Kevin needs to convert 620 millimeters per minute to meters per hour. Which conversion factors should he use?

1 answer:

7 0

Try this option:
1 meter=1000 mm; 1 hour=60 min. From another side: 1 mm=0.001 m. and 1 min=1/60 h.
Using this rule:
620 mm/h=0.06*620 m/h=37.2 meters per hour.

answer: 0.06

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Find an example for each of vectors x, y ∈ V in R.

rjkz [21]

(a) Both conditions are satisfied with x = (1, 0) for $\mathbb R^2$ and x = (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 x or y is the zero vector, or if x = y. For example, in $\mathbb R^2$ , take x = (0, 0) and y = (1, 1). Then

||x + y|| = ||(0, 0) + (1, 1)|| = ||(1, 1)|| = √(1² + 1²) = √2

||x|| + ||y|| = ||(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 x = (1, 0) and y = (0, 1), then

||x + y|| = ||(1, 0) + (0, 1)|| = ||(1, 1)|| = √(1² + 1²) = √2

||x|| + ||y|| = ||(1, 0)|| + ||(0, 1)|| = √(1² + 0²) + √(0² + 1²) = 2

and of course √2 < 2.

Similarly, in $\mathbb R^3$ you can use x = (0, 0, 0) and y = (1, 1, 1) for the equality, and x = (1, 0, 0) and y = (0, 1, 0) for the inequality.

x • y = ||x|| ||y|| cos(θ),

where θ is the angle between the vectors x and y. Both sides are scalar, so taking the norm gives

||x • y|| = ||(||x|| ||y|| cos(θ)|| = ||x|| ||y|| |cos(θ)|

Suppose x = (0, 0) and y = (1, 1). Then

||x • y|| = |(0, 0) • (1, 1)| = 0

||x|| • ||y|| = ||(0, 0)|| • ||(1, 1)|| = 0 • √2 = 0

For the inequality, recall that cos(θ) is bounded between -1 and 1, so 0 ≤ |cos(θ)| ≤ 1, with |cos(θ)| = 0 if x and y are perpendicular to one another, and |cos(θ)| = 1 if x and y are (anti-)parallel. You get everything in between for any acute angle θ. So take x = (1, 0) and y = (1, 1). Then

||x • y|| = |(1, 0) • (1, 1)| = |1| = 1

||x|| • ||y|| = ||(1, 0)|| • ||(1, 1)|| = 1 • √2 = √2

In $\mathbb R^3$ , you can use the vectors x = (1, 0, 0) and y = (1, 1, 1).

8 0

3 years ago

How would I do this problem?

kenny6666 [7]

If it's a geometric sequence then:

$a_1=27;\ a_2=27\\\\r=\dfrac{a_2}{a_1}\to r=\dfrac{27}{36}=\dfrac{3}{4}=0.75\\\\a_{n+1}=a_nr\\\\a_3=27\cdot0.75=20.25\ CORRECT$