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timurjin [86]
1 year ago
6

What is the value of(-4)(-4)(-4)

Mathematics
2 answers:
myrzilka [38]1 year ago
8 0

Answer:

-64

Step-by-step explanation:

-4x-4x-4

Jlenok [28]1 year ago
7 0

Answer:

-64

Please Mark Brainliest If This Helped!

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What are all the shapes that are used to create the net of a cylinder
Ksenya-84 [330]
2 circles and for the part in the middle if you look at it as a net it will be a rectangle just rolled up. That is called a lateral surface or a lateral face
6 0
3 years ago
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
Determine if this relationship forms a direct variation. Verify your answer.
leva [86]

we know that

A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form y/x=k or y=kx

so


we have

a) for x=1 y=0.50--------> y/x=0.50/1------> 0.50

b) for x=2 y=1--------> y/x=1/2-------> 0.50

c) for x=3 y=1.50--------> y/x=1.50/3-----> 0.50

d) for x=5 y=2.50--------> y/x=2.50/5----> 0.50

the value of k is equal to 0.50

so

<span>the relationship forms a direct variation. </span>

the equation is

y=0.50*x


Verify

for x=1

y=0.50*(1)------> y=0.50-----> is correct


for x=2

y=0.50*(2)------> y=1.00-----> is correct


for x=3

y=0.50*(3)------> y=1.50-----> is correct


for x=5

y=0.50*(5)------> y=2.50-----> is correct

4 0
2 years ago
Read 2 more answers
Pls help asap<br>AABC ~ ADEF B. Х 8 А 'C E 3 6. D ' x = [?] Enter​
AleksAgata [21]

Answer:

\boxed{ \tt{x = 4}}

Step-by-step explanation:

\frac{3}{x}  =  \frac{6}{8}  \\ 6x = 24 \\ x = 4

4 0
3 years ago
How many places do you need to move the decimal to the right to write
GenaCL600 [577]

Answer:

7 places.

Step-by-step explanation:

There are seven zeros, so in powers of ten notation it is written as 10^{-7}

2.1 \times 10^{-7}

6 0
2 years ago
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