Calculation: 5x - (x-3) (x + 5)

Calculation: 5x - (x-3) (x + 5)

5x-(x-3)(x+5)
=5x-(x²+2x-15)
=5x-x²-2x+15
=-x²+3x+15

How to calculate 1 / 5x + 1 / 3 times (X-6) = 6?

1 / 5x + 1 / 3 times (X-6) = 6 times 15
3x+5(x-6)=90
3x+5x-30=90
8x=30+90
8x=120
x=15

A triangle has a side length of 6cm. When it is extended by 2cm, its area will increase by 4.2cm. What is the original area of this triangle?

Height 4.2 × 2 △ 2 = 4.2 cm
Area 6 × 4.2 △ 2 = 12.6 square centimeter

Solving equation 18-x = 6

12

If the radius of a circle is increased from 2 decimeters to 5 decimeters, its perimeter will increase () decimeters and its area will increase () square decimeters

Radius 2 decimeter, perimeter = 2 × 3.14 × 2, radius 5 decimeter, perimeter = 2 × 3.14 × 5, perimeter increase: 2 × 3.14 × 5-2 × 3.14 × 2 = 2 × 3.14 × (5-2) = 18.84 (CM), radius 2 decimeter, area = 3.14 × 2 & # 178; radius 5 decimeter, perimeter = 3.14 × 5 & # 178; area increase: 3.14 × 5 & # 178; - 3.14 × 2 &

1.25 times 67.875 + 125 times 6.7825 + 1.25 times 53.875!

1.25×67.875+125×6.7825+1.25×53.875＝125×0.67875+125×6.7825+125×0.53875＝125×（0.67875+6.7825+0.53875）＝125*8＝1000

As shown in the figure, in △ ABC, ∠ ABC = 45 ° AC = 4, h is the intersection of high AD and be, then the length of segment BH is ()
A. 6B. 4C. 23D. 5

Therefore, B is selected

(a1b1+a2b2)^2

Let a = (A1, A2)
b=(b1,b2)
Then a dot multiplied by B = (A1B1 + a2b2)
Module of a = under root (A1 ^ 2 + A2 ^ 2)
Module of B = under radical (B1 ^ 2 + B2 ^)
Reuse (point a multiplied by point B)

It is proved by determinant that the area of the triangle whose vertex is the midpoint of the three sides of the triangle is equal to one fourth of the area of the original triangle q291967968

Let the vector corresponding to the coordinates of point a be a, the vector corresponding to the coordinates of point B be B, and the vector corresponding to the coordinates of point C be c
Because the area of △ ABC is equal to | (C-A) × (B-A) | (the outer two bars represent the module of the outer product of the inner two vectors)
And the delta area corresponding to the midpoint is
|（（a+c）/2）-（（c+b）/2）×（a+b）/2）-（（c+b）/2）|
=1/4|（c-a）×（b-a）| =1/4 s_ ABC
(two bars also represent modules of outer product)

Domain! Given f (2 / x + 1) = lgx, find f (x)
The main problem is how to find the domain of definition. If we use the substitution method to set 2 / x + 1 = t, the range of t should be the domain of definition. But why do we need to find out f (x) = LG2 / (x-1) so that LG2 / (x-1) is meaningful, that is, X is greater than 1?
How to solve this kind of problem? Is it the scope of T?
Is it necessary to consider that x in () is not equal to 0 when calculating the definition field of a given function

In the function f (2 / x + 1) = lgx, when x > 0, t = 2 / x + 1. Under the condition of x > 0, the range of T is (1, + ∞), which is the domain of definition of the function f (x). Your understanding is correct. You may neglect the complement of "x > 0" in the original function f (2 / x + 1) = lgx: Yes, the domain of definition of the function makes all parts of the function meaningful