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Nevermind, I managed to solve it by setting the denominator equal to 0.

Circumference = Diameter(radius x 2) x pi

C = 20 x 2 x 3.141592

C =125.66 cm

C =125.66 / 100

C =1.2566 meters

1.2566 x 582 =731.34 meters travelled.

C =65 x 3.141592 =204.20 cm - circumference of the wheel.

204.20 / 100 =2.042 m - circumference of the wheel.

2 km x 1,000 = 2,000 m

2,000 / 2.042 =979.43 ~ 979 turns of the bicycle wheel.

The grey  half circle on the left  has a radius  of 12cm

Its area is  (1/2)pi(12)^2   =   72pi  cm^2 ≈  226.2 cm^2    (1)

The remaining area is a rectangle  with dimensions 12 * 24  =   288 cm^2  (2)

So  the approximate total area  is  288 + 226.2  =   514.2  cm^2  

The approximate area  of the two black pieces  is  (2) - (1)   =  61.8 cm^2

anyone?

4^x-2  = 2^x

The idea here is to make the bases equal  and then solve for the exponents

(2²)^x-2  = 2^x

(2)^[2(x-2)]  = 2^x

(2)^{2x - 4}  = 2^x

The bases are the same....solve for the exponents

2x - 4  = x        add 4 to both sides.....subtract x from both sides

x  =  4

Let n be a positive integer, and let x larger or equal to -1. Prove, using induction, that (1+x)^n is larger or equal to 1+nx

Look at  "D"  ⇒ Expression 4

³ √ [ 343 x⁹ y z⁴  ]  =

³√343   * ³√x9   *  ³√y   * ³√z⁴   =

7  * x³  * y^1/3  * z^4/3 

a.  We have three half-circles each with a radius  of   6

So....the total crcumference is

3  (1/2) * 2 * pi  (6)   =    18 pi  cm  ≈  56.55 cm

b.  This is an equilateral triangle with sides of 12

Its area is

(1/2)12^2* √3/2   =   144√3/4   =  36√3 cm^2   ≈  62.35 cm ^2    (1)

c. The total area of the three half-circles is

3 (1/2) * pi (6)^2   =  54 pi  cm^2 ≈  169.65 cm^2   (2)

So....the approximate area is  (1) + (2)  ≈  232 cm^2

10000  =    10000 /  [ 1 + 9e^(-1.13t) ]

Divide both sides by 10000

1  =    1 /  [ 1 + 9e^(-1.13t) ]

 1 + 9e^(-1.13t)   =  1

9 e^(-1.13t)  =  0

e^(-1.13t)  =  0

Note that an exponential  can never  = 0

So.....it will  never actually  =  10000

The sum of 5 and √5  is irrational

The product of 5 and 5√5  is irrational

The product of 5 and √5  is irrational

Any irrational number multiplied by a non-zero rational number is equal to an irrational number

2^x=32        Take the log of both sides

x =Log(32) / Log(2)

x = 5, because:

2^5 =32

Yes, because 1/8 and 8 are both rational. Therefore it is B since that would add up to 65/8 or 8 1/8

The only way you can get P(t) =10,000 is to have your denominator = 1 exactly!. Even if t =1,000,000

then you would have: e^(-1,130,000), which of course means: 1 / (e^1,130,000), which is very close to zero, but NOT exactly zero!. In fact e^(-1,130,000) =1.719686672 E-490,753.

The product of A is rational = -8/21

The product of B is also rational =2 x 3 =6

The product of C is irrational =sqrt(7) / 7 =0.377964473........is irrational

The product of D is rational = 7 x .07 =0.49

Both 8 and 1/8 are rational and their sum = 8+1/8 =8 1/8 = 65/8 is rational. Or B, of the 4 choices.

i need help on this too

2. What are the coordinates of the points where the graphs of $f(x)=x^3-x^2+x+1$ and $g(x)=x^3+x^2+x-1$ intersect?  Give your answer as a list of points separated by semicolons, with the points ordered such that their $x$-coordinates are in increasing order. (So "(1,-3); (2,3); (5,-7)" - without the quotes - is a valid answer format.)

