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Simplify the following:
(64/81)^(-2/3)

(64/81)^(-2/3) = (81/64)^(2/3):
(81/64)^(2/3)

(81/64)^(2/3) = (81^(2/3))/(64^(2/3)):
(81^(2/3))/(64^(2/3))

81^(2/3) = (81^2)^(1/3) = (3^8)^(1/3) = 3^2×3^(2/3):
(3^2×3^(2/3))/(64^(2/3))

3^2 = 9:
(9×3^(2/3))/(64^(2/3))

64^(2/3) = (64^2)^(1/3) = (2^12)^(1/3) = 2^4:
(9×3^(2/3))/(2^4)

2^4 = (2^2)^2:
(9×3^(2/3))/((2^2)^2)

2^2 = 4:
(9×3^(2/3))/(4^2)

4^2 = 16:
Answer: |(9×3^(2/3))/(16)

14/25

Yes, thank you very much!!

question:  I still don't get it.

3+1=4

and

3-1=2

y=2 and y=4 are on either side of the centre - the middle is y=3   

so (0,3) must be the 'centre' of your graph,

if the graph was   y = tanx then one centre  is (0,0)  

If you want the centre to be (0,3) it will have to be  y= (tanx) +3

does that help?

Ninja's question is

"Where the heck did you get the 3 from"    :D

y=tanx   is centred on (0,0)     (at least on of the periods is)

so if you wanted points rotationally symmetrical on either side  they would be    

(0-x,0-y) and (0+x,0+y)

so they could be

your points are   (-pi/2 , 2) and (pi/2 , 4)

The y values are 2 and 4 so the centre of them is 3  Hence the graph has be lifted up 3 .

That is where the 3 came from :/

Press "2nd" key first, then "acos", which is the same as cos^-1.

 5 x -2 = -10

5 +(-2) = 3 

Evaluate (54 3^n)/5 where n = -4:
(54×3^n)/(5) = 54/5×1/3^4

54/5×1/3^4 = 54/(5×3^4):
54/(5×3^4)

3^4 = (3^2)^2:
54/(5 (3^2)^2)

3^2 = 9:
54/(5×9^2)

9^2 = 81:
54/(5×81)

The gcd of 54 and 81 is 27, so 54/(5×81) = (27×2)/(5 (27×3)) = 27/27×2/(5×3) = 2/(5×3):
2/(5×3)

5×3 = 15:
Answer: |2/15

log3(x+5)=1-log3(x+3)

Thanks guest, I just want to look too :)

\(log_3(x+5)=1-log_3(x+3)\\ log_3(x+5)+log_3(x+3)=1\\ log_3[(x+5)(x+3)]=1\\ log_3[(x+5)(x+3)]=log_33\\ (x+5)(x+3)=3\\ x^2+8x+15=3\\ x^2+8x+12=0\\ (x+6)(x+2)=0\\ x=-6 \qquad or \qquad x=-2\\ \text{But you cannot find the log of a negative number so -6 is no good}\\ x=-2\\ check\\ LHS=log_33=1\\ RHS=1-log_31=1\\good:)\)

x=-2 just as our first guest already determined. :)

Just flip them from 5/6 to 6/5.

Since it is tan maybe this is what is wanted.  It still doesn't make much sense though.

y=3+tan(x/2)

Oh dear, I did not see the tangent function part.     

Write a tangent function h(x), such that the halfway points on one period are (-pi/2 , 2) and (pi/2 , 4)

Mmm not even sure what this means Ninja....  what is a halfway point and how can you have 2 of them?

Maybe

\(y=3+sinx\)

full graph

Haha Ginger, 

Maybe your comptometer needs repair  LOL

This was an exercise in reading the question properly and you failed. :))

"which of the following is not a possible scenario "

Randall borrowed $850 from his parents to make some repairs on his car. He promised to repay the loan y giving his parents at least $94 from his paycheck each week. Assumming Randall sticks to his promise, which of the following is not a possible scenario for the amount of debt Randall has remaining?

850/94 = 9.0425531914893617     so it will take a maximum of 9 and a bit weeks to repay.

minimum repaid

1 week           94                       maximum  left to be repaid             850-94 = 756

2 weeks        188                       maximum left to be repaid             756-94 = 662

3  weeks       282                       maximum  left to be repaid             662-94 = 568

4 weeks        376                       maximum  left to be repaid            568-94 = 474

5 weeks        376+94 = 470      maximum   left to be repaid            474-94 = 380

6 weeks        470+94 = 564      maximum   left to be repaid            380-94 = 286

7 weeks        564+94 = 658      maximum   left to be repaid            286-94 = 192

8 weeks        658+94 = 752      maximum   left to be repaid            192-94 = 98

9 weeks        752+94 = 846      maximum   left to be repaid            98-94 = 4

A. After 5 weeks, Randall is $125 in debt.       yes this is possible

B. After 6 weeks, Randall is $175 in sdebt.     yes possible

C. After 7 weeks, Randall is $225 in debt.       not possible 

D. After 8 weeks, Randall is $75 in debt.         not possible

So if Randall pays his parents back as aggreed, scenario C and D are not possible

This could have been done a lot quicker but I wanted to show you what was happening.

