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Note that \(1 \over {1 \over3}\) is 3. When \(x\) gets bigger, the value of the expression gets smaller. 

Likewise, \(1 \over -{1 \over 4}\) is -4. As \(x\) gets smaller than \( -{1 \over 4}\), the value of the equation gets bigger, but never is greater than 0. 

So, the 2 ranges that work are \(\color{brown}\boxed{{x<-{1\over4} \space \text {and} \space{x>{1\over 3}}}}\)

it's great you found it out yourself i also got 15 degrees which is a tiny bit of proof your answer is in fact correct

it would really help if you can provide a picture

I don't think there is a point...

ok..

i don't know if this is possible..

x can equal 1 to infinity and it works

You have the right idea guest, here is how you solve it: \(x+y+\sqrt {x^2+y^2}=30\)

Subtarct \(x+y\) from both sides: \(\sqrt{x^2+y^2}=30-x-y\)

Square both sides: \(x^2+y^2=x^2+y^2+2xy-60x-60y+900\)

Subtract \(x^2+y^2\) from both sides:\(0=2xy-60x-60y+900\)

Subsitute \(xy\) for 60:\(1020-60x-60y=0\)

Rewrite as: \(60x+60y=1020\)

Factor the left side: \(60(x+y)=1020\)

Divide both sides by 60: \(x+y = 17\)

From the other equation, divide by \(y\): \(x = {60\over y}\)

Plug into the other equation: \({60\over y}+y=17\)

Multiply by \(y\): \(60+y^2=17y\)

Subtract \(17y\) from both sides: \(y^2-17y+60=0\)

Use quadratic formula, and you find \(y = 12\)

Subsitute into the other equation, and \(x = 5\)

This means that the third side is \(13\), since the triangle must have a perimeter of 30. 

Thus, the side legnths are: \(\color{brown}\boxed{5,12,13}\)

the guest's were both me i forgot to login lol

so draw a line from b to c. ad and bc will intersect at the center of the circle. call that point O. the ratio of angle aob(80 degrees) to a full circle(360 degrees) is the same as the ratio of AB to the cirumference of the circle. in the end, you get 8pi..

i just realizing i was calculating the arc AB because the question on alcumus was asking that

I got it.. angle x is 15 degrees. The perpendicular from D to CB is half of CD. The perpendicular from D to AB and the perpendicular from D to CB form a rectangle. Thus, you can do some 30-60-90 shenanigans, and get that the perpendicular from D to AB (point p) is 2 - sqrt(3) times AP. Thus, with the side ratios I discovered, I got x = 15 degrees. :D :D :D

Do the math if you are a bit confused, i big brained it lol :)

Does 1 look right?   

is 2 the same as 1?

is 3 the same as 2?

is 4 the same as 3?

Where does it go different?

You do not need help with this. There is a calculator on this site if you need to use it.

Maybe first explain your solution, so we can perhaps figure out what's wrong?

Please do not post random answers here. First of all, 9^5 is some big random number. We already know that it is incorrect. 

I think this question has already been posted on this website. 

Anyways, just like EP said, adding them should get your answer 

b + c = (-5)^2 - 2*3 = 19.

When the dolphin re-enters the pool   height (y) = 0

  solve   0 = -16t^2 + 24t    for t     

     another way......note that the dolphin reaches a peak at .75 sec.....  another .75 sec brings the dolphin back down .75 + .75 = 1.5 sec

   Max height occurs at .75 sec       max height = -16(.75)^2 + 24 (.75)   feet

The numerical part is 4*8=32.

32n

Connect  AO,  BO , PO

Tangents drawn from a single point outside a circle are equal

OP = OP

AO = BO

AP = BP

So by SSS.....triangles APO and BPO  are congruent

Then because angle OPA = 30° then angle  APB = 60°

And angle AOB is supplemental to angle APB =  120°

And central angle APB = measure of minor arc AB = 120°

Not as difficult  as it seems if we think about this

sum [ ( sin 5° + sin 10° + sin 15° +  ..... + sin 180° )  + ( sin355° + sin 350° + sin 345° + .... +sin 180° ) ]  =  0

And sin 0° = sin 360°  = 0 

Adding all these, the sum  =  0

Center at   -1,1

(x+1)^2   + (y-1)^2   = r^2         r^2 = distance between center and the point on the circumf     r^2 = ( 3^2 + 2^2) = 13

(x+1)^2 + ( y-1)^2 = 13      expand to get to the form requested ....

x^2 + 2x + 1    +  y^2  -2y + 1   - 13 = 0

x^2   + 2x        +  y^2  - 2y     -11   = 0                 <========= take if from here !