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natita [175]
3 years ago
13

Which point is located on the x-axis?

Mathematics
2 answers:
Rama09 [41]3 years ago
7 0

Answer:

A

Step-by-step explanation:

Fiesta28 [93]3 years ago
5 0

Answer:

C

Step-by-step explanation:

its obvious.... i literally says which axis is X

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Point p’ (1,5) is the image of p (-3,1) under a translation<br><br> khan academy
GarryVolchara [31]

Answer:

<h2>X = \frac{-x}{3}</h2><h2>Y = 5y.</h2>

Step-by-step explanation:

We need to find the translation for which, (-3, 1) becomes (1, 5).

-3 will become 1, if it is divided by -3.

Hence, the translation for x axis is X = \frac{-x}{3}.

1 will become 5, when it is multiplied by 5.

Hence, The translation for y axis is Y = 5y.

6 0
3 years ago
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Simplify completely.<br><br> (w15/w5)4<br><br> Enter your answer in the box.
Vlad [161]

Answer:w40

Step-by-step explanation:

w15/w5 = w10

(w10)^4 = w40

3 0
3 years ago
Can i please get help with this geometry test?
RoseWind [281]

Answer:

im here for the points sorry

Step-by-step explanation:

6 0
3 years ago
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Simplify. Please show work
GuDViN [60]

Answer:

\frac{4b^5}{c^3}

Step-by-step explanation:

\frac{36b^4c^2}{9b^{-1}c^5}

\frac{36b^4c^2}{9b^{-1}c^5}\\\\\frac{4b^{4+1}}{c^{5-2}}\\\\\frac{4b^5}{c^3}

Hope this helps!

5 0
2 years ago
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
d^*(2,10,2\sqrt5)=46
d^*(2,10,-2\sqrt5)=46

So the two points on the surface z^2=xy closest to the point (7,11,0) are (2,10,\pm2\sqrt5).
5 0
3 years ago
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