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dezoksy [38]
2 years ago
10

What is the solution to the system of equations? Use the substitution method. {y=−2x+53x−4y=2

Mathematics
1 answer:
klio [65]2 years ago
8 0

y =  - 2x + 53x - 4y = 2 \\  y  = 51x - 4y = 2 \\  > y = 51 x - 4y \\  > 51x - 4y = 2
y = 51x - 4y \\ 5y = 51x \\ y =  \frac{51}{5} x
Substitute 51/5x into
51x - 4( \frac{51}{5} x) = 2 \\ 51x - 40.8x = 2 \\ 10.2x = 2 \\ x = 2 \times  \frac{10}{102}  \\ x =  \frac{10}{51}  \: aka \: 0.196
Substitute that into
y =  \frac{51}{5}  \times  \frac{10}{51}  \\ y = 2
Hope this helps. Answer should be correct. Unless your quadratic equation is wrongly stated.
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(1 point) Find two unit vectors orthogonal to a=⟨4,5,0⟩a=⟨4,5,0⟩ and b=⟨0,1,−3⟩b=⟨0,1,−3⟩ Enter your answer so that the first no
Anni [7]

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 First vector

V=+ \dfrac{-15i+12j+4k}{\sqrt{385}}

Second vector

V=- \dfrac{-15i+12j+4k}{\sqrt{385}}

Step-by-step explanation:

Given that

Vector a ,  a= 4 i + 5 j + 0 k

Vector b , b= 0 i + 1 j  - 3 k

The orthogonal vector V is given as

V = a x b

V is the cross product of the above two vector which is orthogonal to the vectors a and b.

V=\begin{vmatrix}i & j & k\\ 4 & 5 & 0\\  0& 1 &-3 \end{vmatrix}\\V=i(-15-0)-j(-12-0)+k(4-0)\\V=-15i+12j+4k

The unit vectors

V=\pm \dfrac{-15i+12j+4k}{\sqrt{15^2+12^2+4^2}}\\\\V=+\dfrac{-15i+12j+4k}{\sqrt{385}}\\V=-\dfrac{-15i+12j+4k}{\sqrt{385}}

First vector

V=+ \dfrac{-15i+12j+4k}{\sqrt{385}}

Second vector

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