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r= radius

r^2 x pi 

or

r^2 x 3.14

How do you turn a fraction into a repeting decimal? I need help with this...

NOT every fraction is a repeating decimal!. Examples: 1/4 =0.25 and that is it. You cannot repeat the 5. 3/8=.375 and that is it. But, 1/3=.333333........Here 3 repeats forever. The same with 1/7=0.142857 142857 142857 142857 142857..........these 6 digits repeat forever. And so on.

This is  looks pretty obvious when a graph is plotted:

I deleted my earlier entry because my reasoning was incorrect.  Here is (what I hope is) a correct proof:

lhs = (9/7)^n
rhs = n

When n = 9  lhs = 9.6...   rhs = 9   so lhs > rhs

When n increments by 1 the lhs increments by (9/7)^n(9/7 - 1) = (2/7)*(9/7)^n
The smallest value of this (for n >= 9) is 2.7...,  given when n = 9, and gets larger as n increases.

When n increments by 1 the rhs increments by n + 1 - n or just 1 for all n.

Hence the lhs starts larger than the rhs (where start is n = 9) and increases more rapidly than the rhs as n increases.

i.e. (9/7)^n > n for all n>=9

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p(x)=2x^3+13x^2+17x+-12 has (x + 4 ) as a factor. factor the polynomial into three linear terms.

p(x) = (x+3) (x+4) (2x-1)

Simplify the following:
((15 x^2-9 x-6) (6 x^2-6 x-12))/((3 x-3 x-6) (2 x^2+2 x-4))

3 x-3 x = 0:
((15 x^2-9 x-6) (6 x^2-6 x-12))/(-6 (2 x^2+2 x-4))

Factor 2 out of 2 x^2+2 x-4:
((15 x^2-9 x-6) (6 x^2-6 x-12))/(-62 (x^2+x-2))

The factors of -2 that sum to are 2 and -1. So, x^2+x-2 = (x+2) (x-1):
((15 x^2-9 x-6) (6 x^2-6 x-12))/(-6×2 (x+2) (x-1))

Factor 6 out of 6 x^2-6 x-12:
(6 (x^2-x-2) (15 x^2-9 x-6))/(-6×2 (x+2) (x-1))

The factors of -2 that sum to -1 are 1 and -2. So, x^2-x-2 = (x+1) (x-2):
(6 (x+1) (x-2) (15 x^2-9 x-6))/(-6×2 (x+2) (x-1))

Factor 3 out of 15 x^2-9 x-6:
(6×3 (5 x^2-3 x-2) (x+1) (x-2))/(-6×2 (x+2) (x-1))

Factor the quadratic 5 x^2-3 x-2. The coefficient of x^2 is 5 and the constant term is -2. The product of 5 and -2 is -10. The factors of -10 which sum to -3 are 2 and -5. So 5 x^2-3 x-2 = 5 x^2-5 x+2 x-2 = x (5 x+2)-(5 x+2):
(3×6 x (5 x+2)-(5 x+2) (x+1) (x-2))/(-6×2 (x+2) (x-1))

Factor 5 x+2 from x (5 x+2)-(5 x+2):
(3×6 (5 x+2) (x-1) (x+1) (x-2))/(-6×2 (x+2) (x-1))

(3 (5 x+2) (x-1)×6 (x+1) (x-2))/(-6×2 (x+2) (x-1)) = (x-1)/(x-1)×(3 (5 x+2)×6 (x+1) (x-2))/(-6×2 (x+2)) = (3 (5 x+2)×6 (x+1) (x-2))/(-6×2 (x+2)):
(3×6 (5 x+2) (x+1) (x-2))/(-6×2 (x+2))

6/(-6) = 6/(6 (-1)) = 1/(-1):
(3 (5 x+2) (x+1) (x-2))/(-1×2 (x+2))

Multiply numerator and denominator of (3 (5 x+2) (x+1) (x-2))/(-2 (x+2)) by -1:
Answer: |-(3 (5x+2) (x+1) (x-2))/(2 (x+2))

u=-2i+j+5k

v=4i-3j-3k

The area   =    ll  u x v ll

u x v  =   det   [  i   j   k 

                         2  1  5

                         4  -3 -3 ]      write the first two columns again and take the det

[  i    j    k     i    j

  2   1    5    2  1

  4   -3  -3    4  -3 ]    

[  -3i + 20j  -6k ] - [ 4k -15i - 6j ] = 

[12i + 26j - 10k ]

And the area =  sqrt [ 12^2 + 26^2 + ( -10)^2]  = sqrt (920)  = sqrt (4 * 230) =  2sqrt (230)

first solve for x. what times 8 equals 92?   then just subtract 7 from your answer, and see what you get. 

