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sashaice [31]
4 years ago
12

Point V lies between points U and W on line UW. The distance between U and V is 2x-4, and the distance between V and W is 4x+10.

If UW = 9x – 9, what is UW in units?
Mathematics
1 answer:
gogolik [260]4 years ago
8 0

Answer:

36UNITS

Step-by-step explanation:

If point V lies between U and W, then UV+VW = UW

given parameters

UV = 2x-4

VW = 4x+10

UW = 9x-9

Substituting the given functions in the formula first, we will have;

2x-4+4x+10 = 9x-9

6x+6 = 9x-9

Collect like terms

6x-9x = -9-6

-3x = -15

x = -15/-3

x = 5

Substitute x = 5 into UW.

Since UW = 9x-9

UW = 9(5)-9

UW = 45-9

UW = 36

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Show that W is a subspace of R^3.
musickatia [10]

Answer:

Check the two conditions of Subspace.

Step-by-step explanation:

If W is a Subspace of a vector space, V then it should satisft the following conditions.

1) The zero element should be in W.

Zero element can be different for different vector spaces. For examples, zero vector in $ \math{R^2} $ is (0, 0) whereas, zero element in $ \math{R^3} $ is (0, 0 ,0).

2) For any two vectors, $ w_1 $ and $ w_2 $ in W, $ w_1 + w_2 $ should also be in W.

That is, it should be closed under addition.

3) For any vector $ w_1 $ in W and for any scalar, $ k $ in V, $ kw_1 $ should be in W.

That is it should be closed in scalar multiplication.

The conditions are mathematically represented as follows:

1) 0$ \in $ W.

2) If $ w_1 \in W; w_2 \in W $ then $ w_1 + w_2 \in W $.

3) $ \forall k \in V, and \hspace{2mm} \forall w_1 \in W \implies kw_1 \in W

Here V = $ \math{R^3} $ and W = Set of all (x, y, z) such that $ x - 2y + 5z = 0 $

We check for the conditions one by one.

1) The zero vector belongs to the subspace, W. Because (0, 0, 0) satisfies the given equation.

i.e., 0 - 2(0) + 5(0) = 0

2) Let us assume $ w_1 = (x_1, y_1, z_1) $ and $ w_2 = (x_2, y_2, z_2) $ are in W.

That means: $ x_1 - 2y_1 + 5z_1 = 0 $ and

$ x_2 - 2y_2 + 5z_2 = 0 $

We should check if the vectors are closed under addition.

Adding the two vectors we get:

$ w_1 + w_2 = x_1 + x_2 - 2(y_1 + y_2) + 5(z_1 + z_2) $

$ = x_1 + x_2 - 2y_1 - 2y_2 + 5z_1 + 5z_2 $

Rearranging these terms we get:

$ x_1 - 2y_1 + 5z_1 + x_2 - 2y_2 + 5z_2 $

So, the equation becomes, 0 + 0 = 0

So, it s closed under addition.

3) Let k be any scalar in V. And $ w_1 = (x, y, z) \in W $

This means $ x - 2y + 5z = 0 $

$ kw_1 = kx - 2ky + 5kz $

Taking k common outside, we get:

$ kw_1 = k(x - 2y + 5z) = 0 $

The equation becomes k(0) = 0.

So, it is closed under scalar multiplication.

Hence, W is a subspace of $ \math{R^3} $.

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Need help ASAP!!!!!!!
Temka [501]

D

Step-by-step explanation:

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Find the radius of a circle that has this measurement. d=2.25cm
yuradex [85]
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Step-by-step explanation:

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What is the equation of the line, in slope-intercept form, that is perpendicular to the line with the equation y=6x−1 and passes
Harman [31]
Perpendicular means has slope that multiplies to -1

y=6x-1
slope=6

6 times what=-1
what=-1/6

so

the equation of a line that passes through the point (x1,y1) and has a slope of m is
y-y1=m(x-x1)
given
point (-6,4) and slope is -1/6

y-4=-1/6(x-(-6))
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expand
y-4=-1/6x-1
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4th choice
8 0
3 years ago
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