All Answers 358512 Answers

12 / 18  =    x / 28        [reduce 12 / 18  to simplest form ]

2 / 3   =  x / 28       cross-multiply

28 * 2   =  3x

56  = 3x         divide both sides by 3

56/3   =  x

2x+8y=6    divide through by 2  →  x + 4y  = 3  

-5x+-20y=-15     divide through by -5 →  x +4y = 3  

These equations are exactly the same.....so we will have infinite solution pairs for (x , y)......!!!

easy because  you could take the first one and divide with 2 and it would look like x+4y=3 and also you could take the second one and divide with -5 respectively and it would look like x+5y=3 and then you solve a system  {nl} x+4y=3=x=3-4y {nl} x+5y=3  {nl} and then where ever there is an x at the 2nd equation you replace with 3-4y then solve a 1st grade equation and find y after that replace y on the top equation

Yo!!!! Go and play video games!!.

Give examples of your fractions, so we can help you.

The endpoints will lie below the x axis.....see the graph, here :   https://www.desmos.com/calculator/f4l7hjfy0y

Answered here :   http://web2.0calc.com/questions/solve-the-equation-for-exact-solutions-in-the-interval-0-360-sin2theta-2cos-2theta

Just use the arcsin function.....example :

arcsin (0.5)  = 30” 

Take its "Inverse sine" or "asin" on this calculator. Press "2nd" key first, then you should see "asin" key. Don't forget to press = sign to get your answer.

Yes! You're right about the problem being wrong! I couldn't figure out how to make something dquared using a keyboard and I didn't remember to state that! Thank you for your help!

Here's the second one :

 arcsinx-2arccos sqrt(3)/2 = pi/3           [ arccos sqrt(3/2)  has two values on [0, 2pi] .......pi/6    and 11pi/6 ]

So we have

arc sinx - 2[pi6]   = pi/3

arcsinx - pi/3  =   pi/3       add pi/3 to both sides

arcsinx =   2pi/3

Notice that no solution exists, because the sine's greatest value =  1....so we are trying to find an angle whose sine = 2pi/3 .......  and  2pi/3 > 1, so this doesn't angle exist

Also........if we calculated this for 11pi/ 6, we would have the same problem

arcsinx - 2 [11pi/6]  = pi/3

arcsinx - 11pi/3 =  pi/3

arcsinx = 12pi/3    = 4pi       and this is much greater than 1

So........no solution exists for the second problem

I think it's supposed to be "2sec^2x "  on the right side

(1+sin x)/(1 - sin x) =2sec^2x + 2sec x tan x - 1

Multiply top/bottom  on the left hand side by ( 1 + sinx)

[ 1 + sinx) (1 + sinx)] / [ (1 - sinx) (1 + sin x) ]  = 2sec^2x + 2sec x tan x - 1  

Simplify the LHS and remember that [ 1  = sec^2x - tan^2x]      

[ 1 + 2sinx + sin^2x] / (cos^2x]  =  2sec^2x  + 2secxtanx - [ sec^2x - tan^2x]

[ 1 + 2sinx + sin^2x] / (cos^2x]   = sec^2x + 2secxtanx + tan^2x

And, on the left, we can write :

1/cos^2x + [2sinx]/ [ cosx * cosx]  + [ sin^2x / cos^2x]

sec^2x  + 2[sinx/cosx] * [1/ cosx]  + tan^2x

sec^2x  + 2tanxsecx + tan^2x

sec^2x + 2secxtanx + tan^2x        and this equals the RHS

3x^3/5v^3 * 25v^5/9x

Now, take a look at the expresion! Is it:[3x^3] / [5v^3 * 25v^5/9x] this one? or is it this one:[3x^3/5v^3] * [25v^5/9x]. Or [3x^3]/[5v^3] * [25v^5] / 9x, or what? I await your instructions, Sir!!!.

75 31/50

you guys r geeks

Don't you think that you should be a little more careful about your brackets!!! Just look at this example. It will make life easier for the people who are trying to help you.

Simplify the following:

(35 w^2-1715)/(25 w^2-1225)

Factor 25 out of 25 w^2-1225:
(35 w^2-1715)/25 (w^2-49)

w^2-49 = w^2-7^2:
(35 w^2-1715)/(25 (w^2-7^2))

Factor the difference of two squares. w^2-7^2 = (w-7) (w+7):
(35 w^2-1715)/(25 (w-7) (w+7))

Factor 35 out of 35 w^2-1715:
35 (w^2-49)/(25 (w-7) (w+7))

w^2-49 = w^2-7^2:
(35 (w^2-7^2))/(25 (w-7) (w+7))

Factor the difference of two squares. w^2-7^2 = (w-7) (w+7):
(35 (w-7) (w+7))/(25 (w-7) (w+7))

(35 (w-7) (w+7))/(25 (w-7) (w+7)) = ((w-7) (w+7))/((w-7) (w+7))×35/25 = 35/25:
35/25

The gcd of 35 and 25 is 5, so 35/25 = (5×7)/(5×5) = 5/5×7/5 = 7/5:
Answer: | 
| 7/5

Sorry i ment to say 2arccos not arcc

and thank you very much!

sin^2 (theta)  = 2cos^2(theta)

sin^2(theta)  = 2[ 1 - sin^2(theta)]

sin^2(theta)  = 2 - 2 sin^2(theta)       add 2 sin^2(theta)  to both sides

3sin^2(theta)   = 2     divide both sides by 3

sin^2(theta)  = 2/3      take the positive/negative sqrt 

sin(theta)  =  sqrt (2/3)          and       sin(theta)  =  - sqrt(2/3)

Using the arcsin  we have

arcsin[ sqrt(2/3)]   = about 54.74°   and  about  (180 - 54.74)°  = 125.26°

arcsin [ - sqrt(2/3)]  =  about  - 54.74°   =  (180 + 54.74) ° =  about 234.74°   and about (360 - 54.74)° = about 305.26°

Here's a graph of the intersection points of the two functions........https://www.desmos.com/calculator/8bcyqeldft

[ For your second question, what is "arcc"  ???]

[z^2-6z+9]/[z^2+2z-15]       factor top and bottom

[ ( z -3) (z - 3) ] / [ ( z + 5) ( z - 3) ]       cancel (z - 3)   on top/ bottom

( z - 3) / ( z + 5)

The estimate is: 7.3 billion, or 7.3 X 10^9 people.

a^2-25/a^2+7a+10   factor top and bottom

[ (a + 5) (a - 5) ] / [ ( a + 5) ( a + 2) ]         " cancel" (a + 5)   factors

( a - 5) / ( a + 2)

Where did you come up with such a question? At any rate, if you can understand it, here is the answer:4.256×10^-9 mph  (miles per hour)

Well zero is definitely a number and without it our whole number system would collapse. 

Sir CPhill is still looking for the Roman Zero though.  

How is your quest going Sir CPhill?

I am sure you were told that it is in Sisyphus's boulder so what is the hold up?

5y-5/5y  =

[ 5 (y - 1) ] / [ 5 (y) ] =

(5/5) * [ y - 1] / y  =

[y - 1 ] / y

It is already in "degree" mode. Look at the bottom left-hand corner. You will see "degrees" and "rads".