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x= -14/9 or - 1.5 repeating

f(x) = 6x^7 + 7x^6 

f'(x)  = 42x^6 + 42x^5      set to  0

42x^6 + 42x^5   = 0     factor

42x^5 ( x  + 1)  = 0 

Setting both factors to 0, we have that x = 0   and x = -1

When x = 0,  f(x) = 0

When x = -1, f(x) = 1

So.......the points are    (0, 0)  and ( -1, 1)   →     " b"

Let    (k, m, n), (r, s, t), and (u, v, w)   = (5,-1,2), (7,-4-2), (2,-6,3)

Then....the area can be calculated thusly :

(1/2)absolute value of    det  [ 1     1      1  

                                              r-k  s-m   t-n

                                              u-k  v-m  w-n ]  

(1/2) abs  det  [   1   1   1 

                           2  -3  -4

                          -3  -5  1 ]             write the first two columns again and calculate the determinant

(1/2)  1  1   1   1    1

          2 -3  -4  2   -3

         -3  -5  1 -3   -5

(1/2) abs (  [  1 * -3 *1  +  1* -4* -3 + 1*2*-5] - [ -3*-3*1 +  -5*-4*1 + 1*2*1]  )   = 

(1/2) abs  (  [ -3    + 12   -10 ]  - [ 9 + 20 + 2 ]  ) =

(1/2) abs(-32) =

(1/2 (32) = 

16 sq units

Sorry, for some reason I am unable to see your answer. Would you mind copying and reposting it?

 x+y=0

 8x - 7y = 75

Rearranging the first equation produces  x = - y

Subbing this for x in the second equation, we have

8(-y) -  7y = 75

-8y- 7y  = 75

-15y  = 75    divide both sides by -15

y = -5

And x = -y   so x   = -(-5)  = 5

So  {x, y}   = {5, -5}

4. Given that xy = 3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x+3y/5.

If xy = 3/2  then y = 3/[2x]

So we have

10x + (3/5)[3/(2x)]

10x + 9 / [10x]

10x + (9/10)x^-1      take the derivative  and set to 0 

10  - (9/10)x^-2   =  0       multiply through by x^2

10x^2 - (9/10)  = 0

10x^2  = 9/10     multiply throgh by 10

100x^2 = 9       divide though by 100

x^2  = 9/100     taking the positive root, x = 3/10

And taking the second derivative procuduces a positve result, so this is a minimum

And xy = 3/2.....so........3/10 * y   = 3/2  →   y /10 = 1/2  → y = 5

So {x, y} = { 3/10, 5}     produces a minimum ....where x,y > 0

1.  z(x,y,t) = k*x*y/t   where k is a constant

We know that 5 = k*2*(-3)/(1/2) or 5 = -12k  so k = -5/12  

Hence z(x,y,t) = -5xy/(12t)

2. When x = -12/(5z)^2  we have z = -5*(-12)/(5z)^2*y/(12t)  or  z = y/(5z^2*t)  or t = y/(5z^3)  

     Since z = 2/5  then t = y/(5*2^3/5^3)   or  t =  25y/8

3.  Since f(y) = t we must have f(y) = 25y/8

(check the algebra - I did this quickly!)

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Like:   \(mean=\frac{\int_0^{\pi/w}V(t)dt}{\int_0^{\pi/w}dt}\rightarrow\frac{w}{\pi}\int_0^{\pi/w}V(t)dt\)

See:

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The mean over the whole period is zero.  You might want to do the calculation over a half-period.

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Have a look at this Khan Academy video:  https://www.khanacademy.org/math/in-eighth-grade-math/squares-square-roots/square-roots/v/simplifying-radicals

d=3260km over 4h =

3,260 / 4 =815 km per hour.