Super two math problems, the master is coming

Because angle bac=120 so angle 2 + angle 3 + 120 = 180 angle 2 + angle 3 = 60 (1).

Because angle 1 = angle 2 so 2 * angle 2 + (180 - angle 4) = 1802 * angle 2 - angle 4 = 0 (2).

1) + (2) get.

3 * angle 2 = 60

Angle 2 = 20 because angle 1 = angle 2 so angle 1 = 20

Angle DAC = 120 - Angle 1

Angular dac=100

1) If there is a type A drink x bottle, then a type B (100-x) bottle.

20x+30(100-x)<=2800

40x+20(100-x)<=2800

The solution is 20<=x<=40

So there are 21 scenarios:

Scheme 1, 2, 3, 4 5 ....Twenty twenty-one.

A (bottle) 20 21 22 23 24 ....39 40b (bottle) 80 79 78 77 76 ....61 602) because of the cost) "cost).

Therefore, option 21 is the least costly.

Detailed enough.

The first question, the answer is 100 degrees, right?

Question 1: Angle 4 = Angle 3 = 2 Angle 2

Then the 3 times angle 2 + 120 = 180 degrees (the sum of the 3 angles and the inner angles) then the angle 2 = 20 then dac=100

Question 21) 20x+40(100-x)<=2800, 30x+20(100-x)<=2800

So 60<=x<=80 x is an integer with 21 solutions.

2) Cost = so the bigger the x, the better to take 80

1.Solution: a +3a +a-4=a +2a +(a +a-1)-3=a +2a -3=a(a +a)+a -3=a +a-1-2=-2

2.Solution: (x -y )-2x y-4xy )-2(-y +xy) =x -y -2(x y-xy -xy) +2y -2xy = x -y -2(-3-xy) 2y -2xy

x³-y³+6+2xy²+2y³-2xy²=x³+y³+6=10+6=16

There's nothing wrong with the question, it's too simple.

There are student A and dormitory room B.

then a=4b+20

b-1b-15 then b=6, a=44

There are x dormitory rooms and a number of occupants y

Then 4x+20=y

y 85 then y > 40

In addition, y 8>x-1

then x<7

then x=6 and y=44

We're happy to answer for you!

There are x dormitories and y people, and the equation 4x+20=y 8(x-1)=y can be obtained from the question

The solution is x=7, y=48

Therefore, there are 7 dormitories with 48 people.

15. b^2-ab+4a=0

b^2=ab-4a

Original = (4a 2) [a(b-4)] 1 (a-b) -a b * 4 b

4a) (b-4)(a-b) -4a b 2(4a) (ab-b 2-4a+4b) -4a b 2(4a) [b 2-ab+4a)+4b] -4a b 2(ab) b 2 - 4a huai ran zai b 2

ab - 4a)/b^2

ab - 4a)/(ab - 4a)

1/n =n/mn + m/mn = m+n)/mn=1/(m+n)

m+n)^2 = mn

m^2 + 2mn + n^2 = mn

m^2 + n^2 = mn

n m + m lead n = n 2 mn + m 2 segment min mn = (n 2 + m 2) mn = -mn mn = -1

15. b 2-ab + 4a = 0, shift term, 4a = b (a-b).

2a b) 2*1 (a-b)-a b b 44a 2 b 2(a-b)-4a b 2

4a/b(a-b)

1 m+1 n=1 slow mode empty (m+n).

Denominator, n(m+n)+m(m+n)=mn

m^2+mn+n^2=0

m^2+n^2=-mn。

Thus code infiltration, n m + m n = (m 2 + n 2) mn = -1.

1.Proto-spine pants = 4a2 [b2(a-b)]-4a b24ab [b2(a-b)].

4a/[b(a-b)]

ab-b2)/[b(a-b)]

2.This question is incorrect!

The normal solution should be: from the known, Fushino source (m+n) mn=1 (m+m)m2+mn+n2=0 - (1).

The original is missing = (m2+n2) mn=(-mn) mn=-1, but equation (1) actually has no solution (m,n≠0,m+n≠0).