If the vector a =(k,1), b =(4, k) is collinear and the directions are opposite, then k =() There is something wrong with the answer and option, so I won't write it. But I want to ask you if it is 2 or -2, and the answer makes me confused!

The vector is collinear... k^2=4.k=2 or -2, because the direction is opposite... so k=-2
A=(-2,1)=-2(1,-1/2), b=(4,-2)=4(1,-1/2), so it's obvious.
Glad to answer, liamqy answers
If you have any questions you can ask,

The vector is collinear... k^2=4.k=2 or -2, because the direction is opposite... so k=-2
A=(-2,1)=-2(1,-1/2), b=(4,-2)=4(1,-1/2), so it's obvious.
Glad to answer, liamqy answers
If there's anything you do n' t understand,

If vectors a=(k,1), b=(4, k) are collinear and opposite, then |a-b|= If the vectors a=(k,1), b=(4, k) are collinear and opposite, then |a-b|=

Vector a=(k,1) and b=(4, k) in opposite direction
The slope of the line where the vector is located should be the same.
I. e. k/1=4/k,
Take -2, when k=2, the two vectors are in the same direction.
A-b=(k-4,1-k)=(-6,3)
Then |a-b |=3 nos.5

If the vector a=(-1, x) and b=(-x,2) are collinear and in the same direction, then x=

Vector a=(-1, x) and b=(-x,2) are collinear and in the same direction
Let (-1, x)= k*(-x,2)=(-kx,2k)
-1=-Kx, x =2k
Solution k=√2/2, x=√2
So x =√2

Let f (x)=x2+ax be an even function on R, 1 Fact the value of a 2 Prove by definition that f (x) is an increasing function on (0, positive infinity) Let f (x)=x2+ax be an even function on R, 1 Is the value of the number a 2 Prove by definition that f (x) is an increasing function on (0, positive infinity)

1 Because even function on R is f (-x)=f (x)(-x)2+a (-x)=x2+axx2-ax=x2+ax2ax=0a=02 Let x1> x2>0f (x1)-f (x2)=x1 2-x22=(x1+x2)(x1-x2) Because x1>0x2>0, so x1+x2>0, because x1> x2, so x1-x2>0, so f (x1)-f (x2)>...

What is the condition that a=0 is an even function f (x)=x2+ax+1 What condition is "a=0" is" function f (x)=x2+ax+1 is even function "

A Necessary and Sufficient Condition for "a=0" is a Function f (x)=x2+ax+1 Is Even"
(1) If a=0, then f (x)=x2+1 is an even function, i.e., energy forward and energy backward
(2) If the function f (x)=x2+ax+1 is even, then the axis of symmetry is y, i.e. x=-a/2=0
I. e. that aft can push forward
A necessary and sufficient condition that the front is the back.

A Necessary and Sufficient Condition for "a=0" is a Function f (x)=x2+ax+1 Is Even"
(1) If a=0, then f (x)=x2+1 is an even function, i.e., the energy forward pushes the energy backward.
(2) If the function f (x)=x2+ax+1 is even, then the axis of symmetry is y, i.e. x=-a/2=0
I. e. that aft can push forward
A Necessary and Sufficient Condition of

7.[2011 Mentougou-Module] is a non-zero vector," function is even function "is () of "(A) sufficient but unnecessary condition (B) must 7.[2011 Mentougou-Module] is a non-zero vector, and "function is even function" is () of (A) sufficient but unnecessary condition (B) must

Select (C) Sufficient and Necessary Condition
F (x)=(ax+b) square=(ax) square+2abx+b square a, b is vector
Sufficient: if the function is even, then 2abx=0 is constant, so a is vertical b
Necessary: a vertical b, ab=0, f (x)=(ax+b) square=(ax) square+2abx+b square=(ax) square+b square,
Now the function is even

If the vector a =(k,1), b =(4, k) is collinear and the directions are opposite, then k =() There is something wrong with the answer and option, so I won't write it. But I want to ask you if it is 2 or -2, and the answer makes me confused!