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kaheart [24]
4 years ago
8

How would you go about finding the solution to this system of three equations in three variables? Be specific. For example, you

might want to add the first two equations first, or multiply the third one by two first and use elimination, etc. (No right answer exists.)
2x – y + z = 7

x + 2y – 5z = -1

x – y = 6
Mathematics
1 answer:
kupik [55]4 years ago
3 0
I:2x – y + z = 7
II:x + 2y – 5z = -1
III:x – y = 6

you can first use III and substitute x or y to eliminate it in I and II (in this case x):
III: x=6+y
-> substitute x in I and II:
I': 2*(6+y)-y+z=7
12+2y-y+z=7
y+z=-5

II':(6+y)+2y-5z=-1
3y+6-5z=-1
3y-5z=-7

then you can subtract II' from 3*I' to eliminate y:
3*I'=3y+3z=-15

3*I'-II':
3y+3z-(3y-5z)=-15-(-7)
8z=-8
z=-1

insert z in II' to calculate y:
3y-5z=-7
3y+5=-7
3y=-12
y=-4

insert y into III to calculate x:
x-(-4)=6
x+4=6
x=2

so the solution is
x=2
y=-4
z=-1
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Find u, v , u , v , and d(u, v) for the given inner product defined on Rn. u = (0, −4), v = (5, 3), u, v = 3u1v1 + u2v2
tigry1 [53]

Answer:

=-12

d(u,v)=2\sqrt{31}

Step-by-step explanation:

We are given that inner product defined on R_n

=3u_1v_1+u_2v_2

u=(0,-4),v=(5,3)

We have to find the value of <u,v> and d(u,v)

We have u_1=0,u_2=-4,v_1=5,v_2=3

Substitute the value then we get

=3(0\cdot5)+(-4)(3)=-12

Now, d(u,v)=\left \|v-u\right \|

Using this formula we get

d(u,v)=\left \| (5,7) \right \|=\sqrt{3(5)^2+(7)^2}=\sqrt{75+49}=\sqrt{124}

d(u,v)=2\sqrt{31}

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3 years ago
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Please help me with the question below
jonny [76]

Answer:

a) Angle XBC = 55 because XY is parallel with BD

b) As given, XB = XC, so that triangle XBC is the equilateral triangle.

As XBC is equilateral triangle, angle XBC is equal to angle XCB

Angle XBC = 55° => Angle XCB = 55°

In a triangle, the total value of three corners are equal to 180°

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=> 55° + 55° + Angle BXC = 180 °

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5 0
3 years ago
Read 2 more answers
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