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Katen [24]
3 years ago
13

PLEASE HELP Which expression represents the difference quotient of the function f (x) = negative StartRoot 2 x EndRoot?

Mathematics
2 answers:
natali 33 [55]3 years ago
5 0

Answer:

c is the correct option

Step-by-step explanation:

from,

f'(x) = h >0 <u>f</u><u>(</u><u>x</u><u> </u><u>+</u><u> </u><u>h</u><u>)</u><u> </u><u>-</u><u> </u><u>f</u><u>(</u><u>x</u><u>)</u><u> </u>

h

f(x) = - √2x

f(x + h) = - √(2x + h)

f'(x) = h>0 <u>-</u><u>√(2x + h) - √2x</u>

h

rationalize the denominator

= h>0 <u>-</u><u>√</u><u>(</u><u>2</u><u>x</u><u> </u><u>+</u><u> </u><u>h</u><u>)</u><u> </u><u>+</u><u> </u><u>√</u><u>2</u><u>x</u><u> </u><u> </u><u>(</u><u>-</u><u>√</u><u>(</u><u>2</u><u>x</u><u> </u><u>+</u><u> </u><u>h</u><u>)</u><u> </u><u>-</u><u> </u><u>√</u><u>2</u><u>x</u><u>)</u>

h (-√(2x + h) - √2x)

= h>0 <u>4</u><u>x</u><u> </u><u>+</u><u> </u><u>2</u><u>h</u><u> </u><u>-</u><u> </u><u>4</u><u>x</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

h(-√(2x + h) -√2x)

= h>0 <u>2</u><u>h</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

h(-√(2x+h) - √2x)

= h>0 <u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>2</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

-√(2x+h) - √2x

MrRissso [65]3 years ago
3 0

\\ \sf\longmapsto f(x)=\sqrt{2x}

Now

\\ \sf\longmapsto \dfrac{1}{-(\sqrt{2x+2h}-\sqrt{2x})}

There we plot 2h because if we break root over then it becomes √2h which satisfies f(x)

\\ \sf\longmapsto \dfrac{1}{-\sqrt{2x+2h}+\sqrt{2x}}

Option D

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Answer:

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7 0
3 years ago
A flu epidemic is spreading through a town of 48,000 people. It is found that if x and y denote the numbers of people sick and w
Liula [17]

Answer:

a) The simultaneous equation represented in matrix form, is

[1/3 1/4] [x] = [s]

[2/3 3/4] [y] = [w]

Ax = B

[1/3 1/4] = matrix A (matrix of coefficients)

[2/3 3/4]

[x] = matrix x (matrix of unknowns)

[y]

[s] = matrix B (matrix of answers)

[w]

b) Number of sick people the preceding week = 12005

Step-by-step explanation:

x = Number of sick people in a week

y = Number of people that are well in a week

s = Number of sick people the following week

w = Number of people that are well the following week.

The relationship between these is given as

(1/3)x + (1/4)y = s

(2/3)x + (3/4)y = w

In matrix form, this is simply presented as

[1/3 1/4] [x] = [s]

[2/3 3/4] [y] = [w]

which is more appropriately written as

Ax = B

where

[1/3 1/4] = matrix A (matrix of coefficients)

[2/3 3/4]

[x] = matrix x (matrix of unknowns)

[y]

[s] = matrix B (matrix of answers)

[w]

b) Taking the current conditions as s and w, then the preceding week will be x and y

The number of sick people in this week, s = 13000

The number of people well in this week, w = total population - Number of sick people.

w = 48000 - 13000 = 35000

So, the simultaneous equation becomes

(1/3)x + (1/4)y = 13000

(2/3)x + (3/4)y = 35000

Then we can solve for the number of sick and well people the preceding week.

We can solve normally or use matrix solution.

Ax = B

x, the matrix of unknowns is given by product of the inverse of A (inverse of the matrix of coefficients) and B (matrix of answers)

x = (A⁻¹)B

But, solving normally,

(1/3)x + (1/4)y = 13000

(2/3)x + (3/4)y = 35000

x = 12004.8 = 12005

y = 35995.2 = 35995

Number of sick people the preceding week = x = 12005

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Answer:

{x,y,z} = {-116,28,37}

Step-by-step explanation:

// Solve equation [3] for the variable  z  

 

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__________________________________________________________

// Plug this in for variable  z  in equation [1]

  [1]    5x + 9y + 9•(-2x/5-2y/5+9/5) = 5

  [1]    7x/5 + 27y/5 = -56/5

  [1]    7x + 27y = -56

__________________________________________________________

// Plug this in for variable  z  in equation [2]

  [2]    4x + 9y + 6•(-2x/5-2y/5+9/5) = 10

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  [2]    8x + 33y = -4

__________________________________________________________

// Solve equation [2] for the variable  y  

 

 [2]    33y = -8x - 4

 [2]    y = -8x/33 - 4/33

__________________________________________________________

// Plug this in for variable  y  in equation [1]

  [1]    7x + 27•(-8x/33-4/33) = -56

  [1]    5x/11 = -580/11

  [1]    5x = -580

__________________________________________________________

// Solve equation [1] for the variable  x  

  [1]    5x = - 580  

  [1]    x = - 116

__________________________________________________________

// By now we know this much :

   x = -116

   y = -8x/33-4/33

   z = -2x/5-2y/5+9/5

__________________________________________________________

// Use the  x  value to solve for  y  

   y = -(8/33)(-116)-4/33 = 28

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// Use the  x  and  y  values to solve for  z  

 z = -(2/5)(-116)-(2/5)(28)+9/5 = 37

╦────────────────────────────╦

│Hope this helped  _____________________│    

│~Derelis ____________________________ │

╨___________________________________╨                  

4 0
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