All Answers 358524 Answers

6.8846276

the number is composite since it is devided by 3 and gives the quotient of 73 (a prime factor of 237) and  a remander of 0  

7.6ft^3   =    7.6(1728 in^3)  = 13132.8 in^3

So we have  Volume  =  W * L* H

x * (x + 5) (3x - 1)  = 13132.8       expand the left side

3x^3 + 14x^2 - 5x = 13132.8

3x^3 + 14x^2 - 5x - 13132.8  = 0        this is probably best solved by a computer algebra system

WolframAlpha  gives the solution as about   x =  14.97 in   = about 15 in

So....the width is about 15 in

The length is about [ 15 + 5]  = 20 in

And the height is about [ 3*15 - 1]  =  44 in

[The first answer looks good ]

The answer is (1)- x=~15 inches

Dimensions are: 15, 20, 44 inches.

thank you so much CPhill!

its a prime right 

1.   x^4 -52x + 576  = 0

This has no real solutions.........see the graph, here :https://www.desmos.com/calculator/dniadv9jnx

2.   x^3  = 216

x^3  - 216  = 0       factor  as a difference of cubes

(x -6) (x^2 + 6x + 36)  = 0       

Setting each factor to 0, it's clear that x = 6  is the real solution

Setting x^2 + 6x + 36  to 0    and using the quadratic formula we obtain  two non-real solutions :

-3-3 i sqrt(3)   and  -3+3 i sqrt(3)   [the last answer ]

when you simplifing that it is clearly shown that x=9

\(x*\sqrt{x}-10*\sqrt{x}-3=0\\(x-10)*\sqrt{x}-3=0\\(x-10)*\sqrt{x}=3\text{ (square both sides)}\\(x-10)^2*x=9\\ x=9\)

\(9+\frac{1}{9}=9.1111111111111111\)

x-(-6)=12

--=+

X+6=12

x=12-6

x=6

Ik I'm not a top user, but I'm here to help! :D

I'm fine and you

and how are you

Really? Why'd she quit? 

We have

2A  = A(1 + r/4)^(4t )     divide by A

2 = (1 + r/4)^(4t)          take the log of both sides

log 2  = 4t log ( 1 + r/4)      divide both sides by   4log(1 + r/4)

log 2  / [ 4 * log (1 + r/4) ]    = t        where r is some rate of interest  and t is the time in years

6                       

Lol!! IT WAS SO HORRIBLE  :(

3/4=.75 Just use the calculator here.

There is a key labled "ncr" in the far left of the bottom row.

what is the doublung time if interest is compounded quarterly?

WHAT INTEREST RATE YOU HAVE IN MIND?

Solve for x: -3-10 sqrt(x)+x^(3/2) = 0      Simplify and substitute y = sqrt(x): -3-10 sqrt(x)+x^(3/2) = -3-10 sqrt(x)+sqrt(x)^3 = y^3-10 y-3 = 0: y^3-10 y-3 = 0 The left hand side factors into a product with two terms: (y+3) (y^2-3 y-1) = 0 Split into two equations: y+3 = 0 or y^2-3 y-1 = 0 Subtract 3 from both sides: y = -3 or y^2-3 y-1 = 0 Substitute back for y = sqrt(x): sqrt(x) = -3 or y^2-3 y-1 = 0 Raise both sides to the power of two: x = 9 or y^2-3 y-1 = 0 Add 1 to both sides: x = 9 or y^2-3 y = 1 Add 9/4 to both sides: x = 9 or y^2-3 y+9/4 = 13/4 Write the left hand side as a square: x = 9 or (y-3/2)^2 = 13/4 Take the square root of both sides: x = 9 or y-3/2 = sqrt(13)/2 or y-3/2 = -sqrt(13)/2 Add 3/2 to both sides: x = 9 or y = 3/2+sqrt(13)/2 or y-3/2 = -sqrt(13)/2 Substitute back for y = sqrt(x): x = 9 or sqrt(x) = 3/2+sqrt(13)/2 or y-3/2 = -sqrt(13)/2 Raise both sides to the power of two: x = 9 or x = (3/2+sqrt(13)/2)^2 or y-3/2 = -sqrt(13)/2 Add 3/2 to both sides: x = 9 or x = (3/2+sqrt(13)/2)^2 or y = 3/2-sqrt(13)/2 Substitute back for y = sqrt(x): x = 9 or x = (3/2+sqrt(13)/2)^2 or sqrt(x) = 3/2-sqrt(13)/2 Raise both sides to the power of two: x = 9 or x = (3/2+sqrt(13)/2)^2 or x = (3/2-sqrt(13)/2)^2 -3-10 sqrt(x)+x^(3/2) => -3-10 sqrt(9)+9^(3/2) = -6: So this solution is incorrect -3-10 sqrt(x)+x^(3/2) => -3-10 sqrt((3/2-sqrt(13)/2)^2)+((3/2-sqrt(13)/2)^2)^(3/2) = -6: So this solution is incorrect -3-10 sqrt(x)+x^(3/2) => -3-10 sqrt((3/2+sqrt(13)/2)^2)+((3/2+sqrt(13)/2)^2)^(3/2) = 0: So this solution is correct The solution is: Answer: | | x = (3/2+sqrt(13)/2)^2 =~10.908......

 x+1/x=?=10.908 + 1/10.908=~11

Yes she said that yesterday 

ok ill be like that

Yall stupid

Cyber... Don't be like that.

3x+45=60

Subtract 45 from both sides.

3x=15

Divide by 3 on both sides.

x=5

So, x=5

put interms of x any number and find one by one points on the graph and then cross a line through them 

Here s your graph: