Cube of (2x-1) = 8

Cube of (2x-1) = 8

Cube of (2x-1) = 8 = 2 ^ 3
2x-1=2
2x=3
x=3/2

Given two functions f (x) = 8x ^ 2 + 17x-k, G (x) = 2x ^ 2 + 5x + 4, where k is a real number
(1) For any x ∈ [- 3,3], f (x) ≤ g (x) holds and the value range of K is obtained; (2) for any x ∈ [- 3,3], f (x) ≤ g (x) holds and the value range of K is obtained; (3) for any x 1, x 2 ∈ [- 3,3], f (x 1) ≤ g (x 2) holds and the value range of K is obtained

h(x)=f(x)-g(x)=6x^2+12x-k-4=6(x-1)^2-k-10
H (x) is a parabola with vertices in (1, - K-10) and 1 in the middle of [- 3,3]
(1) Only h (- 3) = 86-k = 120 and 1 / 8

Find the unknown x-40% x = 16.2

X—40%X=16.2
X-0.4X=16.2
0.6X=16.2
X=27

Known: as shown in the figure, △ ABC, CE ⊥ AB, BF ⊥ AC

It is proved that: ∵ CE ⊥ AB is in E, BF ⊥ AC is in F, ∵ AFB = ∵ AEC. ∵ A is the common angle, ∵ Abf ∽ ACE (two equal triangles corresponding to two angles are similar) ∵ AB: AC = AF: AE, ∵ A is the common angle, ∵ AEF ∽ ACB (two equal triangles corresponding to two proportional sides are similar)

The property management department of a residential area charges the residents health fee every month. The charging method is as follows: for the households with 3 or less people, each household charges 5 yuan; for the households with more than 3 people, 1.2 yuan will be added for each person with more than 1 person. An algorithm is designed to calculate the health fee according to the number of people entered. It only needs to draw the program block diagram

According to the meaning of the question, the relationship between the cost y and the number of people n is y = 5 (n ≤ 3) 5 + 1.2 (n − 3) (n > 3)

As shown in the figure, in the triangle ABC, D is the midpoint of AB, e is the point on AC, CE = 1 / 3aC, be = 8cm, be and DC intersect at point O, and calculate the length of OE

Take the midpoint F of be and connect DF, CF and de
So DF is the median line of triangle Abe
DF = AE / 2
Because CE = 1 / 3 AC
So DF = CE. 1
DF / / AE is DF / / CE. 2
It can be obtained from Formula 1 and formula 2
Quadrilateral dfce is parallelogram
So o is the midpoint of parallelogram dfce
OE = EF / 2 = be / 4 = 2

Given point a (1,2) and vector a = (2,3), if vector AB = vector 3a, then the coordinate of point B is?

The vector a = (2,3) can be regarded as
Vector with starting point (0,0) and ending point (2,3)
3a=（6,9）
Then (1,2) + (6,9) = vector ab

It is known that be and CF are the heights of triangle ABC. If BP = AC is intercepted on be and CQ = AB is intercepted on the extended line of CF, is pa perpendicular to AQ?

In right triangle AEB and AFC, angle ACF + angle cab = angle Abe + angle cab, so angle ACF = angle Abe, because AB = CQ, BP = AC, so triangle ABP is equal to triangle QCA, so angle QAC = angle APB, because angle APB = angle AEB (right angle) + angle cap; angle QAC = angle QAP + angle cap, so angle QAP = angle AEB is straight

Given that the sum of the distances from a point P (x, y) on the ellipse X & # 178; / (K & # 178; + 11) + Y & # 178; / K & # 178; = 1 to two foci F1 and F2 is equal to 12, find K

Using the definition of ellipse
The sum of distances from P to F1 and F2 = 2A = 12
In this question, a & # 178; = K & # 178; + 11
So 36 = K & # 178; + 11
k²=25
k=±5

In RT △ ABC, the bisector of angle c and the bisector of angle B intersect at point E, connect AE, and calculate the degree of ∠ AEB
Urgent demand, there must be a high reward

Because the angular bisector of angle c intersects the outer angular bisector of angle B at point E, connecting AE
So AE is the bisector of angle A
Because ∠ C = 90 degree
Therefore, CBE = 90 ° + 1 / 2 ∠ B
∠CAE=90°+1/2∠A
So ∠ AEB = 360 °～ CBE - ∠ cae-90 ° = 45 °