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y = 4log₅(3-x) + 2   .........set y =0

0 = 4log₅(3-x) + 2   .........subtract 2 from both sides

-2 = 4log₅(3-x)       ..........divide by  4

-1/2 = log₅(3-x)        ....in exponential form, we have

5 ^-1/2 = 3-x               .....or........

1/√5 = 3 -x              .......add x to both sides, subtract 1/√5 from both sides

x = 3 - 1/√5  =  [3√5 - 1]/√5 

So we have.......  ( [3√5 - 1]/√5 , 0 ) 

2N= Ne^.10t       where N is the original population

divide by N on both sides

2 = e^.10t         take the ln of both sides

ln2 = lne^.10t    and by a log property we can write

ln2 = (.10t) lne        and lne = 1   ....so....

ln2 = .10t            divide both sides by .10

(ln2 / .10) = t ≈ 6.93 hrs.

(1-cosx)(cscx+cotx)

Well..it could be alot of things....let's see

(1-cosx) (1/sinx + cosx/sinx)

(1-cosx)(1 + cosx)/sinx

(1-cosx)² / sinx

(sinx)² / sinx

(sinx)

And there you go......

Provided that there are no parentheses in this problem, the order of operations would be as follows:

Division/multiplication first, then addition/subtraction.

Divide 5/25

= 0.2

Multiply by 100

= 20

Subtract answer from 25

= 5

If you meant to do the subtraction first, you would need to put parentheses around it:)

25-5=20

20/25=.8

.8*100=80

$$\left(\dfrac{3}{x^2}-4x^3\right)^{10}$$

   $$\mbox{The (r+1)th term will be }\quad 0\le{r}\le10\quad r\in \mathBb{Z}$$      

$$10Cr\times \left(\frac{3}{x^2}\right)^r\times (-4x^3)^{10-r}\\


=10Cr \times 3^r\times(-4)^{10-r}\times \dfrac{1}{x^{2r}}\times (x^3)^{10-r}\\

=10Cr \times 3^r\times(-4)^{10-r}\times x^{-2r}\times x^{30-3r}\\

=10Cr \times 3^r\times(-4)^{10-r}\times x^{-2r+30-3r}\\

=10Cr \times 3^r\times(-4)^{10-r}\times x^{30-5r}$$ 

You want the constant term so 30-5r=0,   r=6  

so the term we want is 

$$10C6\times 3^6\times (-4)^4 \times x^0$$

etc (correct assuming I didn't make any stupid errors)

Thanks Sportslover,

Anon, 

Can you give an example of what you need to do.  Your question is a bit vague.

No, Im assuming your meaning with logs...You can only put log base 10s or natural logs into your calculator

so your best bet is to convert it into exponential form or use your log laws !

From my understanding your talking about using 60 degrees as a reference angle

clockwise : 0degrees-60=300degrees  

or 90degrees-60=30degrees

counterclockwise :0 degrees +60=60degrees

or 90degrees+60150degrees

does this make sence?

here this might help :) this is what i used !

A decimal point (5.9)

okay then.

yea its a supervised exam haha no im done all my course work these are just types of questions they didnt cover very well so i wanna see how others do it os see if they have tricks :)

do you have to pass a supervise test for that?

Are you getting us to do all your homework for you?

You have a lot of questions up here.

haha ya thanks ! i hope so to ! I have a diploma june 24th

You are very welcome.

I see you only joined yesterday.

Welcome to Web2.0calc forum.  I hope that you learn lots and have fun here.

no that totally helps thanks !

y-3x=5 | 2x=3y+6

y=3x+5

|

|

V

2x=3(3x+5) + 6

2x=9x+15+6

     -9x

-7x=21

-(-7x=21)

7x = -21

x = -3

y-3(-3)=5

y-(-9)=5

y+9 = 5

  -9    -9

y= -4

So, y= -4 and x = -3

Let's check it:

y-3x=5 

(-4) - 3(-3) = 5

-4 - (-9) = 5

-4 +9 = 5

5=5 ✔

$$(3y-2x)^5$$

the highestr power of y is going to be 5 followed by 4,3,2,1,0

The forth largest is the power of 2 so you need the term with y² in it.  Which will be

$$(3y)^2...$$

The whole term will be

 $$5C2\times (3y)^2 \times (-2x)^3 = 10\times9y^2\times -8x^3 = -720x^3y^2$$

Now if you don't understand, tell me what is troubling you and I will attempt to explain better!

I would love for you to explain it thanks !

That is what I am saying.  you should ask people to explain.  And give thumbs up when you are totally satisfied that you have been taught or at lest told what you need to know.

would you like me to try and explain or are you happy with just the answer?

probably , but im not sure how to

weren't you supposed to be able to do it without a calculator?

it was a multiple choice question melody :)

Alan, I strong suspect that this was supposed to be done without a calculator!

Although, since the Sportslover gave you a thumbs up, maybe I am wrong.