\(x^3-x^2+x+1 = x^3+x^2+x-1\\ -x^2+1 = x^2-1\\ 2=2x^2\\ x^2=1\\ x=\pm1\\ f(1)=1-1+1+1=2\\ f(-1)=-1-1-1+1=-2\\ \)

so the points of intersection are     (-1,-2); (1,2)

check:

1. Suppose that the graph of a certain function, y=f(x), has the property that if it is shifted 20 units to the right, then the resulting graph is identical to the original graph of y=f(x).  What is the smallest positive  a  such that if the graph of \(y=f\left(\frac x5\right)\) is shifted   \(a\)   units to the right, then we know that the resulting graph is identical to the original graph of \(y=f\left(\frac x5\right)\)?

I wasn't sure so I experimented, I decided to use a sine graph with a wavelength of 20

y=sin nx

the wavelength is   2pi/n = 20

so  n = pi/10

so my experimenting graph is

\(  y=sin(\frac{\pi }{10}x)\\   y=sin(\frac{\pi }{10}x)=sin(\frac{\pi }{10}(x-20))\\ let \;\;\frac{x}{5}\;\;be\;\;substituted\;\;for\;\;x\\   y=sin(\frac{\pi }{10}\frac{x}{5})=sin(\frac{\pi }{10}(\frac{x}{5}-20))=sin(\frac{\pi }{10}(\frac{x-100}{5})) \)

So the smallest a that we know with certainty is 100.

Here is the graph i was playing with.

I know, I put your expressions into the forum latex function.

If you do not know already how to do that you should learn.

You just open the latex box  ( which is on the ribon) and copy your latex into it and then hit ok.

Just delete the dollar signs that surounds the code.

You rewrote them both correctly 

I can't even read that question.  I will rewrite it.   

1. Suppose that the graph of a certain function, \(y=f(x)\), has the property that if it is shifted 20 units to the right, then the resulting graph is identical to the original graph of \(y=f(x)\).  

What is the smallest positive  \(a\)   such that if the graph of \(y=f\left(\frac x5\right)\) is shifted \(a\) units to the right, then we know that the resulting graph is identical to the original graph of \(y=f\left(\frac x5\right)?\)

2. What are the coordinates of the points where the graphs of

\(f(x)=x^3-x^2+x+1\)     and       \(g(x)=x^3+x^2+x-1\)     intersect?  

Give your answer as a list of points separated by semicolons, with the points ordered such that their \(x\)-coordinates are in increasing order.

[So      (1,-3); (2,3); (5,-7)       is a valid answer format. ]

125 electrons gained.

Call the radius  of the cylinder and sphere, r

And the height  of the cylinder  = 2r

So

60  =  pi * r^2 * 2r

30 =  pi * r^3

30/pi  =  r^3

[30/pi]^1/3  =  r

So...the volume of the sphere is

(4/3)pi [ (30/pi)^1/3 ]^3  =

(4/3) pi * 30/pi  =

40 cm^3

We could  also  see this by the forllowing ratio

Vsphere      =   (4/3)pi *r^3              (4/3)                4                 2            40

_______          __________    =      _____  =     ______  =      __    =    ___

V cylinder         pi * r^2 * 2r                2                  6                  3            60

If we took off  1  inch from each stick, we'd have sticks of length

7, 14, and 16

For these to be the lengths of the sides of a triangle, the sum of the smallest two sides must be greater than the third side.

7 + 14 > 16

21 > 16          this is true, so  7, 14, and 16 can be the sides of a triangle.

So.....we want to know the smallest integer  n  such that....

(8 - n) + (15 - n)   is NOT greater than   17 - n

(8 - n) + (15 - n)  ≤  17 - n

23 - 2n  ≤  17 - n

23  ≤  17 + n

6  ≤  n

n  ≥  6

We want to know the smallest integer  n  such that  n ≥ 6  .

The smallest number that is greater than or equal to 6 is  6 .

The smallest piece that needs to be cut from each of the sticks is 6 inches long.

\(2\log{(a -2b)} = \log{a} + \log{b} \)

We can write

log (a - 2b)^2  = log ( a * b)

And we can solve

(a - 2b)^2  =  ab

a^2 - 4ab + 4b^2  = ab

a^2 - 5ab + 4b^2  = 0     factor

(a - 4b) ( a - b)  =  0

Setting  each factor to 0  and solving we have that

a  = 4b         and  a  = b

Assuming  a, b   are both positive, the first solution is the only one where all logs are defined for real numbers

So

a  / b  =

[4b]/ b  =

4

First one....the scale factor is  2

Second one

n/6  =  5/3    multiply both sides by 6

n  =  6 * 5/3

n = 30/3

n  = 10