Which system of equations has no solution?
2 answers:
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(a) Both conditions are satisfied with <em>x</em> = (1, 0) for and <em>x</em> = (1, 0, 0) for
:
||(1, 0)|| = √(1² + 0²) = 1
max{1, 0} = 1
||(1, 0, 0)|| = √(1² + 0² + 0²) = 1
max{1, 0, 0} = 1
(b) This is the well-known triangle inequality. Equality holds if one of <em>x</em> or <em>y</em> is the zero vector, or if <em>x</em> = <em>y</em>. For example, in , take <em>x</em> = (0, 0) and <em>y</em> = (1, 1). Then
||<em>x</em> + <em>y</em>|| = ||(0, 0) + (1, 1)|| = ||(1, 1)|| = √(1² + 1²) = √2
||<em>x</em>|| + ||<em>y</em>|| = ||(0, 0)|| + ||(1, 1)|| = √(0² + 0²) + √(1² + 1²) = √2
The left side is strictly smaller if both vectors are non-zero and not equal. For example, if <em>x</em> = (1, 0) and <em>y</em> = (0, 1), then
||<em>x</em> + <em>y</em>|| = ||(1, 0) + (0, 1)|| = ||(1, 1)|| = √(1² + 1²) = √2
||<em>x</em>|| + ||<em>y</em>|| = ||(1, 0)|| + ||(0, 1)|| = √(1² + 0²) + √(0² + 1²) = 2
and of course √2 < 2.
Similarly, in you can use <em>x</em> = (0, 0, 0) and <em>y</em> = (1, 1, 1) for the equality, and <em>x</em> = (1, 0, 0) and <em>y</em> = (0, 1, 0) for the inequality.
(c) Recall the dot product identity,
<em>x</em> • <em>y</em> = ||<em>x</em>|| ||<em>y</em>|| cos(<em>θ</em>),
where <em>θ</em> is the angle between the vectors <em>x</em> and <em>y</em>. Both sides are scalar, so taking the norm gives
||<em>x</em> • <em>y</em>|| = ||(||<em>x</em>|| ||<em>y</em>|| cos(<em>θ</em>)|| = ||<em>x</em>|| ||<em>y</em>|| |cos(<em>θ</em>)|
Suppose <em>x</em> = (0, 0) and <em>y</em> = (1, 1). Then
||<em>x</em> • <em>y</em>|| = |(0, 0) • (1, 1)| = 0
||<em>x</em>|| • ||<em>y</em>|| = ||(0, 0)|| • ||(1, 1)|| = 0 • √2 = 0
For the inequality, recall that cos(<em>θ</em>) is bounded between -1 and 1, so 0 ≤ |cos(<em>θ</em>)| ≤ 1, with |cos(<em>θ</em>)| = 0 if <em>x</em> and <em>y</em> are perpendicular to one another, and |cos(<em>θ</em>)| = 1 if <em>x</em> and <em>y</em> are (anti-)parallel. You get everything in between for any acute angle <em>θ</em>. So take <em>x</em> = (1, 0) and <em>y</em> = (1, 1). Then
||<em>x</em> • <em>y</em>|| = |(1, 0) • (1, 1)| = |1| = 1
||<em>x</em>|| • ||<em>y</em>|| = ||(1, 0)|| • ||(1, 1)|| = 1 • √2 = √2
In , you can use the vectors <em>x</em> = (1, 0, 0) and <em>y</em> = (1, 1, 1).
Answer:
Age of son = 6 years
Age of man = 5×6 = 30 years
Step-by-step explanation:
<u>GIVEN :-</u>
- A man is 5 times as old as his son. (In Present)
- 4 years ago , the man was 13 times as old as his son
<u>TO FIND :-</u>
- The present ages of the man & his son.
<u>SOLUTION :-</u>
Let the present age of son be 'x'.
⇒ Present age of man = 5x
4 years ago ,
Age of son = (Present age of son) - 4 = x - 4
Age of man = (Present age of man) - 4 = 5x - 4
The man was thirteen times as old as his son. So,
Now , solve the equation.
- Open the brackets in R.H.S.
- Take 5x to R.H.S. and -52 to L.H.S. Also , take care of their signs because they are getting displaced from L.H.S. to R.H.S. or vice-versa.
- Divide both the sides by 8
<u>CONCLUSION :-</u>
Age of son = 6 years
Age of man = 5×6 = 30 years
Answer:
UV=29
Step-by-step explanation:
In right triangles AQB and AVB,
∠AQB = ∠AVB ...(i) {Right angles}
∠QBA = ∠VBA ...(ii) {Given that they are equal}
We know that sum of all three angles in a triangle is equal to 180 degree. So wee can write sum equation for each triangle
∠AQB+∠QBA+∠BAQ=180 ...(iii)
∠AVB+∠VBA+∠BAV=180 ...(iv)
using (iii) and (iv)
∠AQB+∠QBA+∠BAQ=∠AVB+∠VBA+∠BAV
∠AVB+∠VBA+∠BAQ=∠AVB+∠VBA+∠BAV (using (i) and (ii))
∠BAQ=∠BAV...(v)
Now consider triangles AQB and AVB;
∠BAQ=∠BAV {from (v)}
∠QBA = ∠VBA {from (ii)}
AB=AB {common side}
So using ASA, triangles AQB and AVB are congruent.
We know that corresponding sides of congruent triangles are equal.
Hence
AQ=AV
5x+9=7x+1
9-1=7x-5x
8=2x
divide both sides by 2
4=x
Now plug value of x=4 into UV=7x+1
UV=7*4+1=28+1=29
<u>Hence UV=29 is final answer.</u>
