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murzikaleks [220]
3 years ago
13

Which unit should be used to measure mass of pencil?

Mathematics
2 answers:
SCORPION-xisa [38]3 years ago
6 0

Answer:

grams :)

Step-by-step explanation:

I looked it up

alexandr402 [8]3 years ago
4 0

Answer:

Grams

Step-by-step explanation:

Pencils are pretty light, so they should be measured in grams.

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Find the minimum and maximum of f(x,y,z)=x^2+y^2+z^2 subject to two constraints, x+2y+z=4 and x-y=8.
Alika [10]
The Lagrangian for this function and the given constraints is

L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-4)+\lambda_2(x-y-8)

which has partial derivatives (set equal to 0) satisfying

\begin{cases}L_x=2x+\lambda_1+\lambda_2=0\\L_y=2y+2\lambda_1-\lambda_2=0\\L_z=2z+\lambda_1=0\\L_{\lambda_1}=x+2y+z-4=0\\L_{\lambda_2}=x-y-8=0\end{cases}

This is a fairly standard linear system. Solving yields Lagrange multipliers of \lambda_1=-\dfrac{32}{11} and \lambda_2=-\dfrac{104}{11}, and at the same time we find only one critical point at (x,y,z)=\left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right).

Check the Hessian for f(x,y,z), given by

\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}

\mathbf H is positive definite, since \mathbf v^\top\mathbf{Hv}>0 for any vector \mathbf v=\begin{bmatrix}x&y&z\end{bmatrix}^\top, which means f(x,y,z)=x^2+y^2+z^2 attains a minimum value of \dfrac{480}{11} at \left(\dfrac{68}{11},-\dfrac{20}{11},\dfrac{16}{11}\right). There is no maximum over the given constraints.
7 0
3 years ago
HELP please! ‍♂️<br><br> Last one is 0.5 &lt; x &lt; 1.5<br><br> And none
Ket [755]

1.5 is lesser than 0.5. Oh, wait nevermind I meant to say -1.5 is lesser than 0.5

3 0
2 years ago
What's the answer to 16-12+12x17
daser333 [38]
16-12+12x17= 208 u should use a calcuator
4 0
2 years ago
Read 2 more answers
PLS HELP THIS IS DUE TODAY
Anastasy [175]

Answer:

Draw a C plane and plot the dots if the coordinates

4 0
3 years ago
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A company determined that the marginal​ cost, C' (x )of producing the xth unit of a product is given by C' (x )= x^3- 6x. Find t
Rashid [163]

Answer:

C(x) = \frac{x^4}{4}-3x^2+3,000

Step-by-step explanation:

The marginal cost function, C'(x), is the derivate of the cost function, C(x).

Therefore, we can obtain the cost function by finding the integral of the marginal cost function:

C(x) = \int\ {C'(x)} \, dx \\C(x) = \int\ {(x^3-6x)} \, dx \\C(x) = \frac{1}{4} x^4-3x^2+a

Where 'a' is a constant and represents fixed costs. If fixed costs are $3,000, the cost function is:

C(x) = \frac{x^4}{4}-3x^2+3,000

3 0
3 years ago
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