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EleoNora [17]
3 years ago
11

Sorry I suck at math but PLEASE HELP solve these 3!!!​

Mathematics
1 answer:
Leto [7]3 years ago
8 0

Answer:

1. x = 10

2. x = 8

3. x = 7

Step-by-step explanation:

1. Sum of Angles in a triangle = 180

(6x + 3) + (5x - 7) + 74 = 180

11x - 4 + 74 = 180

11x + 70 = 180

Subtract 70 from both sides

11x = 110

Divide both sides by 11

x = 10

2. (8x + 1) + (7x + 14) + 45 = 180

15x + 15 + 45 = 180

15x + 60 = 180

Subtract 60 from both sides

15x = 120

Divide both sides by 15

x = 8

3. We first have to get the 3rd angle: 180 - Exterior angle = 3rd angle

180 - 118 = 62

Now we can find the value of x

10x + (6 + 6x) + 62 = 180

16x + 68 = 180

Subtract 68 from both sides

16x = 112

Divide both sides by 16

x = 7

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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}.
7 0
3 years ago
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