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3 years ago

Please help with this

Mathematics

1 answer:

dexar [7]3 years ago

5 0

Answer:

13) 1,11 14) 1,2,11,22 15) 1, 2, 4, 7, 14, 28 16) 1, 2, 4, 8, 16, 32 17) 1, 2, 3, 6, 7, 14, 21, 42

Step-by-step explanation:

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mote1985 [20]

Answer: The equation that represents the other equation is $y=\dfrac{1}{3}x+5$ .

The solution of the system is (3,6).

Step-by-step explanation:

Linear equation: $y=mx+c$ , where m= slope

c = y-intercept.

In the first table, the y-intercept = 5 [ y-intercept = value of y at x=0.

Slope for first table = $\dfrac{y_2-y_2}{x_2-x_1}=\dfrac{6-5}{3-0}=\dfrac{1}{3}$

The equation that represents the first table:

$y=\dfrac{1}{3}x+5$

So, the equation that represents the other equation is $y=\dfrac{1}{3}x+5$ .

Also, the solution of the system is the common point (x,y) that satisfy both equations in the system.

Here, x=3 and y=6 is the common value in both tables.

So, the solution of the system is (3,6).

6 0

3 years ago

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

Number 16: complete the table

soldier1979 [14.2K]

Answer:

7.5

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Todd is training to bike in the Turkey Trot Bike Race and rode his bike 7 ¾ miles per hour for 2 2/3 hours. How many miles did T

trapecia [35]

(7 3/4 mi/h)*(2 2/3 h) = 20 2/3

Todd rode 20.667 miles.
_____
A calculator that works with mixed numbers is a handy thing.

5 0

3 years ago

3-7= 4-6= 3+-6= 4--1= -5+-5= -3+5= -4-6=

kipiarov [429]

Answer:

Answers are below

Step-by-step explanation:

3 - 7 = -4

4 - 6 = -2

3 + -6 = -3

4 - - 1 = 5

-5 + -5 = -10

-3 + 5 = 2

-4 -6 = -10

Hope this helps!!

4 0

3 years ago

Please help with this​

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