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grin007 [14]
2 years ago
7

How many different triangles can you make if you are given the measurement for two angles in the length of a side that is NOT sh

ared?

Mathematics
2 answers:
brilliants [131]2 years ago
6 0

Answer:

Only 1 triangle can be made from the given measurements.

Step-by-step explanation:

A triangle is a shape with three sides and three angles. It can be classified either according to its sides or by its angles.

From the given question, applying one of the theorems of triangle, we have:

          35^{0} + 52^{0} + x^{0} = 180^{0} (the sum of angles in a triangle)

               85^{0} + x^{0} = 180^{0}

                    x^{0}   = 180^{0} -  85^{0}

                  x^{0}   = 95^{0}

Therefore, the third angle of the triangle ought to be 95^{0}, if the given side was shared.

But the length of a side is not shared, thus only one triangle can be made due to some restrictions regarding its sides and angles.

Irina18 [472]2 years ago
3 0

Answer:

  One

Step-by-step explanation:

Given two angles, you know all three angles. Only one triangle will have those angle measures and the given side length.

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Find the first four term of (1 – 2y) –4​
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7 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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