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Total costs  =  500 +  8 * 8x   = 

 Fixed cost + (hrs per day)( hourly wage rate) (x)  =

  500 + 64x      where x is the number of workers we need

Total Revenue  = 

(hrs per day) (number of widgets/hour)(sales price/widget) (x)  =

8 * 5 * 3.10 * x  =  124x

And we want to solve this

124x >  500 + 64x      subtract  64x from both sides

60x > 500

x > 500/60

x > 8.33  workers

So...we need 9 workers (at a minimum )

Thanks for helping

3x^2 + 7x  + c  = 0

If this has  2 real roots, the discriminant  [ 7^2 - 4(3)c ] is > 0

So

49 - 12c   >  0

49 > 12c

49/12  > c  means the same thing as

c < 49/12 

c < 4 + 1/12

And since  c must be a positive integer value then the possibilities for c  are given by

0 < c < 4 + 1/12

So....c  = 1, 2, 3, 4

And the product of these  =  24

Since  the numerator is a polynomial, it won't have any domain restrictions

The domain restrictions  will occur when an x value makes the denomiantor = 0

So...to find these values, set the denominator to 0  and solve for x

x^2 + 2x  - 3    = 0     factor

(x + 3) ( x - 1)  = 0

Set each factor to  0  and solve for x

x + 3  = 0                 x - 1  = 0   

x  = -3                       x  = 1

So  x cannot be  -3    or   1

My bad, read it wrong! Thanks, EP!

We have 0.22mm for 8cells, so for one cell it is: \(\frac{0.22}{8}=0.0275mm\)

Let

x1  =      [( 2m  + √ [ 4m^2  - 4(1)(m^2 + 2m + 3 ) ]  ) / 2 ]

x2  =  [( 2m  - √ [ 4m^2  - 4(1)(m^2 + 2m + 3 ) ]  ) / 2 ]

(x1)^2  + (x2)^2  =

2 * [4m^2 + 4m^2  - 4m^2 - 8m - 12 ] / 4  =

[4m^2 - 8m - 12 ] / 2  =

2m^2 - 4m - 6

The m value that minimizes this is   4 / (2 * 2)  =  4 / 4   = 1

So...the minimum value of  (x1)^2 + (x2)^2  = 2(1)^2 - 4(1) - 6  =  -8

-5   is the REAL part      -i is the IMAGINARY part    -i = - sqrt(-1)

20/200         (x+20)/(x+200)           x+20

We have these points on the graph  ( 4 , 1400)   and  (6 , 2100)

The slope between these points is  [ 2100 - 1400 ] / [ 6 - 4 ] = 700 / 2  = 350 / 1

This means that, for every additional student, the tuition cost is $350 more

Depends are where they are sourced from.......multiple ways......try to find one.  Thanx 

We have these points on the graph  ( 0, -2)  and  ( 4, 0)

The slope is   [ 0 - -2 ] / [ 4 - 0 ]  =   2 / 4   =   1 / 2

Slope is rise/run  =   700/2 = 350      student      $350

Ok, there.

How?

Please post questions DIRECTLY on the forum, rather than giving URLs !

Two points on the graph  are (0, -3)  and (-3, 0)

The slope  between these are  [ 0 - -3 ] / [ -3 - 0 ]  =  3 / -3   = -1

So....the equation of the line is

y   = -1(x - 0) - 3    

y =  -x - 3

f(x) =y=1/2 x -2    So the slope = 1/2

Please post questions DIRECTLY on the forum, rather than giving URLs !

y=-2/2x-3

Thank you, that was fast!

If the polynomial  has one real root, the discriminant [ b^2 - 4ac ] must  = 0

So   a  =1, b  = c+ 1 ...so we have

(c + 1)^2  - 4(1)c    simplify

c^2 + 2c + 1  - 4c  = 0

c^2 - 2c + 1  = 0      factor the left side

(c - 1)^2  = 0         take the square root of both sides

c - 1  = 0

c  = 1

The horse's heartbeat  can be modeled by the regression equation

y = 38.26x  - 5.25     (1)

Because the slope  of y = 65x    is  greater than that of (1), the cow has the faster heartbeat per minute = 65 beats per minute

The heartbeat of the horse per minute  is ≈ 38 beats per minute

JULIUS

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Acres in each section =

8 / [3 3/5]  =

8 / [ 18 /5]  =

[ 8 /1] / [ 18/5]  =

8/1  *   5 /18  =

40 / 18  =

20/ 9   =  2 + 2/9  acres  in each section

.01 = .15x  -250    add 250 to both sides 

250.01   =.15x     divide both sides by .15

250.01 / .15  =   x  ≈ 1666.73