All Answers 358236 Answers

-2 + 4   =

-2 + [ 2 + 2]  =

-2 + 2 + 2  =

0 + 2  =

2

I'm assuming that the 8 inches is the diameter....if so.....we have

V = pi (r^2) * h         in tis case   r = 4 in and h = 85 x 12 = 1020 in   ....  so we have

V = pi (4 in)^2 (1020 in)  = about 51270.79 cubic inches

We can prove this by mathematical induction......

First.....the sum of the first n positive integers = [n][n+1]^2/ 2

Therefore, the [sum of the first n positive integers]^2  = [n]^2 [n =1]^2 / 4

So....we want to prove that

1^3 + 2 ^3  + 3^3 + .....+ n^3  =   [n]^2 [n =1]^2 / 4

Show that it is true for n = 1

1^3   =  [1]^2 [2)^2 / 4  = [1][4]/4 = 1

Assume it's true for k.....that is......

1^3 + 2^3 + 3^3 + ....... + k^3 =  [k]^2 [k + 1]^2/ 4

Show that it is true for k + 1    .... that is.........

1^3 + 2^3 + 3^3 + .......+ k ^3 + [k + 1]^3   =  [k + 1]^2 [k + 2]^2 / 4

 So we have.......

1^3 + 2^3 + 3^3 + .......+ k ^3 + [k + 1]^3   = [k]^2 [k + 1]^2 / 4  + [k + 1]^3

1^3 + 2^3 + 3^3 + .......+ k ^3 + [k + 1]^3   =  [k + 1]^2 ( k^2 + 4[k + 1] ) / 4

1^3 + 2^3 + 3^3 + .......+ k ^3 + [k + 1]^3   =  [k + 1]^2 ( k^2 + 4k + 4 ) / 4

1^3 + 2^3 + 3^3 + .......+ k ^3 + [k + 1]^3   =  [k + 1]^2 [k + 2]^2 / 4

Which is what we wanted to prove

7.93*.23=1.8239

1.8239

1.8239

This is true. Here is the Algebra involved:

Solve for x over the real numbers:
x^3+(x+1)^3 = (2 x+1)^2

Expand out terms of the left hand side:
2 x^3+3 x^2+3 x+1 = (2 x+1)^2

Expand out terms of the right hand side:
2 x^3+3 x^2+3 x+1 = 4 x^2+4 x+1

Subtract 4 x^2+4 x+1 from both sides:
2 x^3-x^2-x = 0

The left hand side factors into a product with three terms:
x (x-1) (2 x+1) = 0

Split into three equations:
x-1 = 0 or x = 0 or 2 x+1 = 0

Add 1 to both sides:
x = 1 or x = 0 or 2 x+1 = 0

Subtract 1 from both sides:
x = 1 or x = 0 or 2 x = -1

Divide both sides by 2:
Answer: | 
| x = 1 or x = 0 or x = -1/2

Example: .25= .25 X 100/100=25/100 divide both sides by 25=1/4

Volume of what?

800 - 600=200 profit on the first purchase.

1,200 - 1,000=200 profit on second purchase

200 + 200=$400 profit she made on the two deals.

3=3x+4-6

3=3x+(-2)

3+2=3x

5=3x

x=5/3=1 2/3

duh, 2

GO TO YOUR BOOKS!!.

7658 = (7 x n 2) + (6 x n 1) + (5 x n 0)

Solve for n:
7658 = 20 n

7658 = 20 n is equivalent to 20 n = 7658:
20 n = 7658

Divide both sides of 20 n = 7658 by 20:
(20 n)/20 = 7658/20

20/20 = 1:
n = 7658/20

The gcd of 7658 and 20 is 2, so 7658/20 = (2×3829)/(2×10) = 2/2×3829/10 = 3829/10:
Answer: | 
| n = 3829/10

n=382.9

1.25^2=1.5625

72=2 X 2 X 2 X 3 X 3=6sqrt(2)

All Solutions for x:

