Urgent for a junior high school math problem, urgent for a junior high school math problem

There are x number of students.

If the last 1 student gets 2.

4x+3=6(x-1)+2

7=2x x= impossible, discarded.

If the last 1 student gets 1.

4x+3=6(x-1)+1

2x=8x=4

Apple 4*4+3=19

If the last 1 student doesn't:

4x+3=6(x-1)

2x=9 x= impossible, discarded.

So there are 4 students, 19 apples.

4 students, 19 apples.

3+(6-1)] 6-4)=4 (people), 6 per person, the last student can only get 1, if it is 2 or 0, the number of students calculated is not a whole number.

The number of apples: 4 4 + 3 = 19 (pcs).

or 6 (4—1)+1=19 (pcs).

You're tying the question wrong!!

I'm wrong. Why didn't I count to the end??!

Let the number of students be x and the number of apples be y

y=4x+3

y=6(x-1)+k,k=0,1,2 (k is an integer) solves the equation x=(9-k) 2,x integer, so k=1x=4, y=19

Solution: Because DFC= AEB

So dfa= AEC

Because of AD BC

So ace= daf

So ADF CAE

In RT ACD, AD = 8 and DC = 6 F is the midpoint.

So AC=10 df vertical AC

So df=af=5

Because of the ADF CAE

So ae ac ce=df ad af

So ae= ae vertical bc bc = 2ce=

Area = (AD+BC)*AE2=1025 16

Scheme 1, even mo, no, no=mo=50 meters.

ao=bo=100 meters.

mao=∠nbo=30°

moa=∠nob=∠mon=60°

Then the round answer am=bn=50 3 meters.

MN arc length = 2 50 6 meters.

The length of the road built between the ab rooms is meters.

Solution 2: Connect to the PO

mao=∠nbo=30°

ao=bo=100 meters.

ap=bp=100 2 Wu San 3 m.

The length of the road built between AB is AP+BP=231 meters.

The first option is more economical in the cost of colangiac.

a. Calculate the arc length, fortune and line segment. It is known that the radius is fifty, and a to o is one hundred late rots, and tangent to the circle forms a right triangle. The length of the available line segment.

The length of the arc only needs to be angled, because the triangle is a right-angled triangle, and the right-angled side is half of the hypotenuse, so the inner corner of the circle is sixty degrees, and because it is congruent, the other angle is also sixty degrees, so the opposite angle of the arc is one hundred and eighty minus one hundred and twenty, which is sixty degrees. Method B starts with the same as A, and then partially solves it using trigonometric functions, code leaks, or similar.

Trapezoidal Balance Excavation AEFD Trapezoidal EBCF

ad=6,bc=10

Dress the stool ef = ad·bc = 60 (similar figures, proportional formula intersecting the nucleus fork multiplication), ef = 60 = 2 15

Circumference ratio = similarity ratio.

Circumference ratio = 6:2 15 = 15:5

According to the intersecting string theorem, there is Pa*Pb=PC*Pd

So 4*3=6*pd

So pd=2

According to the cutting line theorem, there is EA 2=ED*EC

So, let's let ea=x

So, 20=x*(x+6+2).

x^2+8x-20=0

x+10)*(x-2)=0

So x1=-10 and x2=2

Because x>0

So x=2, so ea=2

So PE=PD+ED=2+2=4

So pe=4

1. Use the area of the triangle twice to make the area of the triangle ABC = 1 2AB multiplied by AC = 1 2BC multiplied by AD