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zavuch27 [327]
3 years ago
7

What is the perimeter of the triangle? units Stuck? Watch a video or use a hint. Report

Mathematics
1 answer:
Snezhnost [94]3 years ago
4 0
The answer to your question is 30
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Four points are labeled on the number line. K L M N + 0.5 0 Which point best represents ? colm ? Point 2 Point M Point K Point N
MakcuM [25]

Answer:

point k and point n

K L M N

Step-by-step explanation:

6 0
3 years ago
Set A contains 35 elements and set B contains 22 elements. If there are 40 elements in (A ∪ B) then how many elements are in (A
Leona [35]

Answer:

17

Step-by-step explanation:

The computation of the number of elements are in (A ∩ B) is shown below;

Given that

Set A contains 35 elements

And, set B contains 22 elements

Now if there are 40 elements in (A ∪ B)

So, the number of elements are in (A ∩ B) is

= 35 + 22 - 40

= 17

5 0
3 years ago
Help me, please!! I wasn't there for the lesson.
liberstina [14]
The first option would be the right answer. Martha has 33, where jackson has 22
6 0
3 years ago
Read 2 more answers
• The shortest side of a triangle is 8 centimeters long. The longest side is 8 cm longer than the
gizmo_the_mogwai [7]
The shortest side is 9, longest is 17 of the triangle IS NOT a 90 degree triangle
5 0
3 years ago
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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