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Let the time it takes to hike downhill =T

Distance = Speed x Time

5T =3(T+40), solve for T

5T = 3T + 120

5T - 3T = 120

2T = 120

T =120 / 2

T =60 - minutes, or 1 hour, to hike downhill.

Distance =1 x 5 mph=5 miles.

ProMagma,

Could you please tell me what the formula for the volume of a sphere is?

You can look it up, that is ok :)

The easiest way to work out probablilities with 2 dice is to draw up a 6 by 6 grid. 

Die 2 is down the side

Die 1 is acros the top and 

The red ones are the combinations that work. There are 20 of them

So

P(of a 6 OR a 1)  = 20/36 = 5/9

Hi Guest, I hope you acturally see my answer, I'd have more confidence if you were a member.

If you do see it it would be nice if you leave a comment saying so :)

verify by using induction

5^n+2*3^n+5 is divisible by 8

Step 1

Prove true for n=1

LHS=\(5^1+2*3^1+5=5+6+5=16=2*8 \)

Which is divisable by 8.  

So true for n=1

Step 2

Assume true for n=k     Where k is a  positive integer.

so

\(5^k+2*3^k+5 =8M \qquad where \;\;k \in N\;\;\;and\;\;\;M \in N\;\;(natural \;number)\\ \text{Prove that this will be true for n=k+1}\\ \text{That is, prove}\\ 5^{k+1}+2*3^{k+1}+5 = 8G\qquad G\in N\)

\(LHS=5*5^{k}+2*3*3^{k}+5\\ LHS=5^{k}+4*5^{k}+2*3^{k}+4*3^{k}+5\\ LHS=[5^{k}+2*3^{k}+5]+4*5^{k}+4*3^{k}\\ LHS=[8M]+4(5^{k}+3^{k})\\ \)

Now 5 to the power of any positive integer will have 5 as the last digit so 5^k will be an odd number.

3^1=3, 3^2=9, 3^3 ends in 7, 3^4 ends in 1 and so the pattern repeats.

So 3^k will have a last digit of 3,9,7, or 1 so     3^k will be an odd number.

An odd number +an odd number = an even number and all even numbers are divisable by 2 

so

\(5^k+3^k=2H \qquad H \in N\\ \therefore\\ LHS=8M+4*2H\\ LHS=8M+8H\\ LHS=8(M+H)\\ LHS=8G\\\)

So if the expression is a multiple of 8 for n=k then it is also a multiple of 8 for n=k+1

Step 3

Since the espression is a multiple of 8 for n=1 it must be a multiple of 8 for n=2, n=3 .....

Hence the expression is a multiple of 8 for all positive interger values of n

QED

Hey....I know who that "chick" is.....

"Mr. "Green Jeans" from "Captain Kangaroo".....before surgery, of course

OK...no prob....!!!

You may have missed your calling, JB. You should have been a philosopher. I’m adding your philosophical thought to my toolbox of aphorisms.

Right, of course. Thanks man!

No....a rational function is one that has a polynomial in the numerator and the denominator....note that

(x + 3)  / (x^2 - 2)    would be an example....but...

 (x^2+2x+3)/3  is actually just

(1/3)x^2  + (2/3)x  +  1       and this is just a quadratic function.....

Those are pretty cool...especially the "Stuck In Space" one....!!!!

This equation is a straight line which is in slope-intercept form.

You can observe easily that slope = -3, y-intercept = -4 

the graphing? do it yourself.

Mmm

I think you might mean

\(\dfrac{5}{\frac{1}{6} + \frac{1}{x}+1}\qquad \text{firstly x cannot = 0}\\ =5 \div [\frac{1}{6} + \frac{1}{x}+1]\\ =5 \div [\frac{x}{6x} + \frac{6}{6x}+\frac{6x}{6x}]\\ =5 \div [\frac{7x+6}{6x} ]\\ =5 \times[\frac{6x} {7x+6}]\\ =\dfrac{30x} {7x+6}\\\)

Good explanation of what Max?

Anyway it is copied directly from here:

"Calculus".

Isn't that awesome? LOL

Good explanation :)

1)

f(b,3) = 2b - 3(3)^2 + 7  = 2b - 20

2b - 20 = 90

2b = 110

b = 55

2)

b(3) = 2 - 3^3 = -25

a(3) = 3(3) - 7 = 2

a(b(3)) - b(a(3))

= a(-25) - b(2)

= 3(-25) - 7 - (2 - 2^3)

= -75 - 7 + 6

= -76

Let f(x) = 5x+3 and g(x)=x^2-2. What is g(f(-1))? 

\(f(x)=5x+3\\ f(-1)=5*-1+3=-2\\~\\ g(x)=x^2-2\\ g(f(-1))=g(-2)=(-2)^2-2=2\)

and

If f(x)=16/(5+3x), what is the value of [f^-1(2)]^-2?

\(f(x)=\frac{16}{5+3x}\\ let\;\; y=\frac{16}{5+3x} \qquad x\ne-\frac{5}{3}\\ 5y+3yx=16\\ 3yx=16-5y\\ x=\frac{16-5y}{3y}\\ f^{-1}(x)=\frac{16-5x}{3x}\qquad x\ne0\\ f^{-1}(2)=\frac{16-5*2}{3*2}\\ f^{-1}(2)=\frac{6}{6}\\ f^{-1}(2)=1\\ [f^{-1}(2)]^{-2}=1\\ \)

Here is a graph:

You need to show working ProMagma.

Answers without working are usually of no use. Not it the asker is trying to learn anything anyway.

If you do not know how to do it and you get the answer from somewhere else then you should site your source, 

Guest, if you want more info than you need to come back and ask for it. 

Ah.  Trick question?

You gave me 3 numbers to work with, but the answers say "squared."  I guess I am finding the area of the walls in total.

The area formula is \(bh\).  In this problem, \(b=12 \) and \(h=9\).  Each wall is \(108 ft^2\).

The room is a square (12 by 12), so all the walls' areas are the same.  So, \(108 ft \cdot\ 4 ft= \boxed{432 ft ^2}\).  The answer choice is \(\boxed{A}\).

(x^5-1)/3, find f^-1(-31/96)

Write  

y = (x^5-1)/3       

3y  = x^5 - 1

3y + 1  =  x^5

(3y + 1)^(1/5)  = x        swap x and y

(3x + 1 )^(1/5)  = y  = f^-1 (x)

So

f -1 (-31/96)  =    ( 3 (-31/96)  + 1 )^(1/5)  = (3/96)^(1/5)  = .03125^(1/5)  = .5 =  1/2  !!!!!!

\(3\) is a prime!  So, just \(3 \cdot\ 1\).   

(x^(2)+8x-20)/(x^(2)-4)   factor

[ (x +10) ( x - 2) ] /  [ ( x - 2) (x + 2 ) ]       "cancel" the (x - 2)  factor

( x + 10 )  / ( x + 2 )

So a function (?) question.  Also, it doesn't specify whether it should be an integer or not.  I'm guessing you mean an integer.  

We are told that \(x = x\) and \(y=3\).  Plug it in.

\(x\)#\(3\)\(=\dfrac{x^2}{3}\)

Also, \(x\)#\(3=9\).  Plug that in too.

\(9= \dfrac{x^2}{3}\) .  Now multiply both sides by 3 to get rid of the fraction...

\(27=x^2\).  So \(x = \sqrt{27}\).  It is also equal to \(- \sqrt{27}\).  THAT DOES NOT LOOK LIKE A NICE NUMBER.  

Unless I made a mistake which is 100% possible.