A few questions in the first year of mathematics, please come in with math masters!

2x+y-4/5=0

If yes|2x+3|+(2x+y-4 5)=0.

2x+3=2x+y-4/5

or 2x+3=-(2x+y-4 5).

3. Two cases: if C is between AB, then AC AB-BC 24-8 16AM=AC2=8

If C is outside of AB, AC AB + BC 24+8 32AM AC 2 16

Let's stop here.

1.Let =x°, =90-x°

x/90-x=3/2

2x=3(90-x)

5x=270

x=54, so =54° =90-x=90-54=362Because .|2x+3|+(2x+y-4 5)=0 so |2x+3|and (2x+y-4 5) are inverses to each other.

2x+3|=-(2x+y-4/5)

Because x is not positive or negative, and absolute values are non-negative.

So go to the absolute value sign and add a negative sign in front of it.

i.e. -(2x+3)=-(2x+y-4 5).

2x-3=-2x-y+4/5

3=-y+4/5

y=19/5

Y = 19 5 generations - (2x+3)=- (2x+y-4 5) is solved to -3=-3

So |2x+3|and (2x+y-4 5) are not inverse to each other, |2x+3|and (2x+y-4 5) are both 0

When|2x+3|=0

2x+y-4 5=0.

The solution is x=-3 2

Substitute x=-3 2 into 2x+y-4 5=0

y=19/5

So x-y=-53 10

3.Because ab = 24 cm, bc = 8 cm

So ac=16

Because m is the midpoint of the line segment AC.

So am=mc=1 2ac=8

1.If the angle and the angle are mutually reciprocated, then +=90° and angle : angle =3:2 so =3*90 5=54° =90-54=36°

2.|2x+3|+(2x+y-4 5)=0 because |2x+3|=0 so 2x+y-4 5=0 |2x+3|=0 x=-2/3 y=19/5 x-y=-67/15

3.When C is inside AB, AC=24-8=16 AM=1, 2AC=8cm, and when C is outside AB, AC=24+8=32 AM=1, 2AC=16cm

1.Let =3x, =2x

The sum of the reciprocal descriptions is 90°, that is, 3x+2x=90°

x=18° then =54°, =36°

2.I guess this question should be less squared, and if the question is correct, there are several cases:

1) 2x+3=0 and 2x+y-4 5=0, x,y2)2x+y-4 5<0, and 2x+3=2x+y-4 53)2x+y-4 5<0, and 2x+3=-(2x+y-4 5)3The result was 8 or 16 (with units) and the upstairs process was quite right.

1. The angle is 54 degrees, and the angle is 36 degrees.

2, x=-2 3, y=19 5, x-y=-4 and 7 153, when c is between a and b.

AM = 8 when C is outside AB.

am=16

1) If A discharges 2x liters of water, then B discharges x liters of water. Bucket A has 400-2x litres of water left, and bucket B has 150-x litres left.

400-2x=4(150-x)

So: x=100

2x=200

Namely. 200 liters for A and 100 liters for B.

4) Set the circumference of the tree x feet.

3x+4=4x-3

So x=7 so the rope length: 3*7+4=25 feet.

The circumference of the tree is 7 feet.

1) If you buy A and B, set up x for A and 50-x for B.

1500x+2100(50-x)=90000x=250/6

Doesn't match the topic. If you buy A and C, set up to buy X A and C 50-x.

1500x+2500(50-x)=900000x=35 If you buy B C, set up B x and C 50-x.

2100x+2500(50-x)=900000x=350/4

Doesn't match the topic. Therefore, when buying A and C, 35 A and 15 C are purchased.

2) Set up the purchase of X X and Y.

x≥0y≥0

1500x+2100y≤90000

Find the maximum value of z=150x+250y. Draw a linear program, and the results will come out.

6) Set the rectangle long x cm and wide y cm.

xy=(x-15)*(y+6)

x-15=y+6

x = 25 cm.

y = 4 cm.

Question 1. Solution: Let the water released by B be x, then the water released by A is 2x400-2x=4(150-x).

solution, x=100, 2x=200

Answer: Slightly. Question 2.

Solution: Let the length of the circle around the tree be y, and the length of the rope be x

3y+4=x

4y-3=x

solution, x=25, y=7

Answer: Slightly.

1 question. Let B put x liters of water in bucket, and the equation 400-2x=4 (150-x) is solved to get x=100, that is, A puts 100 2=200 liters of water, and B puts 100 liters of water.

4 questions. From the inscription, it is known that the circumference around the tree is 4 + 3 = 7 feet, and the length of the rope is 3 7 + 4 = 25 feet.

5 questions. Let the purchase quantity of A, B and C be xyz, and the system of equations is 1500x+2100y=90000, x+y=50 or 1500x+2500z=90000, x+z=50 or 2100y+2500z=90000, y+z=50, respectively, and finally there are only two schemes, namely 25 units of A and 25 units of B or 35 units of A and 10 units of C.

6 questions. Let the length x width y, the system of equations (x-15) (y+6) = xy, x-15 = y+6, and the solution is x=29 and y=8

7 questions. Let A walk x every hour and B walk y every hour, and the system of equations is 2x+, 2y+3(x+y)=36, and the solution is x=6,y=

11 questions. Option 3

12 questions. Set the front door through x people and the side door through y people, 2 (x + 2 y) = 560, 4 (x + y) = 800, the solution is x = 120, y = 80

Let the velocity of A be x and the velocity of B y, then ax+ay=bx-by

Get x y=(a+b) (b-a).

A and B set out from two places at the same time, and if they go in the opposite direction, they will meet in a hour; If they go in the same direction, then B hour A catches up with B. Then the velocity of A is the velocity of B ((a+b) (b-a)).

Set the itinerary to be in units "1".

v A + v B = 1 a

v A - v B = 1 b

v A = (1 A + 1 B) 2 = (A + B) 2 ABV B = (1 A - 1 B) 2 = (B - A) 2 ABV A V B = [(A+B) 2AB] [(B-A) 2AB] = (B + A) (B-A).

Solution: Let the distance between the two places be s meters, the speed of A is x meters per second, and the speed of B is y meters per second.

The meaning of the title is: x > y

s/(x+y)=a ……1

s/(x-y)=b ……2

Derived from the formula:

x/y=(a+b)/(a-b)

Let A and B be separated by x, the velocity of A is y, and the velocity of B is z, and x (y+z)=a so x=a(y+z) yb=zb+x Substituting x is yb=zb+a(y+z), that is, y(b-a)=z(a+b) So y z=(b-a) (a+b).

Let the speed of A be x, B be y, then a(x y) b(x y), you should be able to make it.

In the first formula, the landlord replaces x with -1 and multiplies both sides of the equation by -1 at the same time, and the answer is obtained