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Dennis_Churaev [7]
3 years ago
15

Factor. Check by multiplying. x^2 - 3x

Mathematics
2 answers:
bazaltina [42]3 years ago
7 0
Both term has factor of x

So that would be x² - 3x = x·x - 3·x = x( x-3 )

You can check by distribute x back to x-3:

x( x-3 ) = x·x - 3·x = x² - 3x. So our answer is correct.

Final answer: x( x-3 )

Hope this helps.
serious [3.7K]3 years ago
7 0
X^2-3x

x(x)- 3x

=x(x-3)
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Y=x-3 what is my answer
ale4655 [162]

Answer:

x = 3 or y = -3

Step-by-step explanation:

To find x, substitute y with a 0.

0=x-3. Minus the x on both sides and x will equal 3

To find y, substitute x with a 0.

y=0-3. 0-3 equals -3 which means that y will equal -3

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What is 7/8 divided by 11/16
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Find the point(s) on the surface z^2 = xy 1 which are closest to the point (7, 11, 0)
leonid [27]
Let P=(x,y,z) be an arbitrary point on the surface. The distance between P and the given point (7,11,0) is given by the function

d(x,y,z)=\sqrt{(x-7)^2+(y-11)^2+z^2}

Note that f(x) and f(x)^2 attain their extrema, if they have any, at the same values of x. This allows us to consider the modified distance function,

d^*(x,y,z)=(x-7)^2+(y-11)^2+z^2

So now you're minimizing d^*(x,y,z) subject to the constraint z^2=xy. This is a perfect candidate for applying the method of Lagrange multipliers.

The Lagrangian in this case would be

\mathcal L(x,y,z,\lambda)=d^*(x,y,z)+\lambda(z^2-xy)

which has partial derivatives

\begin{cases}\dfrac{\mathrm d\mathcal L}{\mathrm dx}=2(x-7)-\lambda y\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dy}=2(y-11)-\lambda x\\\\\dfrac{\mathrm d\mathcal L}{\mathrm dz}=2z+2\lambda z\\\\\dfrac{\mathrm d\mathcal L}{\mathrm d\lambda}=z^2-xy\end{cases}

Setting all four equation equal to 0, you find from the third equation that either z=0 or \lambda=-1. In the first case, you arrive at a possible critical point of (0,0,0). In the second, plugging \lambda=-1 into the first two equations gives

\begin{cases}2(x-7)+y=0\\2(y-11)+x=0\end{cases}\implies\begin{cases}2x+y=14\\x+2y=22\end{cases}\implies x=2,y=10

and plugging these into the last equation gives

z^2=20\implies z=\pm\sqrt{20}=\pm2\sqrt5

So you have three potential points to check: (0,0,0), (2,10,2\sqrt5), and (2,10,-2\sqrt5). Evaluating either distance function (I use d^*), you find that

d^*(0,0,0)=170
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So the two points on the surface z^2=xy closest to the point (7,11,0) are (2,10,\pm2\sqrt5).
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attashe74 [19]

Answer:

x = -3/14

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