Thanks Max, I am told your interpretation is correct :)

Hi Lisa,

How do you simplify (1+( sqrt 2/2) / (sqrt 2/2)

so you question is

How do you simplify (1+( sqrt 2/2)) / (sqrt 2/2)

or better still

How do you simplify(1+( (sqrt 2)/2)) / ((sqrt 2)/2)

This is how I would do it

\(\frac{1+\frac{\sqrt 2}{2}}{\frac{\sqrt 2}{2}}\\ =(1+\frac{\sqrt 2}{2})\div{\frac{\sqrt 2}{2}}\\ =(1+\frac{\sqrt 2}{2})\times{\frac{2}{\sqrt 2}}\\ \qquad \text{I want a rational denominator so if I multiply the bottom by sqrt2 }\\ \qquad \text{but what I do to the bottom I must do to the top as well because I am really multiplying by1}\\ =(1+\frac{\sqrt 2}{2})\times{\frac{2}{\sqrt 2}}\times\frac{\sqrt2}{\sqrt2}\\ =(1+\frac{\sqrt 2}{2})\times{\frac{2\sqrt2}{2}}\\ =(1+\frac{\sqrt 2}{2})\times\sqrt2\\ =\sqrt2(1+\frac{\sqrt 2}{2})\\ =\sqrt2+\frac{2}{2}\\ =\sqrt2+1\\\)

Now, if you still do not quite understand Lisa, make sure you say so :)

Sequence {10, 4, 5, 7, 4} satisfies mean, median and mode.

Solve for x:
(log(x + 5))/(log(3)) = 1 - (log(x + 3))/(log(3))

Subtract 1 - (log(x + 3))/(log(3)) from both sides:
-1 + (log(x + 3))/(log(3)) + (log(x + 5))/(log(3)) = 0

Bring -1 + (log(x + 3))/(log(3)) + (log(x + 5))/(log(3)) together using the common denominator log(3):
-(log(3) - log(x + 3) - log(x + 5))/(log(3)) = 0

Multiply both sides by -log(3):
log(3) - log(x + 3) - log(x + 5) = 0

log(3) - log(x + 3) - log(x + 5) = log(3) + log(1/(x + 3)) + log(1/(x + 5)) = log(3/((x + 3) (x + 5))):
log(3/((x + 3) (x + 5))) = 0

Cancel logarithms by taking exp of both sides:
3/((x + 3) (x + 5)) = 1

Take the reciprocal of both sides:
1/3 (x + 3) (x + 5) = 1

Multiply both sides by 3:
(x + 3) (x + 5) = 3

Expand out terms of the left hand side:
x^2 + 8 x + 15 = 3

Subtract 15 from both sides:
x^2 + 8 x = -12

Add 16 to both sides:
x^2 + 8 x + 16 = 4

Write the left hand side as a square:
(x + 4)^2 = 4

Take the square root of both sides:
x + 4 = 2 or x + 4 = -2

Subtract 4 from both sides:
x = -2 or x + 4 = -2

Subtract 4 from both sides:
x = -2 or x = -6

(log(x + 5))/(log(3)) ⇒ (log(5 - 6))/(log(3)) = (i π)/(log(3)) ≈ 2.8596 i
1 - (log(x + 3))/(log(3)) ⇒ 1 - (log(3 - 6))/(log(3)) = -(i π)/(log(3)) ≈ -2.8596 i:
So this solution is incorrect

(log(x + 5))/(log(3)) ⇒ (log(5 - 2))/(log(3)) = 1
1 - (log(x + 3))/(log(3)) ⇒ 1 - (log(3 - 2))/(log(3)) = 1:
So this solution is correct

The solution is:
Answer: |x = -2

Give the trains an arbitrary length of, say, 240 meters each. Since train A overtatakes train B, it means A is moving faster than B. Or, A - B =240 meters divided by 24 seconds =10 meters per second.
Since train C is also moving faster, which is the same speed as A, than train B, it appears to be moving at C + B meters per second. But, since A and C have the same speed, then A + B =meters per second.
Since it takes C 8 seconds to pass B, then A + B =240 meters divided by 8 seconds =30 meters per second.
Since we know from the above A - B =10 meters per second, it follows from both equations that:
A=20 meters per second and B = 10 meters per second. And, since A and C are travelling at the same speed, then A = C = 20 meters per second, which twice as fast as train B. Moreover, If A and C are travelling at the same speed, or 20 meters per second, their combined speed is 20 + 20 = 40 meters per second, when they pass each other.
Finally, since the A and C trains are 240 meters long, it therefore takes: 240 / 40 = 6 seconds to pass each other.
MODERATORS:Please verify my reasoning!!.

Sorry I wasn't clear enough. Here's what I'm really asking for:

(5 / (x-3) ) + ( 4 / (x+3) ) =  (9x+2) / ( (x-3) (x+3) )

Find "S"

4.95 (twitter.com/customkkk-

\(\frac{15}{400}*100=3,75%\)%

Divide the price$  by the kg amount

56/100    or simplified: 23/50