Square root and arithmetic square root, such as √ 9, is it to find the square root of 9 or its arithmetic square root

Square root and arithmetic square root, such as √ 9, is it to find the square root of 9 or its arithmetic square root

There are two square roots of a positive number. One is the arithmetic square root, the other is the negative square root. The positive one is the arithmetic square root

Given that X and y are real numbers and y = ∫ x-3 + ∫ 3-x + 4, find the square root of YX

Since both X and y are real numbers, the representation of Y determines that the domain of X is x-3 ≥ 0 and 3-x ≥ 0, so x = 3
Substituting it into y = 4, so YX = 4 * 3 = 12
Its square root is ± 2 √ 3

Given that the arithmetic square root of 12-n (n is greater than or equal to 0) is an integer, find the value of N satisfying the condition

3 8 11 12

The area of a triangle is 5 / 12 square meters, the bottom is 1 / 2 meters, how many meters is its height

Area of triangle s = base * height / 2
Therefore, 5 / 12 = 1 / 2 * height * 1 / 2
So the height is 5 / 3

12 and 4 / 5 times 34.5 plus 12.8 times 65.5 (simple calculation) 30%: x = 2 / 5:0.25 (solving the equation)

12 and 4 / 5 times 34.5 plus 12.8 times 65.5
=12.8×34.5+12.8×65.5
=12.8×（34.5+65.5）
=12.8×100
=1280
30%: x = 2 / 5: 0.25
0.3/x=0.4/0.25
0.4x=0.3×0.25
x=0.3×0.25÷0.4
x=0.1875

In RT △ ABC, ad is the middle line on the hypotenuse BC, ad = 3, try to find the length of BC? If AB = 4, find the area of △ ABC
If AB = 4, finding the area of △ ABC is the second question

In RT △ ABC, ad is the middle line on the hypotenuse BC, ad = 3, try to find the length of BC? If AB = 4, find the area of △ ABC
In ∵ RT △ ABC, ad is the median line on the hypotenuse BC
∴AD=BD=DC=3
∴BC=6
And ∵ AB = 4 (known) according to Pythagorean theorem
∴AC=2√5
Triangle area s = 1 / 2Ab = 1 / 2 × 4 × 2 √ 5 = 4 √ 5

Let a > 0, b > 0, C > 0, and a + B + C = 1, prove the inequality: 1A + 1b + 1c ≥ 9

It is proved that: ∵ a > 0, b > 0, C > 0, and a + B + C = 1, ∵ 1A + 1b + 1C = a + B + Ca + A + B + CB + A + B + CC = 3 + (Ba + AB) + (Ca + AC) + (CB + BC) ≥ 3 + 2BA · AB + 2ca · AC + 2CB · BC = 3 + 2 + 2 + 2 + 2 = 9 (if and only if a = b = C, take "=") (end of the proof)

A trapezoid, after its bottom is shortened by 5 cm, becomes a square, and the area is reduced by 20 square cm. What is the area of the original trapezoid?

20 × 2 / 5 = 8 (CM), (8 + 5 + 8) × 8 / 2, = 21 × 8 / 2, = 84 (square cm); answer: the original trapezoid area is 84 square cm

If two of the equations 1 / x + 1 = absolute value (lgx) are x1, x2
A.x1*x2

The graphs of the functions y = 1 / x + 1 and y = | lgx | are given respectively,
Let 0 < X1 < 1 < X2, then | lgx1 | > lgx2 |,
∴-lgx1＜lgx2,
That is lgx1 + lgx2 < 0,
∴0＜x1x2 ＜1,
So choose B

There is a rectangular carpet, which is 4.8 m long and 3 m wide. Now put it in a room of 4 m long and 3.6 m wide. Please cut it into two congruent figures with the same shape and area, so that it will just cover the room

Let the original rectangle be ABCD, draw a picture AB, CD is long, 4.8m, ad, BC is wide, 3M, then take a point E from AB, 0.8m away from a point, make a vertical line from e to DC, take EF = 0.6m, then make a vertical line from F to BC, that is, make a line parallel to AB, take FG = 0.8m, then make a vertical line from G to DC, take 0.6m, repeat this process, you will get a good