\(\begin{array}{rcl} \cosh{(x)} &=& \dfrac { e^x + e^{-x} } {2} = 0 \\ \dfrac { e^x + e^{-x} } {2} &=& 0 \\ e^x + e^{-x} &=& 0 \\ e^x + \dfrac{ 1 } { e^{x} } &=& 0 \qquad | \qquad \cdot e^x\\ e^{2x} + 1 &=& 0 \qquad | \qquad \boxed{~ \text{Euler's identity: } e^{i\pi} + 1 = 0 ~}\\ e^{2x} + 1 &=& e^{i (\pi \pm 2k\pi) } + 1 \\ e^{2x} &=& e^{i(\pi \pm 2k\pi)}\\ 2x &=& i(\pi \pm 2k\pi) \\ x &=& \dfrac{ i(\pi \pm 2k\pi) } {2}\\ \mathbf{x} & \mathbf{=}& \mathbf{ \dfrac{ i\pi }{2} \pm k\pi i } \qquad k \in N \qquad | \qquad \dfrac{ i\pi }{2} - \pi i = -\dfrac{ i\pi }{2} \\ \mathbf{x} &\mathbf{=}& \mathbf{ -\dfrac{ i\pi }{2} \pm k\pi i \qquad k \in N } \\ \end{array}\)

Please don't use innapropriate language on this website. Children use this site, and your not the only person in the world who has feelings. If you want to be like that, go on another website please.

remeber that any thing that involed -1 (multiplying or dividing) will make the number negitive. so the answer is -2.2 

Divide both sides by pi:   R^2 = A/pi

Take the square root of both sides R = (A/pi)^0.5

Mathematically, there is also R = -(A/pi)^0.5, but if, A and R refer to circles, with R as the radius, then only the positive solution is valid.

There aren't any Real values for x that satisfy this equality:

The following Imaginary numbers satisfy it:  i(2n-1)*pi/2  where n is an integer.

what values of x give cosh (x) =0 ?

\(\begin{array}{rcl} \cosh{(x)} &=& \dfrac { e^x + e^{-x} } {2} = 0 \\ \dfrac { e^x + e^{-x} } {2} &=& 0 \\ e^x + e^{-x} &=& 0 \\ e^x + \dfrac{ 1 } { e^{x} } &=& 0 \qquad | \qquad \cdot e^x\\ e^{2x} + 1 &=& 0 \qquad | \qquad \boxed{~ \text{Euler's identity: } e^{i\pi} + 1 = 0 ~}\\ e^{2x} + 1 &=& e^{i\pi} + 1 \\ e^{2x} &=& e^{i\pi}\\ 2x &=& i\pi \\ x &=& \dfrac{ i\pi } {2} \end{array}\)

I'll assume this is \(17332.50=Ko(1+0.35)^6\)

If so, then divide both sides by 1.35^6 to get Ko = 17332.5/6.05344514  or Ko ≈ 2863.25

.

35% of 80

\(\begin{array}{rcl} 80 \cdot 35\ \% &=& 80 \cdot \frac{35}{100} \\ &=& 35 \cdot \frac{80}{100} \\ &=& 35 \cdot \frac{8}{10} \\ &=& \frac{35\cdot 8}{10} \\ &=& \frac{7\cdot 8}{2} \\ &=& 7\cdot 4 \\ &=& 28 \end{array}\)

Box A contained 579 paper clips. Box B contained 756 paper clips. After Kevin moved some paper clips from Box B to Box A, Box A contained twice as many paper clips as Box B.

b) How many paper clips did Kevin move from Box B to Box A?

Box A = 579 paper clips

Box B = 756 paper clips

Kevin moved some paper clips from Box B to Box A:

\(\begin{array}{rcl} A+x &=& 2\cdot ( B-x ) \\ A+x &=& 2B-2x \\ 3x &=& 2B - A \\ x &=& \dfrac{2B-A}{3}\\ \hline \\ x &=& \dfrac{2\cdot 756- 579}{3} \\\\ x &=& \dfrac{933}{3} \\\\ x &=& 311 \\ \end{array}\)

Kevin moved from Box B to Box A 311 paper clips

a) How many paper clips did Box B contain in the end?

\(\begin{array}{rcl} B -x &=&756 - 311 = 445 \end{array}\)

Box B did contain in the end 445 paper clips