If the positions of the corresponding points of real numbers a, B and C on the number axis are shown in the figure, the absolute value of B + 2 + the absolute value of a-C is obtained A is a positive number, B, C is a negative number, B is greater than C

If the positions of the corresponding points of real numbers a, B and C on the number axis are shown in the figure, the absolute value of B + 2 + the absolute value of a-C is obtained A is a positive number, B, C is a negative number, B is greater than C

b+2>0,a-c>0
Absolute value of B + 2 + absolute value of a-C = B + 2 + a-C
b+20
Absolute value of B + 2 + absolute value of a-C = - B-2 + a-C

As shown in the figure, the positions of a, B and C on the number axis are shown in the figure, then the absolute value of a + B - the absolute value of a + C - the absolute value of C-B is equal to

From the number axis: a + B

As shown in the figure, in the triangle ABC, the angle bisector ad; be; CF intersects at point h, makes Hg through point h, is perpendicular to AC, and the perpendicular foot is g, then the angle ah
Acute triangle B
F H D
A, e, G, C is the triangle in the middle, Hg is the vertical line, ad, CF, be are the three middle lines

The title is not complete! The question is angle ah? What triangle ABC is

The side length of equilateral triangle ABC is 6. Ad is the middle line on the edge of BC, M is the moving point of AD, e is a point on the edge of AC. if AE = 2, what is the minimum value of EM cm?

Take the point e 'on AB so that AE' = AE = 2, connect CE 'to ad at M' ∵ AB = AC = 6, ad is the center line of BC, ∵ bad = ∵ CAD (three lines of isosceles triangle in one) and ∵ AE '= AE, am' = am '≌ AE'm' ≌ AEM '≌ e'm' = em ', ≌ em' + cm '= e'm' + cm '= e'c, ≌ when m and M' coincide, EM + cm is the smallest (two points

There are 200 trees to be planted in the school, a quarter of which are planted in the lower grades, and the ratio of trees planted in the middle and higher grades is 2:3. How many trees are planted in the higher grades?

Junior grade = 200 * 1 / 4 = 50
Remaining = 200-50 = 150
Senior students: 150 * 3 / (2 + 3) = 90

In the fifth grade, 150 trees were planted in spring,

150x (1-20%) / 6 / 5 = 150x0.8x5/6 = 120x5 / 6 = 100 trees

The school carried out tree planting activities. 180 trees were planted in grade five. The number of trees planted in grade five was one fifth more than that in grade six

180÷（1 + 1/5）
＝180 ÷ 6/5
= 150
In the sixth grade, 150 trees were planted

560 trees should be planted in three classes of grade five. According to the number of students in three classes (47 in class one, 45 in class two and 48 in class three), tasks should be assigned to each class
How many trees should each plant?

Per capita: 560 / (47 + 45 + 48) = 4 (trees)
So class 1 planting trees 4 * 47 = 188
Class 2 45 * 4 = 180
Class 3 48 * 4 = 192

The school divided the task of planting 360 trees into two classes of grade 6 according to the number of students. There were 47 students in class 1 and 43 students in class 2. How many trees were planted in each class
What's the formula

Class 1: 47 * (360 / 90) = 188
Class 2: 43 * (360 / 90) = 172

The school assigned the task of 220 trees to two classes of grade 6 according to the number of students. There were 46 students in the first class and 42 students in the second class. How many trees were planted in each class?

Class 1 = 220 ÷ (46 + 42) × 46 = 115 people
Class 2 = 220-115 = 105