Answer:
Step-by-step explanation:
Given:
u = 1, 0, -4
In unit vector notation,
u = i + 0j - 4k
Now, to get all unit vectors that are orthogonal to vector u, remember that two vectors are orthogonal if their dot product is zero.
If v = v₁ i + v₂ j + v₃ k is one of those vectors that are orthogonal to u, then
u. v = 0 [<em>substitute for the values of u and v</em>]
=> (i + 0j - 4k) . (v₁ i + v₂ j + v₃ k) = 0 [<em>simplify</em>]
=> v₁ + 0 - 4v₃ = 0
=> v₁ = 4v₃
Plug in the value of v₁ = 4v₃ into vector v as follows
v = 4v₃ i + v₂ j + v₃ k -------------(i)
Equation (i) is the generalized form of all vectors that will be orthogonal to vector u
Now,
Get the generalized unit vector by dividing the equation (i) by the magnitude of the generalized vector form. i.e
Where;
|v| =
|v| =
=
This is the general form of all unit vectors that are orthogonal to vector u
where v₂ and v₃ are non-zero arbitrary real numbers.
Let x be the 1st odd number, and x+2 the second odd consecutive number:
(x)(x + 2) = 6[((x) + (x+2)] -1
x² + 2x = 6(2x + 2) - 1
x² + 2x = 12x +12 - 1
And x² - 10x - 11=0
Solve this quadratic expression:
x' = [+10 +√(10²- 4.(1)(-11)]/2 and x" = [+10 -√(10²- 4.(1)(-11)]/2
x' = [10 + √144]/2 and x" = [10 - √64]/2
x' = (10+12)/2 and x" = (10-12)/2
x = 11 and x = -1
We have 2 solutions that satisfy the problem:
1st for x = 11, the numbers at 11 and 13
2nd for x = - 1 , the numbers are -1 and +1
If you plug each one in the original equation :(x)(x + 2) = 6[((x) + (x+2)] -1
you will find that both generates an equlity
Formula of the line
y=mx+b.
For parallel line slope m is the same, so m= -2.
y=-2x+b.
We need to find b,using the point (3,1).
1= - 2*3+b
b=7
y = -2x +7
Answer is A. y = -2x +7.
Answer:
yes
Step-by-step explanation:
because side a (4) and side b (5) added together are greater or equal to side c (8)
Answer:
ok
Step-by-step explanation:
whats the question?