Yes it would be 4604
hope this is helpful to you! :D
Since the whole squares that are surrounding 12 are 9 and 16, and their square roots are 3 and 4, we know that the square root of 12 must be in between 3 and 4.
Assuming that your own ticket is already secured.
From the 5 friends you can only choose 3.
The order in which these friends are chosen is not relevant. So this is a Combination problem.
Where the order is relevant the problem becomes a Permutation problem!
So the answer is 5 combination 3. = ⁵C₃
⁵C₃ = (5 * 4 * 3) / ( 3 * 2 * 1) = 10
You can take 10 groups of any 3 of your 5 friends.
Passes through (3, 0) and is perpendicular to 2x - y = 5
Turn our given equation into y=mx+b form.
2x - y = 5
Subtract 2x from both sides.
-y = -2x + 5
Divide both sides by -1.
y = 2x - 5
So, our slope is 2.
However, because our equation is perpendicular, we must find the negative reciprocal of 2.
Slope₁ = 2
Slope₂ = -1/2
y - y₁ = m(x - x₁) (3, 0)
Plug everything in.
y - 0 = -1/2(x - 3)
y = -1/2x + 3
~Hope I helped!~
Answer with explanation:
For, a Matrix A , having eigenvector 'v' has eigenvalue =2
The order of matrix is not given.
It has one eigenvalue it means it is of order , 1×1.
→A=[a]
Determinant [a-k I]=0, where k is eigenvalue of the given matrix.
It is given that,
k=2
For, k=2, the matrix [a-2 I] will become singular,that is
→ Determinant |a-2 I|=0
→I=[1]
→a=2
Let , v be the corresponding eigenvector of the given eigenvalue.
→[a-I] v=0
→[2-1] v=[0]
→[v]=[0]
→v=0
Now, corresponding eigenvector(v), when eigenvalue is 2 =0
We have to find solution of the system
→Ax=v
→[2] x=0
→[2 x] =[0]
→x=0, is one solution of the system.
