I can't figure it out! Square of 280 (1 + x) = 403.2

I can't figure it out! Square of 280 (1 + x) = 403.2

(1+X)^2=1.44
1+X=±1.2
X = 0.2 or - 2.2

Understanding the definition of function limit of X → X0
Is δ in the definition of function limit in the centreless neighborhood of x0?

No, it changes with epsilon

2X of 1 × x + 2x of 3 × 5. + 2x of 2005 × 2007 = 2006

1 × 3 / 2x + 3 × 5 / 2x +. + 2005 × 2007 / 2x = 2x × 1 / 2 (1-1 / 3 + 1 / 3-1 / 5 +... + 1 / 2005-1 / 2007) = x (1-1 / 2007) = 2006X / 2007 = 2006X = 2007. This is my conclusion after meditation. If you can't ask, I will try my best to help you solve it

The number whose square equals 2.89 is
The value range of the approximate number 6.50 is ()
A. 6.475 less than or equal to a less than 6.505 b.6.40 less than or equal to a less than 6.50
C. 6.495 less than a less than or equal to 6.505 d.6.50 less than or equal to a less than 6.505

The number whose square equals 2.89 is ± 1.7

Solution equation: (16 + x): x = 13:11

(16+x)：x=13：11
13x=11(16+x)
13x=11*16+11x
13x-11x=176
2x=176
x=176÷2
x=88

Given a column number A1, A2, A3 ，an，… Among them, A1 = 0, A2 = 2A1 + 1, A3 = 2A2 + 1 ，an+1=2an+1，… . then the individual digits of a2004-a2003 are ()
A. 2B. 4C. 6D. 8

∵a1，a2，a3，… ，an，… If the number of digits in a2004 is 7 and the number of digits in A2003 is 3, then the number of digits in a2004-a2003 is 7-3 = 4

2X ^ 2 + 14x + 49 = 169, find X

2x^2+14x+49=169
2x^2+14x-120=0
x^2+7x-60=0
(x+12)(x-5)=0
x+12=0,x-5=0
x1=-12,x2=5

The following linear programming Max z = 2x1 + x2 {3x1 + 5x2 ≤ 15 {6X1 + 2x2 ≤ 24 {x1, X2 ≥ 0} is solved by graphic method and simplex method respectively

Simplex: Standard Type: Max z = 2x 1 + x 2 + 0 x 3 + 0 x 4 ST: 3x 1 + 5 x 2 + x 3 = 156 x 1 + 2 x 2 + x 4 = 24 CJ → 2100cb base B x 1 x 2 x 3 x 40

Factorization 4x & #178; - 48xy + 144y & #178; + X & #178; - 6y + x-6xy

4x² - 48xy + 144y² + x² - 6y + x - 6xy= (4x² - 48xy + 144y²) + (x² - 6y + x - 6xy)= 4(x² - 12xy + 6²y²) + (x² - 6y + x - 6xy)= 4(x - 6y)² + [(x -...

Let m and n be two points on the radius op of the ball center O, and NP = Mn = OM, respectively, through N, m and o as perpendicular lines, intercept the ball on the surface of OP to obtain three circles, then the area ratio of these three circles is: ()
A. 3，5，6B. 3，6，8C. 5，7，9D. 5，8，9

Let n, m, o be perpendicular to the surface of OP respectively, and the radius of the three circles be R1, R2, R3, and R, then: R12 = R2 − (23R) 2 = 59r2, R22 = R2 − (13R) 2 = 89r2, R32 = R2 − (23R) 2 = R2 ℅ R12: R22: R32 = 5:8:9 ℅ the area ratio of the three circles is 5, 8, 9, so D is selected