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The topic is inscribed angles.  I have to find the either the arcs or the angles of a circle.  As far as I know none of the problems are supposed to be decimal answers.

not really sure what this is but,,, a tip i would give is to simplify the equations to get plain x so idrk... 

maybe that's why you got it wrong, but i don't know...just simplify for example the 14x=64


please correct me if i am wrong, and i am sorry i wasn't much help :') but i hope this fixes your problem :"))

Part A

The degree of the numerator  is equal to the degree of the denominator, so the horizontal asymptote is

y  =  a / 1     which we know is  -4

a / 1  = -4

a  =  -4

There is a vertical asymptote at  x =  3

So we know when  x = 3 ,    x + c  =  0

3 + c  =  0

c  =  -3

Now we know:     f(x)  =  (-4x + b)/(x - 3)

There is an x-intercept at  (1, 0)  so we know

f(1)  =  0

(-4(1) + b)/(1 - 3)  =  0

(-4 + b)/( -2 )  =  0

-4 + b  =  0

b  =  4

Altogether:     f(x)  =  (-4x + 4)/(x - 3)

n / √n  >   [ n + 1 ]  / √[n + 1]

n / √n   - [ n + 1 ] / √[n + 1]   > 0

n √[ n + 1] - [n + 1]√n 

__________________  >  0

          √n * √[n + 1]

n √[n + 1]  - [n + 1]√n  >  0       

n √[n + 1]  > [ n + 1]√n        square both sides

n^2 [ n + 1]  > [n + 1]^2 * n

n^2 [ n + 1] - [n + 1]^2*n > 0

n [n + 1]  [ n - (n + 1 ) ]  > 0

n (n + 1] [ - 1] >  0

n ( n + 1) < 0

This is only true on   -1 < n < 0

So....no poistive  integers make this true

Also.....see the graph here : https://www.desmos.com/calculator/tmsohselsc

Q(n)  is never greater than Q(n + 1)

What about this? I'm not sure if it counts as an algebraic rule!

\(y=\max\left(-\left|2x-3\right|+2,1\right)\)    on the unbroken domain  (0, 2)

Thx, I got it now, aced my test also :D

Pattern I went with made it so easy for me,

I can just take x=7cot(39)

divide the 7 out right away so that x over something = cot(theta)

x/7 = cot39

Then just set trig function what side it represents, in this case adj/op

x/7 = a/o

and it matches up every time x=a and 7=o adjusted to theta 

Good so far thanks 

Yeeeeeeeeeeeeessssssssssssssssssssssssssssssssss!!!!!!!!!!!!!!!!!!!!!!

heh I'm game

I'm in lol

Just a pattern I noticed: 

x=3 sec 42

let it be in that form: 

x= r sec z (angle)

sec z= r/f 

where f is the adjacent side. (Changes depends on the trigonometric  function etc) 

while r is the hypotenus 

so r=hyp. 

in this case: 

x= 3 sec 42

r=3 

so x is the adjacent according to the trigonometric function sec hyp/adj

Don't hesitate to ask questions. :)

Ok, sure. I'm in.

Since the correct clock is: 60 / 50 =1.2 or 20% faster for every minute of the slower clock, it therfore follows that in 3 hours or 180 minutes on the slower clock will be: 180 x 1.20  =216 minutes or 3 hours and 36 minutes on the regular clock.

Thanks for your help.

Can you recommend anything to help me expand as partial fractions, please?

"how do you find the inverse Laplace transform of 5/(s-1)(s2+4)"

Assuming the function of interest is  \(f(s)=\frac{5}{(s+1)(s^2+4)}\)

First expand as partial fractions:  \(f(s)=\frac{1}{s-1}-\frac{s}{s^2+4}-\frac{1}{s^2+4}\)

Then do the inverse Laplace transforms of these (the first term wil result in an exponential, the second will have the form of a cosine, and the third the form of a sine).

4x = 3y     so   x = 3/4 y

7y = 5 z    so   z = 7/5  y

3/4 y - y + 7/5 y = -46

y=- 40

x = 3/4 y = -30

z= 7/5 y =-56

Okay. Great. Thanks for answering. Have a great day.

You're welcome, CR !    Keep after it........

Thank you EP

I see where i messed up thx

You can use either method....whichever the Q asks you t use....