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1. Inverse variation

Let the price of bananas  = P   and the quantity  = Q  (in pounds )

Then, the quantity we buy  will be greater when the price decreases and  less when the price increases

This can be modeled by the function

Q  = k / P          where k is the variation constant

2. 

Discontinuity  resulting in a vertical asymptote at  x  = 5 ⇒ f(x)  =   1  / ( x - 5 )

[Occurs when the denominator of a rational function cannot be "factored out " ]

Discontinuity that results in a "hole" at   x = 5 ⇒  f(x)  = ( x^2 - 25) / (x - 5)

[Occurs when the denominator of a rational function can be "factored out " ]

A vertical  asymptote goes through the x axis and is parallel to the y axis

1.  (r - c) (3)  =   r(3)  - c(3)  =

r(3)  =  (3)^2+5(3)+14   = 38

c(3) =  (3)^2−3(3)+4  =    4

So   r(3)  - c(3)   =   38  - 4   =    34

The store makes a profit of $3400 in the third month of business

2.  (p * c) (2)   =  p(2) * c(2)  =

p(2)  = 100−2(2)   =  96

c(2)  = 50+5(2)  =  60

So

p(2) * c(2)   =   96 * 60  =  $5760  when the price per customer  = $60

Yep 8)

A coat that is 3025 sq ft..... Juan is a BIG dude...probably gets his clothes from Omar the tent maker.

All that paint for one coat? What size coat does he wear!?

If it is a square, then the area is 55 x 55 sq ft = 3025 ft^2

3025 ft^2 / 275 ft^2/gal = 11 gallons of paint (for one coat).

thank you very much for the support

First one:

(3X^2 +5)(x-2) = 3x^3 - 6x^2 +5x - 10 

Second one is written incorrectly, I think you meant f/g(x)

f(x) = 2x^2-5x-3  =  (2x+1)(x-3)     

then   f/g =   (2x+1)(x-3)  /  (x)(2x +1)          x cannot equal 0 or -1/2  because the denominator would be ZERO

              Simplify to     (x-3)/x                       (the (2x+1)'s  'cancel out')

x in months, r(x) & c(x) in $100 
r(x)=x²+6x+10 
c(x)=x²-4x+5 

(r-c)(x) = r(x)-c(x) = (10x+5) (×100) 
(r−c)(4) = (10(4)+5)(100) = 4500 
profit of $4500 

f(x)=4x²-5 
g(x)=x+3 
(f•g)(x) = f(x)•g(x) = (4x²-5)(x+3) 
= 4x³+12x²-5x-15 

f(x)=3x²+10x-8 = (3x-2)(x+4) 
g(x)=3x²-2x = x(3x-2) 
(f/g)(x) = f(x)/g(x) = [(3x-2)(x+4)]/[x(3x-2)] 
= (x+4)/x, x≠0,2/3

I count 5

4 at right angles to the square end and one more.

Well for starters you stop giving other people thumbs down for no reason !!

How many of the factorials from 1! to 100! are divisible by 9?

9! up to 100! all have 9 as one of the factors. so that is 100-9+1=92   (assuming 100! id included)

6! = 1*2*3*4*5*6 = 2*4*5*18 = 2*4*5*2*9 

so 6!, and above all have 9 as a factor,  so I will include 6!, 7!, and 8!    92+3=95 so far.

5!=1*2*3*4*5  9 is not a factor.

So 95 of them have 9 as a factor

If n cuts are made in a pie, what is the largest number of pieces that may be obtained?

I draw some pictures and decided that the number is maximum if all the former cuts have been crossed (not at any point of intersections.

This is the pattern that I found.

Now I looked to find the pattern,, with each cut the increaded number of peices  is 1 more than last time.

i mean 2+2=4, 4+3=7, 7+4=11

it will be a quadratic function. 

So the maximum number of peices if the cake is cut linearly n times is  (n^2+n+2)/2

\({5x-2\over x+3} = 5 - {17 \over x+3}~\\ \qquad 5(x+3)=5x+15\\ LHS\\={5x-2\over x+3}\\={5(x+3)-15-2\over x+3}\\={5(x+3)-17\over x+3}\\= {5(x+3)\over x+3}-{17\over x+3}\\= 5-{17\over x+3}\ \)

This is not really what you were expected to do though.

