Sn = 1 / 4 (an + 1) &# 178;, the general term formula for an

Sn = 1 / 4 (an + 1) &# 178;, the general term formula for an

4x^3-6x-16
4x^3-6x-16=0

4x^3-6x-16
=2(2x^3-3x-8)
x1=1.899034800720710.

It is known that a quadratic function has a minimum value. Is the minimum value y or x
It is known that the quadratic function y = ax2-4x-13a has a minimum value of - 17, then why is a = 17 the value of Y instead of X

Your question is about the concept of function
In the quadratic function y = ax2-4x-13a, X is the independent variable and Y is the function value,
By "function has minimum value", we mean that function has minimum value, that is, y has minimum value
As for calculation, the algorithm of ordinate in vertex coordinate formula is used

If we know that the perimeter of a square is x and its circumcircle area is y, then the functional relation of Y with respect to X is

It is known that the side length of a square is x / 4, and the diagonal length is (√ 2 / 4) X,
The radius of circumcircle is (√ 2 / 8) X,
Y=π（√2/8）^2x^2
Y=(π/32)X^2

If a / C * B / C > A / C + B / C, and ABC is not equal to 0, then ()
A.a+b>c B.a+b=c C.a+b

A. Because a is, BC is wrong, so a is. The delivery speed of timely delivery time is accelerated, and Si ahui Dongguan is improved

Given the set a = {x | x2-16 ＜ 0}, B = {x | x2-4x + 3 ＞ 0}, find a ∪ B, a ∩ B

∩ a = {x | - 4 ＜ x ＜ 4}, ∩ B = {x | - 4 ＜ x ＜ 4}, ∪ a = R

5x-4 = (2x + 5) × 2 to solve the equation
5x-4=(2x+5)×2
6 X-29 + 3 9 + 3x
And these questions
1. 6 X-29 + 3 9 + 3x = 35
2. 3x + 4x + 5 / 6 = 10 / 12 + 7x + 86 + 2-10
3、4*（x-2）+20x=300
4. 1-2 (x-1 / 6) = 1 / 3

（1）：
5x-4=(2x+5)×2
5x-4=4x+10
x=14
(2)：
(x-29)/6+(9+3x)/3
=[(x-29)+2(9+3x)]/6
=(x-29+18+6x)/6
=(7x-11)/6

Find the maximum distance from the point on the circle x ^ 2 + y ^ 2-8x-4y + 16 = 0 to the straight line 4x + 3Y + 10 = 0

8.4
Let's change the equation to (x-4) ^ 2 + (Y-2) = 4 to determine that the radius of the center (4,2) is 2
Then, the distance between the straight line and the center of the circle | 4 × 4 + 3 × 2 + 10 | / (√ 3 & sup2; + 4 & sup2;) = 6.4 is larger than the radius 〈 the maximum distance between the straight line and the circle is 6.4 + 2 = 8.4, and the minimum distance is 6.4-2 = 4.4

How to remove calculator rounding
My calculator is kadio

The easiest way is to reset all the settings
The steps are as follows:
Shift + CLR + 3 + equal
complete

(1) If 2a-b = 5, then the value (2) of polynomial 6a-3b is known, then the value (3) of 10-2a + 3b is known. When x = 1, the value (2) of polynomial 6a-3b is known
(1) If 2a-b = 5, then the value of polynomial 6a-3b (2) is known, and 2a-3b = 5, then the value of 10-2a + 3b is (3). If x = 1, then the value of 2aX + BX is 3, then if x = 2, ax + BX is just a number

The first is 15, the second is 5, and the third is 6