Finding the solution set of inequality (X-2) (x + 3) > 0

Finding the solution set of inequality (X-2) (x + 3) > 0

This is less than 0 when x is negative and greater than 0 when x is integer

The solution set of inequality (X-2) (x-3) > 0 is

x> 3 or X

What is the law of the number of squares 1 23 45 right triangles 0 48 12 16 if you want to get 100 right triangles, you should draw () squares?

Square n
Triangle M
M = 4 (n-1) is the general formula
When m = 100, then n = 26

88 as shown in the figure: it is known that a (x1, Y1), B (X2, Y2) are the points of hyperbola y = m / X in the first quadrant, connecting OA and ob
(1)  prove: Y1 & lt; OA & lt; Y1 + m / Y1 (2) cross point B, make BC perpendicular to point C, and find the area of triangle BOC

In RT △ OAE, a is in the first quadrant; OA is hypotenuse, AE = Y1, so OA > Y1, because a is on y = m / x, M / Y1 = X1 = OE, so Y1 + m / Y1 = AE + OE > OA. So Y1 < OA < Y1 + my1.. 2, s △ BOC = 1 / 2oC × BC, because B is on the first quadrant, so OC = X2, BC = Y2, so s △ BOC = 1 / 2x2y2, because B is on y = m / x, x2y2 = m, so s △ BOC = 1 / 2m,

Let z = ∫ TF (x ^ 2 + y ^ 2-T ^ 2) DT, where f (x) has a continuous derivative, and find; ^ 2Z /; X; y
Let z = ∫ TF (x ^ 2 + y ^ 2-T ^ 2) DT, where f (x) has a continuous derivative, find; ^ 2Z /; X; y. the upper limit is the root, and the lower limit of x ^ 2 + y ^ 2 ~ is 0

[0 ∫ [0 8747; [0 - - [0 [0 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\- > 0] UF (U & # 178;) Du = ∫ [0 -

Divide the 1-meter-long paper tape into sections, and each section is 1 meter long (), which is () meters. If 12 pieces of sugar are divided into 3 piles, each pile is 12 pieces of sugar (), and each pile has ()?

If 12 grains of sugar are divided into three piles, each pile is 1 / 3 of 12 grains of sugar, and each pile has (4)?

As shown in the figure, if the coordinates of the parabola vertex are p (1,3), then the function y decreases with the increase of the independent variable x, and the value range of X is ()
A. x＞3B. x＜3C. x＞1D. x＜1

∵ the coordinates of the vertex of the parabola are p (1,3), ∵ the axis of symmetry is x = 1, and ∵ the opening of the parabola is downward, ∵ the value range of the function y decreases with the increase of the independent variable x is x > 1

Given the function f (x + 1) = the square of X + 2x, find f (x)

Law one
f(x+1)=x²+2x=x²+2x+1-1=(x+1)²-1
So f (x) = x & # 178; - 1
Method 2
Let x + 1 = t, then x = T-1
Then f (T) = (t-1) & # 178; + 2 (t-1) = T & # 178; - 2T + 1-2t-2 = T & # 178; - 1
So f (x) = x & # 178; - 1

(X)　＝x2－2x　＋　2a(a2-2ab-b2)-b(2a2 ab-b2)
loga[(x2^2-ax2)/(x1^2-ax1)]>0f(x)=x^5/5-ax^3/3 (a 3)x a^2

An = 2A (n-1) (N 2) / [n (n 1)] (n ≥ 2, n ∈ n *) so A5 = 8, a7 = 16, find A1 and common ratio Q, so (x) = x2-2x + 2, all items in the formula are vectors, the same below

As shown in the figure, in the triangle ABC, the angle ACB = 90 ° and D is the point on BC. Make de perpendicular to AB, intersect AC at point F, and CD ^ 2 = de multiplied by DF. Then point D is the midpoint of AB? Explain the reason

The conclusion of the original problem should be that "point D is the midpoint of BC". This paper proves the following. By examining △ DCF and △ DEB, we know that both triangles are right triangles from ∠ ACB = 90 ° de ⊥ AB, and ∠ CDF = ∠ EDB, then △ DCF ∽ DEB, we get CD / de = DF / DB, or CD · DB = de · DF, and from the known CD & sup2; = de · DF, then CD · D