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miv72 [106K]
3 years ago
10

Here is a riddle: “I am thinking of two numbers that add up to 5.678. The difference between them is 9.876. What are the two num

bers?”
The riddle can be represented with two equations. Write the equations
Mathematics
2 answers:
Svetradugi [14.3K]3 years ago
6 0

Answer:  

X + Y = 5.678

X - Y = 9.876

The two numbers: X = 7.777   Y = -2.099

Step-by-step explanation: solve by elimination

X + Y = 5.678

X - Y = 9.876 . Add the two equations so the y cancels. Solve for x

2x = 15.554 .  Divide both sides by 2

X = 7.777    Substitute in the original equations to find Y

7.777 + y = 5.678 . Subtract 7.777 from both sides

Y = - 2.099

check in the second equation:

7.777 -(-2.099) = 9.876

7.777 + 2.099 = 9.876 <u>True</u>

Bumek [7]3 years ago
4 0

Answer

:X + Y = 5.678

X ± 9.876 = Y

Step-by-step explanation:

needed points

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Fraction before painting = 5/6

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Lowest common factor of 6 and 30 = 30

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Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1
Varvara68 [4.7K]
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be x+y+z=1.

You're minimizing d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2} subject to the constraint f(x,y,z)=x+y+z=1. Note that d(x,y,z) and d(x,y,z)^2 attain their extrema at the same values of x,y,z, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

Take your partial derivatives and set them equal to 0:

\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

Adding the first three equations together yields

2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

The squared distance is then d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43, which means the shortest distance must be \sqrt{\dfrac43}=\dfrac2{\sqrt3}.
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3 years ago
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