All answers to the new observation of mathematics in Grade Seven

Look at the answer

Explain sentences in English
1.Everyone in my class _ joined_ in The sports meeting__ 2.The man provides Lots of money_ for_ the poor people._______ 3.Scientists are looking for mays to _ reduce_ The air pollution________ 4.Nobody _ is_ interested_ In this book_____ 5.Do you agree with me?_

1.= Everyone in my class (took part in) the sports meeting.
2.= The man (donates) lots of money (to) the poor people.
3.= Scientists are looking for mays?(many ways) to (decrease) the air pollution.
4.= Nobody likes this book.
= Nobdy has/takes interest in this book.
5.Do you approve me?

It is known that the image of the parabola y = AX2 + BX + C (a > 0) passes through points B (12,0) and C (0, - 6), and the axis of symmetry is x = 2. (1) find the analytical formula of the parabola; (2) point D
On the line AB and ad = AC, if the moving point P starts from a and moves uniformly along the line AB at a speed of 1 unit length per second, and the other moving point Q starts from C and moves uniformly along the line CB at a certain speed, is there a certain time when the line PQ is vertically bisected by the straight line CD? If there is, the time t (seconds) and the motion speed of point Q at this time are requested; if not, Please explain the reason; (3) under the conclusion of (2), is there any point m on the line x = 1 that makes △ MPQ an isosceles triangle? If so, ask for the coordinates of all points M. if not, please explain the reason

1. From the title, we know that 144A + 12b + C = 0
c=-6
-b/2a=2
The solution is a = 1 / 16, B = - 1 / 4, C = - 6
The parabolic equation is y = 1 / 16x ^ 2-1 / 4x-6
2. Let the velocity of Q be V, a (- 8,0), D (2,0), P (- 8 + T, 0) (- 8) can be obtained

Given a + B = 2, A2 + B2 = 8, find the value of 7th power of a + 7th power of B

The solution is a = 1-radical 3, B = 1 + radical 3 or B = 1-radical 3, a = 1 + radical 3, then the result is 1136

The known function f (x) = LNX + x2
The function f (x) = LNX + x ^ 2 is known. ① if the function g (x) = f (x) - ax is an increasing function in its domain, the value range of real number a is obtained. ② under the condition of ①, if a > 1, H (x) = e ^ 3x-3ae ^ x, [0, LN2], the minimal direct of H (x) is obtained. ③ Let f (x) = 2F (x) - 3x ^ 2-kx (KX? R) if the function f (x) has two zeros m, n (0 < m < n), and 2x = m + n, Is the tangent at f (x0)) parallel to the x-axis? If yes, find out the tangent equation; if not, explain the reason

1,f(x)=lnx+ x^2
x>0
g(x)=f(x)-ax=lnx+ x^2-ax
g`(x)=1/x+2x-a>0
1/x+2x>a
1/x+2x>=2√2x(1/x)=2√2
a

On factorization
The detailed solution of 3a3 + 6A & # 178; B-3A & # 178; c-6abc = 3A (A-C) (a + 2b)

3a3＋6a²b-3a²c-6abc=3a3-3a²c＋6a²b-6abc=3a²(a-c)+6ab(a-c)=(3a²+6ab)(a-c)
=3a(a-c)(a+2b)

If the image of a quadratic function is known to pass through points (- 1, - 22), (0, - 8), (2,8), the expression of the quadratic function can be obtained

Let the quadratic function be y = ax ^ 2 + BX + C, because it passes (0, - 8), so C = - 8
The quadratic function becomes y = ax ^ 2 + bx-8
Then bring in two points
a-b-8=-22
4a+2b-8=8
Two style simultaneous
We get a = - 2, B = 12
Then the quadratic function is y = - 2x ^ 2 + 12x-8

-3/20(x-4)²+3=0

-3/20(x-4)²+3=0
-3/20(x-4)²=-3
(x-4)²=20
x-4=±√20
x1=4+2√5 x2=4-2√5

We know the quadratic function of x y = x square-2x-a, 1) if there are two intersections between this function and X axis, find the value range of A
It is known that the quadratic function y = xsquare-2x-a of X, 1) if there are two intersections between the function and X axis, the value range of a is obtained; 2) if the abscissa of the two intersections of the function and X axis are X1 and X2, and satisfy 1 / 2 of X1 + 2 / 2 of x2 = two-thirds of minus, the value of a is obtained

(1) Delta = 4 + 4A > 0
a>-1
(2) The quadratic equation x ^ 2-2x-a = 0 is known from the first question a > - 1
Weida theorem X1 + x2 = 2, x1x2 = - A
1/x1+1/x2=(x1+x2)/x1x2=-2/a=-2/3
a=3

Derivation of summation formula of equal ratio sequence
First item A1, common ratio Q
a(n+1)=anq=a1q^(n
Sn=a1+a2+..+an
q*Sn=a2+a3+...+a(n+1)
qSn-Sn=a(n+1)-a1
Sn=a1(q^n-1)/(q-1)
It seems that there is no problem
But many of the formulas I've given are the same
Sn=a1(1-q^n)/(1-q)
Are these two results the same? It seems that the numerator and denominator can be transformed by multiplying by negative 1 at the same time, but is this feasible? I've been a little confused since I haven't studied mathematics for many years. Recently, I used this formula when I was learning how to calculate the present value of ordinary annuity, but the teacher didn't talk about the derivation of this formula and asked for the advice of high-level people
Suppose that the first ratio sequence is a / (1 + R), a / (1 + R) ^ 2,. AA / (1 + R) ^ n, how to find the sum?

First, it's OK to multiply the numerator and denominator by - 1 at the same time
Let an = A / (1 + R) ^ ncommon ratio q = 1 / (1 + R); the first term A1 = A / (1 + R)
Sn=a1(1-q^n)/(1-q)=A/(1+r)*[1-（1/1+r）^n]/[1-（1/1+r）]=A/r *[(1+r)^n-1]/(1+r)^n

All answers to the new observation of mathematics in Grade Seven