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Yuliya22 [10]
2 years ago
13

Nothinggdweiuge3fgjhdsgfweyugfehfbnwefvcejqyfjeqhvfv?

Mathematics
1 answer:
pentagon [3]2 years ago
3 0

Answer:

hi, need help with anything?

Step-by-step explanation:

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Please help! Thank you!!
vampirchik [111]
Y=-2/3x+2 i believe!!
7 0
3 years ago
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Food Mart<br>job<br>Tuesday to Friday<br>Pay: £56 for each day in pounds​
AURORKA [14]
You do 4 x 56 = 224. It’s just multiplication.
4 0
3 years ago
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If you add one-third of a number to the number itself, you get 16.what is the number?
GuDViN [60]
The answer to this is 12.


One third of 12 is 4.

12+4=16

Comment if you have ANY questions!

Hope this helps!
7 0
3 years ago
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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
What is it called in cubic inch of a cylinder with a height of 4in and a base radius of 5in to the nearest tenths place?
Maru [420]

Answer:

1,256.8in

Step-by-step explanation:

Given data

Height h =4in

Radius r= 5in

The expression for the volume of cylinder is

V=πr²h

Substitute

V=3.142*4²*5

V=3.142 *16*25

V=1,256.8in³

6 0
3 years ago
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