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irakobra [83]
3 years ago
10

Which system of equations has no solution?

Mathematics
2 answers:
allochka39001 [22]3 years ago
6 0

Answer:

B.

{ \binom{7x - 2y = 9}{7x - 2y = 13} }

tatuchka [14]3 years ago
6 0
B has no solution ! Hope this helps, sorry if I’m with wrong
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The perimeter of the garden is 10x + 8. find the missing side length
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Subtract the sum of the known sides from the perimeter to find the length of the missing side.
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Absolute value of _3
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Read 2 more answers
Find an example for each of vectors x, y ∈ V in R.
rjkz [21]

(a) Both conditions are satisfied with <em>x</em> = (1, 0) for \mathbb R^2 and <em>x</em> = (1, 0, 0) for \mathbb R^3:

||(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 \mathbb R^2, 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 \mathbb R^3 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 \mathbb R^3, you can use the vectors <em>x</em> = (1, 0, 0) and <em>y</em> = (1, 1, 1).

8 0
3 years ago
A man is five times as old as his son. Four years ago the man was thirteen times as old as his son.
skad [1K]

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,

=> 5x - 4 = 13(x - 4)

Now , solve the equation.

  • Open the brackets in R.H.S.

=> 5x - 4 = 13x - 52

  • 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.

=> 13x - 5x = 52 - 4

=> 8x = 48

  • Divide both the sides by 8

=> \frac{8x}{8} = \frac{48}{8}

=> x = 6

<u>CONCLUSION :-</u>

Age of son = 6 years

Age of man = 5×6 = 30 years

8 0
2 years ago
Find UV (Pic Stated Below)
natta225 [31]

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>

7 0
3 years ago
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