Let a 1, a 2, a 3, a 4 be linearly related, and a 2, a 3, a 4, a 5 be linearly independent. Let a 1 be linearly expressed by a 2, a 3, a 4

Let a 1, a 2, a 3, a 4 be linearly related, and a 2, a 3, a 4, a 5 be linearly independent. Let a 1 be linearly expressed by a 2, a 3, a 4

Because A2, A3, A4, A5 are linearly independent,
Thus, a 2, a 3 and a 4 are linearly independent
Moreover, A1, A2, A3, A4 are linearly correlated, so there are constants K1, K2, K3, K4 which are not all zero,
Let k1a1 + k2a2 + k3a3 + k4a4 = 0 (1)
It is easy to know that K 1 is not zero, otherwise a 2, a 3, a 4 are linearly correlated from (1)
So (1) can be changed into
a1=(-k2/k1)a2+(-k3/k1)a3+(-k4/k1)a4
The proof is complete

7x ^ 2 + 14x-3600 factorization

=7(x+1)^2-3607
=7[(x+1)^2-3607/7]
=7 (x + 1 + radical (3607 / 7)) (x + 1-radical (3607 / 7))

Given x ~ n (μ, σ ^ 2), randomly select the sample of n = 14, and calculate the probability that the absolute value of the difference between the sample mean and the population mean is less than 1.5,
（1）σ^2=25
(2) σ ^ 2 is unknown, but s ^ 2 = 17.26

X[i]~N（μ,σ^2） i=1、2、3…… fourteen
Then the average value y ~ n (μ, σ ^ 2 / 14)
1)σ^2=25 Y~N（μ,25/14）
Then p (| Y - μ|

If the left side of the quadratic equation of one variable x + MX + m + 3 = 0 is a complete square expression, the value of M is obtained

Square of X + MX + m + 3 = 0
x^2+mx+m+3=0
(x+m/2)^2-m^2/4+m+3=0
Complete square form
-m^2/4+m+3=0
m^2-4m-12=0
m=6,m=-2

In the trapezoid ABCD, AD / / BC, the diagonal line AC is perpendicular to BD and O ∠ DBC is equal to 30, and the trapezoid median line intersects with AB and DC at the point Mn
Draw your own picture

Because the diagonal AC is perpendicular to o, DBC is equal to 30,
oa=1/2ad,oc=1/2bc,
oa+oc=ac=1/2(ad+bc)=mn.

The x power of mathematical problem 9 * the x power of 3 = the X + 4 power of X
9 to the power of X * 3 to the power of x = x + 4 to the power of X
The nth power of 2 = 3, the nth power of 3 = 4, find the nth power of 6
The nth power of 2 = 3, the nth power of 3 = 4, find the nth power of 36
As of today

1.
9^x*3^x=x^(x+4)
Is there a problem with the title?
two
2^n=3,3^n=4
So 6 ^ n = (2 * 3) ^ n = 2 ^ n * 3 ^ n = 3 * 4 = 12
three
2 ^ n = 3,3 ^ n = 4, then 36 ^ n = (6 * 6) ^ n = (6 ^ n) * (6 ^ n) = 12 * 12 = 144

Compare 1111 out of 1111 with 1111 out of 11111

1111 of 11111

The minimum value of the absolute value - 1 of the function y = 2x + 4x is___
Also, I mainly want to know what a quadratic function with absolute value is like?
The minimum value of y = 2x ^ 2 + 4 │ x │ - 1
I think the book I bought has the following solution: y = 2 (│ x │ + 1) ^ 2-3
=⑴2（x+1）^2-3，x》0
⑵2（x-1）^2-3，x《0
From the image (the image is drawn on the book, but not here), we can see that when x = 0, the minimum value of Y is - 1
From (1) 2 (x + 1) ^ 2-3, X "0, (2) 2 (x-1) ^ 2-3, X" 0 "
It seems that when x = 1 or - 1, the value of Y is not only - 3,
Also, some of the images on the book are dotted lines and some are solid lines. Why?

It seems that you have to work harder in mathematics. Your book has made it very clear. Y = 2 (| x | + 1) ^ 2-3 = (1) 2 (x + 1) ^ 2-3 (x > 0 or x = 0) (2) 2 (1-x) ^ 2-3 (x0 or x = 0). Its image is a part of the image of quadratic function y = 2 (x + 1) ^ 2-3, and the image of independent variable x satisfying the condition (x > 0 or x = 0)

When a motor is connected to a circuit with a voltage of 220 V, the current is 4a. If the coil resistance of the motor is 4 Ω, the heat generated per minute is ()
A. 52800 coke B. 3840 coke C. 4800 coke D. 14080 coke

The heat produced by the coil resistance of motor per minute: q = i2rt = (4a) 2 × 4 Ω × 60s = 3840j

If passing through point a (4,2), two straight lines can be made to be tangent to circle C: (x-3m) ^ 2 + (y-4m) ^ 2 = 5 (M + 4), and the value range of M can be obtained
If two straight lines can be made through point a (4,2) and tangent to circle C: (x-3m) ^ 2 + (y-4m) ^ 2 = 5 (M + 4), then point a is on the edge of circle C_______ (inside / outside / above), the value range of M

Point a is outside circle C
If two straight lines are tangent to circle C, then AC ^ 2 > 5 (M + 4)
5(m+4)>0
m>-4
(3m-4)^2+(4m-2)^2>5(m+4)
5m^2-9m>0
-4