You were expected to do algebraic division or synthetic division

This is an example of what you were supposed to do :)

Angle OBQ  = 90 ⇒  a radius meeting a tangent forrns a right angle

Angle CBO  = 90 - 2x  ⇒  ( m ∠OBQ - m∠ CBQ = m ∠ CBO)

Angle OCB  = Angle CBO ⇒  In triangle OCB....OB = OC...so the angles opposite these sides are also equal

And in triangle  DOC, DO = OC.....so angle CDO = angle DCO  = x

So  angle BCD   =  angle OCB + angle ACD  =  90 - 2x + x  =  90 - x

And  minor arc  DB  =  2 * m∠ DCB  =  180 - 2x  [ an inscribed angle is measures1/2 of its intercepted arc ]  =  angle DOB

So......major arc DB  =   (360) - ( 180 - 2x)  =  180 + 2x

And an angle external to a circle and formed by two secants, is equal to one half the difference of the intercepted arcs.  This means that   

y  = (1/2) [ measure of major arc DB -  measure of minor arc DB ]

y  = (1/2) [  (180  + 2x)  - (180 - 2x ) ]

y  =  (1/2)  ( 2x + 2x)

y = (1/2) (4x)

y  =  2x

A)  10 x 0.78^6 =2.25 m

B) Sum = F x [1 - R^N) / [1 - R]

    Sum = 10 x [1 - 0.78^4] / [1 - 0.78]

    Sum = 28.63 m

C} Sum = F / [1 - R]

    Sum = 10 / [1 - 0.78]

    Sum = 45.45 m

Any that are not divisible by 3 or 5

7, 8, 11, 13, 14, 16, 17, 19, 22, 23, 26, 28, 29, 31, 32, 34, 37, 38, 41, 43, 44, 46, 47, 49, 52, 53, 56, 58, 59

If I didn't miss any....I get 29

2)In the figure with four circles below, let A1 be the area of the smallest circle, let A2 be the area of the region inside the second-smallest circle but outside the smallest circle, and so on. If A1 = A2 = A3 = A4 then find the ratio of the largest radius to the smallest radius.

We can imagine these circles to be concentric..the area between them would still be the same

Without a loss of generality....let the smallest radius   = 1

And let the next largest radius  = b

And we don't need to worry abut "pi" ...  it wll  "cancel"  in the compuations

The  Area  of the second smallest circle -  Area of smallest circle = Area of smallest circle

b^2  -  1^2  =  1^2

b^2   =  1^2 + 1^2

b^2  = 2

b =  √2

Call the radius of the next-to-largest circle  "c"  ..... so...

Area of this circle - Area of circle with radius "b"  =  Area of smallest circle

c^2  -  2  =  1

c^2 =  3

c^2  =   3

c = √3

A call the radius of largest circle  = "d"

Area of this circle - Area of circle with radius "c"   =  Area of smallest circle

d^2  - 3  =  1

d^2  =  4

d  = 2

So.... the ratio of the largest circle's radus to the smallest is just   

2 / 1    =   2

Thanks Chris :)

Let's look at the 6 x 6 case,  first

Notice that the number of  "vertical" segments   is   6(7)

And the number of "horizontal" segments  is 6(7)

This implies that the number in a 10  x 10 array should be

2 (10) (11)  =    220 

I'm assuming  that when the problem says  " 2s to the right and left  and the return to its original position "  that  the total period is 4 seconds.....

So.... the period is 4 seconds   ....

The amplitude is 3

The  sine fuction seeks to model this best

In the form

y  =  Asin (Bx)

A  =  amplitude  = 3

To find B, we need to solve this

B   =  2pi / period   =  2pi/ 4   =  pi/2

So....our function becomes

y  = 3sin ( (pi/2) * x )

Here's the graph  ....two periods are shown :  

https://www.desmos.com/calculator/skrl7kbslk

Thanks, Guest!  That is correct!

30! =  2^26 * 3^14 * 5^7 * 7^4 * 11^2 * 13^2 * 17 * 19 * 23 * 29. Count the base numbers only.

I see now! Thanks for the explanation, CPhill!

i believe there is a "formula" for this

We have a pattern of   3 x 3  "boxes"

The possible rectangles  =  [ 3(4) / 2] * [ 3(4)/2]   = [ 6 ] * [ 6 ]   =   36