ABCD so that point d falls at point F of BC side. AB = 8cm, BC = 10cm are known?

From the properties of folding, we can get: ad = AF = BC = 10, in RT △ ABF we can get: BF = af2 − AB2 = 6, ｜ FC = bc-bf = 4, let CE = x, EF = de = 8-x, then in RT △ ECF, ef2 = EC2 + CF2, that is, X2 + 16 = (8-x) 2, the solution can get x = 3, so CE = 3cm

Let two of the equations 4x & # 178; - 7x-3 = 0 be X1 and x2. Solve the equation and find the values of the following expressions (x ① - 3) (x ② - 3)

According to Weida's theorem, X1 + x2 = 7 / 4, x1x2 = - 3 / 4.. so (x1-3) (x2-3) = x1x2-3 (x1 + x2) + 9 = - 3 / 4-3 × 7 / 4 + 9 = 3

As shown in the figure, EF is the median line of trapezoidal ABCD, ah bisects ∠ DAB intersects EF in M, and extends DM intersects AB in n

It is proved that: if ∵ EF is the median line of ladder ABCD, ∵ EF ∥ AB, ∵ EMA = ∵ Nam, ∵ ah bisects ∵ DAB, ∵ EAM = ∵ Nam, ∵ EAM = ∵ EMA = ∵ Nam, ∵ EA = em, ad = 2ae can be obtained, and EM ∥ ab, e is the midpoint of AD, ∵ m is the midpoint of DN, ∵ EM is the median line of △ Dan, ∵ an = 2em = 2ae, ad = an can be obtained