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jenyasd209 [6]
3 years ago
7

For 3 and 4 find the slope of the line that passes through the given points

Mathematics
1 answer:
maks197457 [2]3 years ago
5 0
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Based on the graph, which inequality is correct for a number that is to the right of -3?
RSB [31]
Positive infinity is going after three hope this helps
8 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
Y^2 x^3-7x = ysin(x+y)
d1i1m1o1n [39]

Answer: -54

Step-by-step explanation: 53 + 1 = 54

3 0
3 years ago
Select the sets of data where it would be better to use a histogram than a
Contact [7]

Answer:

Weight of students in your class

Daily rainfall in Miami over a month

Step-by-step explanation:

A histogram is applied when the data is discrete/ continuous to present a visual understanding of the data by showing the number of data set fitting is a specified range.

A dot plot utilizes data which has only one measured variable, where the variable can be organized into categories.

8 0
3 years ago
For the members of a neighborhood meeting group, the probability of randomly selecting a member who likes to cook is 42%, the pr
Monica [59]

Answer: Probability of randomly selecting a member who likes to cook or likes to sew = 58%

Step-by-step explanation:

Since we have given that

Probability of a randomly selecting a member who likes to cook=P(C) = 42%

Probability of a randomly selecting a member who likes to sew =P(S)= 23%

Probability of a randomly selecting a member who likes to both cook and sew

P(C\cap S)=7\%

As we know the "Probability rules " :

P(C\cup S)=P(C)+P(S)-P(S\cap C)\\\\P(C\cup S)=42+23-7\\\\P(C\cup S)=58

So, Probability of randomly selecting a member who likes to cook or likes to sew = 58%



5 0
3 years ago
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