A junior 2 math problem, all the big Gs and Js help

x=2,y=1

Because the sum of the two sides is greater than the third side, so.

The bottom edge is 1, and the waist length is 2

Solving the system of equations yields x=2, y=1

When the waist length is 2, the triangle three sides are 2, 2, 1 (to meet the triangle trilateral requirements) When the waist length is 1, the triangle three sides are 1, 1, 2 (do not meet the triangle trilateral requirements), so the triangle side is 2, 2, 1

x+2y=4

3x+y=7

Solve the system of equations.

x=2 y=1

This is not true when x is the base and y is the waist.

So y is the base equal to 1 and the waist is long to 2

I didn't study well, but I think that's what I did.

x+2y=4

3x+y=7

x=3*(4-2y)+y=7

x=?y=?

Should it be a waiting?

So it's either xxy or xyy

Solution: x+2y=4 gives x=4-2y

Bring x=4-2y into the equation of 3x+y=7.

Get. y=1, then x=2

Solving the equation x+2y=4, 3x+y=7 gives x 2, y 1 can have two cases: 1 is waist 1 bottom 2 or waist 2 bottom 1, and because the first case does not conform to the trilateral relationship of the triangle, the answer is 221

x=2, y=1, only the bottom edge is 1, and the waist length is 2 (the sum of the two sides is greater than the third side).

You didn't get the guide right.

To do this, find a little E on AD to make AE equal to AC, that is, AE=AC, because P is the outer angle of the triangle ABC A point on the bisector of DAC, and AE=AC, AP is a common line, then the triangle AEP and the triangle APC are equal, then AC=AE, PC=PE, so PB+PC=PB+PE>BE=AB+AE=AB+AC

So PB+PC>AB+AC

Extend the CP intersection AD or the extension of AD at the point E, because the point P is the outer angle of the triangle ABC A point on the bisector of DAC has AC=AE, so AB+AC=BE;

Because the point P is a point on the bisector of the DAC at the outer angle of the triangle ABC, the point P bisects CE, so there is PC=PE, then PB+PC=PB+PE;

Because PB+PE>BE, PB+PC>AB+AC

It's usually a triangle where the sum of the two sides of the column is greater than the third side, and at the end of the day, the other sides are eliminated.

1. A company divides soldiers into 8 groups in a mission. If the actual number of people allocated to each group is 1 more than the predetermined number, then the total number of fighters will exceed 100 people; If each group is actually allocated 1 less than the predetermined number, then the total number of fighters will be less than 90 people. Determine the number of fighters assigned to each group.

Suppose the number of fighters assigned to each group is x, 8 (x+1) > 100

8(x-1)<90

Solution, 100-60-6*3-6x<0

x "The ** of oranges is between kilograms and yuan.

Solution: Let the single-digit digit of this two-digit number be x, then the ten-digit number is x+1, and according to the title, 30 10*(x+1)+x 42

The solution results in 20 11 x 32 11

Because the value of x in this problem must be a positive integer, x=2

x+1=3 that is, this two-digit number is 32 solutions: let the children's book be x, according to the meaning of the question.

4(x-1)+1 3x+4 4(x-1)+3 is obtained.

5 x 7 so x = 6

3x+1=19 i.e. five children and nineteen toys.

Let the time for the ship to leave and reach a safe distance is x

Then 20 roots 10 40 = x = roots 10 2 h

The time it takes for b to reach a is x0

x0=100/40=5/2 h

Because the root 10 2 < 5 2, there will be a typhoon.

Let the initial encounter time be t, let b move to y-axis y0, and a move to x-axis x0 (0(100+y0-20 root 10) 40=x0 20---1).

And (x0,0) is on the circle, and the coordinates of the center of the circle are (0,y0-20 root 10).

The circular equation is (x-0) 2+[y-(y0-20 roots 10)] 2=(20 roots 10) 2

Bring in x0 coordinates.

x0 2 + (y0-20 root 10) 2 = 4000 ---2).

1) (2) The solution is x0 2-80x0+1200=0

x0-20)(x0-60)=0

x0=20 or x0=60

t1 = 1h (rounded).

t2=60/20=3h

Therefore, the initial time for encountering a typhoon is 3 hours

Assuming encountered, let the time be t

At this time, with A as the origin, the ordinate of the typhoon is 100-40t, and the abscissa of the ship is 20t

100-40t) 2+(20t) 2=(20 roots, 10) 210000-8000t+1600t 2=400 2=4000t 2-4t+3=0

t-1)+(t-3)=0

t=1 or 3 so t=1, the eye of the typhoon is 60 nautical miles away and the ship is 20 nautical miles away.

The distance (AD) of the ship from west to east from point A to point D after t seconds is 20 tons of nautical miles;

The distance (BC) from point B to point C after t seconds is 40 tons nautical mile (BC);

Then the distance (ca) from point A at this time is (100-40 t) nautical miles;

As can be seen from the figure, the triangle ADC is a right triangle.

ad^2+ac^2=cd^2

20t) 2+(100-40t) 2=(20 roots, No. 10) 2400t, 2+10000-8000t+1600t, 2=40002000t, 2-8000t+6000=0

t^2-4t^2+3=0

t-1)(t-3)=0

t=1 or t=3

And because of "the time when the ship first encountered a typhoon".

It is the earliest time to encounter a typhoon.

Then take t=1