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shtirl [24]
3 years ago
11

Is the relationship shown by the data linear? If so, model the data with an equation.

Mathematics
1 answer:
slega [8]3 years ago
4 0
Which ones are the x and which ones are the y
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On a stopwatch, the tip of the second hand moves 2 cm in 15 seconds
Agata [3.3K]
The given 2 cm is the length of the arc that the second hand had traveled for 15 seconds. Since the second hand completes a whole revolution in 60 seconds, it traveled approximately 1/4 of its circumference for 15 seconds. The length of the second hand is the radius of the circle. 
                            2 cm = (1/4) x 2π x (L)
Solving for L gives an answer of 1.273 cm. Thus, the answer is the second choice, 1.3 cm.
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3 years ago
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3 years ago
Any Netflix suggestions to watch?​
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Answer:

stranger things

Step-by-step explanation:

6 0
2 years ago
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
For each equation choose a value for x and then solve
ss7ja [257]

Answer:

Step-by-step explanation:

For each equation choose a value for x and then solve to find the corresponding y value that makes that equation true.

5 0
3 years ago
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