2 math problems in the first year of junior high school in a hurry !!

1.Let the side length of the square be x, then there is:

x+2)(x+2)-x*x=36

The solution is x=82The bottom edge of the folded tin box is 2a-2x in length, a-2x in width, and x in height, so its volume is (2a-2x)*(a-2x)*x

1。Let the side length of the square be x, so we can get the square area of x squared, when the area of the square increases by two centimeters, the area of the square is (x+2) squared, and the square of the equation (x+2) - the square of x = 36 can be listed

The solution is x=82Volume = (a-2x)(2a-2x)x

1.Let the side length of the square row be x, and find x=8 according to the meaning of the question (x+2) 2-x 2=36

2.Width = a, so length = 2a, so the bottom of the tin box is long = 2a-2x, width = a-2x height = x, so its volume = (2a-2x) (a-2x) * x The second question can be seen as long as you draw a diagram.

1.Let the side length of this square be x, and according to the title, it is: (x+2)(x+2)-x x=36 to solve x=8 cm

2.The length of the iron box is 2a-2x, the width is a-2x, and the height is x, so the volume (that is, the volume) of the iron box =

2a-2x)×(a-2x)×x

If the side length is x, then the current side length is x+2

x+2)*（x+2)=x*x+36

Solution x=8

3/3 x - 5/5 y = 4

If we replace y with x, then we can find a definite solution:

Remove the denominator, get: 5x-3x

60x=30 If it is correct, then it is possible to find the solution of a positive integer (or some other solution within a definite range), otherwise it is an infinite number of solutions.

Also remove the denominator, get: 5x-3y

60x=123y/5

It can be seen that y needs to be a multiple of 5 ,......In this way, each set of solutions that meet the requirements is found.

If g and d are the same point, then bh=bg=

(1) AB is parallel to CD, so the angle A + angle ACD = 180°, the angle ACD = 80°

2) CB bisects the angle ACD, so angle 1 = angle 2 = 40°

3) The sum of the inner angles of the triangle is 180°, so the angle abc = 180° - angle a - angle 1 = 40°

Solution: (1) Because ab cd and angle a=100°, so acd=80°

2) Because CB is the angular bisector of ACD, ACB= BCD=40°

3) a=100° acb=40 so abc=40° done.

AB CD angle ABC = angle 1, triangle ABC inner angle and ACD ---

There are 2 more answers: Yes.

Obtained from (a+b-c) c=(a-b+c) b=(-a+b+c) a. a+b)/c=(a+c)/b=(b+c)/aabb*b

acc*caba*a

bcc*caca*a

bcb*b by .

b*ba*aacbc

babac(ab)

When b≠a, a+b

c, when ab, c

a or c2a

This is useful

The same goes for - gotcha.

The same goes for - gotcha. When a=b

c, the answer is:

When abc = 2a. For.

When A+BC, for.

Answered. It took two minutes to write a question, but twenty minutes to write it down, oh my God.

(1) It will take X days for A to complete this project, and it will take Y days for B to complete this project.

x=2/3y ①

30+10)/x+30/y=1 ②

Substitute : 40 (2 3y)+30 y=160 y+30 y=1

90/y=1

y=90 substitution: x=90 2 3=60

Solution: x=60

y=90A: It takes 60 days for A to complete this project alone, and 90 days for B.

2) Teams A and B cooperate to complete this project, which requires 1 (1 60 + 1 90) = 36 (days).

The fee is: 36 (10,000 yuan).

10,000 yuan) A: The construction cost of the project budget is not enough, and an additional budget of 10,000 yuan is required.

The line l2: y=2x+8 when y=0, x= -4, so the coordinates of g (or b) are (-4,0).

The straight line l1: y=-x+2 when x= -4, y=6, so the coordinates of point c are (-4,6), so bc=6

Therefore, the longitudinal coordinates of point d are equal to 6, that is, the line l2: y=2x+8, y=6, and find x= -1, that is, the coordinates of point d are (-1,6).

1.Therefore, cd is equal to the absolute value of the abscissa difference between the two points, that is, cd = -1 - (-4) = 32, and the second question should be the intersection of the line L2 and the y axis at the point m.

1) g is the intersection point of L2 and the X axis, so Y=0 is substituted into L2: Y=2X+8 to get X=-4, so the coordinates of point C are (-4, Y), and point C is on L1, and point C is substituted into L1: Y=-X+2 to get :

y=6, so the coordinates of point c are (-4,6);

At this time, the coordinates of point D are set to (x,6) and it is on the line L2, so it satisfies the equation of L2 y=2x+8, and the coordinates of point D are substituted to obtain x=-1, so the coordinates of point D are (-1,6);

So: dc length is 3, bc length is c and d the ordinate is 62) bcme can not reach the quadrilateral, you take a good look is not bad is required is the area of the triangle bce.

Solution: e(2,0)b(-4,0) is known

So c(-4,6)d(-1,6)a(-1,0) so dc=|-1-(-4)|=3 bc=|0-6|=6 If the L1 and Y axes are compared to the M point, the CME is on the same straight line, and the BCE is a right triangle.

If the L2 and Y axes are compared to the M point, the extension of Cd is compared to the M axis, and the extension of Cd is compared to H, and the Y axis is compared to Ne(2,0) M(0,8).

Straight line em:y=-4x+8, so h(,6).

be=|-4-2|=6 ， ch=| ，mn=|8-6|=2s quadrilateral, bcme=s, trapezoidal bche+s, triangle, mchs, trapezoidal bche=bc*be 2=18

S triangle mch=ch*mn2=

So the s quadrilateral bcme = 18+

1) First of all, the equations of L1 and L2 should be solved to obtain point F as (-2,4), and then substituted Y=0 into L1 and L2 respectively to obtain E(2,0), G(-4,0), because B coincides with point G, that is, x=-4 of point C, substitute X=-4 into L1, obtain Y=6, that is, point C is (-4,6), according to the y of point D of the rectangle is 6, substitute L2 to obtain X=-1, that is, the coordinates of point D are (-1,6). The above can be obtained with a CD length of 3 and a BC of 6.

2) Since the intersection of the straight line L1 and the Y axis at the point M, and the extension intersection at the point of the X axis is E, finding BCME is actually just finding the area of the BCE triangle, and the area (if there is no mistake in the question) is 6*6 2=18.

Solution: Since the line l2:y=2x+8 intersects the x-axis at the point g, the point g(-4,0).

And because the point b coincides with the point g.

So point b(-4,0).

Let the point c(-4,y), because the point c is on the straight line l1:y=-x+2, so y=4+2=6

So point c(-4,6).

So bc=6

Let the point d(x,6), because the point d is on the line l2:y=2x+8, so 6=2x+8, x=-1

So point d(-1,6).

So dc=3

1) G(-4,0) is obtained from the straight line L1, and X=-4 is brought into the straight line Y=-X+2 by the BC X-axis to obtain C(-4,6), so BC=6

Substituting y=6 into y=2x+8 from the cd y-axis yields d(-1,6), so cd=3

2, Question 2 I feel you are typing wrong, it should be L2 and Y axis intersect with mThen find the coordinates of each point and divide it into triangles.

1. There is the intersection point of the line L2 and the x-axis, b(-4,0), and c(-4,6)d(-1,6)a(-1,0).

So dc=ab=3, bc=6

2.Which point is required for the second question? Is there anything wrong with that?

The area of the quadrilateral BCME, point m at **?

Can't find it!