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Easiest  to put these in exponential form and then evaluate......

8x^(3/2)   [  8x^(1/2)  - 7x^(-1/2) ]  =

8 *8 * x^(3/2) * x^(1/2)   -  8 * 7 * x^(3/2) * x^(-1/2)   =

64 x^(3/2 + 1/2)  -  56x^(3/2 -1/2)  =

64x ^2  - 56x

(h,k) = center = (7,-10)

(x-7)^2 + (y+10)^2 = r ^2    r is equal to the y magnitude = 10

(x-7)^2 + (y+10)^2 = 100 

Here is a graph:

https://www.desmos.com/calculator/be4d5xjlg8

b^3/4  * b^1/2  =  b ^{3/4 + 1/2}  = b ^5/4

^{Can you do the second one now?}

so wheres the answer?

Second one....good job, so far

As you found  QR and ST are equal so they span the same arcs...let the arcs they span  = x

 We have this equation 

360   -  QS - RT   =   QR  + ST

360  - 34 -  98  = x  +  x

228 = 2x     divide by 2

114°   = x  = measure of arc ST = measure of arc QR

And the measure of the span of arc XT  = (1/2)(arc ST)  = (1/2)(114°)  = 57°

The only thing we need to do is  to find  r^2.....this is given by

(-6 - -3)^2  + (-8 - 5)^2   = r^2

(-3)^2 + (-13)^2   =    r^2

9  + 169   = r^2

178  = r^2

So  the equation using the center and r^2  is

( x - - 3)^2 + ( y - 5)^2  = r^2

(x + 3)^2  + (y - 5)^2  = 178 

The distance from the center (-3,5) to the point (-6, -8) is the radius of the circle

r = sqrt ((-3 - -6)²+(5- -8)²= sqrt 3^2 + 13^2) = 13.34

(x-h) ^2  + (y-k)^2 = r^2

(x+3)^2 + (y -5)^2 = 13.34^2

(x+3)^2 + (y-5)^2 = 178

Aopshelp, if you want to type text in latex, do \text{sample}

What is your aops name by the way?

Here you go   ....

https://web2.0calc.com/questions/halp-asap_2

x^2+y^2- 1/4x + 1/4y=1/32   rewrite as

x^2  - (1/4)x  +  y^2  + (1/4)y  = 1/32     complete the square on x and y

Take  1/2 of 1/4  =1/8     square it  = 1/64     add it twice to  both sides

[ x^2 - (1/4)x + 1/64] + [ y^2 + (1/4)y + 1/64 ] =  1/32 + 1/64 + 1/64

Factor the terms in both brackets  and simplify the right side

( x - 1/8)^2  + ( y + 1/8)^2  = 1/32 + 2/64

(x - 1/8)^2  + (y + 1/8)^2   =   1/32 + 1/32          [ 1/32 + 1/32 = 2/32  = 1/16  ]

(x - 1/8)^2 + ( y + 1/8)^2 =  1/16

The center is   (1/8, - 1/8)

The raduis  = sqrt (1/16)   =   1/4

You got the first one correct....

I'll look at the second one in just a bit

(x + y)^2  = 31                                        ( x - y)^2  =  x^2 - 2xy + y^2     =  

x^2 + 2xy + y^2  = 31                                                 (x^2 + y^2) - 2xy

x^2 + y^2  = 31 - 2xy                                                  (x^2 + y^2) - 2(6) 

x^2 + y^2  = 31 - 2(6)                                                 (x^2 + y^2)  -12      (2)

(x^2 + y^2) = 31 -12 =  19      (1)

Sub (1)  into (2)    and we have that

(x - y)^2   =   (x^2 +y^2) - 12  =  (19) - 12   =   7

Last one

Equal segments drawn from the center of the circle meet chords BC and CD at right angles

This means that the two chords are equal and thus, they span the same arcs....so....like before...

9x - 53  = 2x + 45     add 53 to both sides  and subtract 2x form both sides

7x = 98     divide both sides by 7

x  = 14

So  each chord spans an arc of  9(14)  - 53  = 73°

Then BAD  will span  an arc  =  360 - 73 - 73   = 214°

Hint: Expand everything

Then use substitution

I thought I was supposed to some special way to solve the arc ones. So I just solve the arc and line questions the same way? 

Fourth one

LM and NP  are equal distances from the center of the circle....so....they have equal length

Then.....they also span the same arcs

This implies that

(8x - 56)°  = (5x + 22)°

8x - 56  =  5x + 22          add 56  to both sides, subtract 5x from both sides

3x =  78      divide both sides by 3

x  =  26

So  LM spans an arc of   8(26) - 56  = 208 - 56  = 152°

And NP spans the same arc

So  LP  spans an arc  = 360 - LM - NP - MN =

(360 - 152  -152  - 39)°   =    17°

Third one

Since chords  DE  = GF....equal chords are the came distance from the center of the circle

This implies that  HJ  = JI   ....so....

3x + 20  =  15x  - 64         add 64 to both sides, subtract 3x from both sides

12x  = 84    divide both sides by 12

x  = 7

So JI   = 15x - 64  =  15(7) - 64  =  105 - 64  = 41

Second one.....same thing.....the chords LM and NP have the same length

So

41 - 2x  = 7x  +  5         add  2x to both sides.....subtract 5 from both sides

36  = 9x      divide both sides by  9

4   = x

So 

LM  = 41 - 2x   =   41 - 2(4)   =  41  - 8   =   33

I'll try a few

In the first one.....these chords have the same length....so....this implies that

(7x - 39)°  = 87°

7x - 39   = 87      add 39 to both sides

7x  =  126        divide both sides by 7

x  = 18

3(x^2 + 2x -3) - 4(4x^2 - 7x +5)  =

3x^2  + 6x - 9    -  16x^2 + 28x  -20  =

-13x^2 + 34x - 29

Ther's an algebraic solution which isn't difficult....but.....the easiest way to solve this is to divide 189  by 3  =

63 ....this is the middle odd integer

Add 2  to 63  = 65    ....this is the largest odd integer

Subtract 2 from 63   = 61......this is the smallest  odd integer

Answered here :

https://web2.0calc.com/questions/help_42954

 P(−1, 2) and Q(7, 8)  

The midpoint of  PQ   will be the center of the ciricle

So

( [ -1 + 7 ] /2 , [ 8 + 2 ] /2 )  = [ 6/2 , 10/2)  = (3 , 5)  = (h, k)

And the distance from this point to either P or Q   is  the radius, r 

And we can find r^2 as

(7 -3)^2  + ( 8 - 5)^2  = r^2

4^2 + 3^2  = r^2

16 + 9  = r^2

25 = r^2

So....the equation of the circle is

(x - h)^2  + (y - k)^2  = r^2

( x - 3)^2 + (y - 5)^2   = 25

The form of the equation of a circle is r²=(x-a)²+(y-b)², where r is the radius and (a,b) is the center.

The length of the line PQ is the diamiter of the circle, so plug (-1,2) and (7,8) into the distance formula

\(d=\sqrt{(-1-7)^2+(2-8)^2}\)

\(d=\sqrt{(-8)^2+(-6)^2}\)

\(d=\sqrt{64+36}\)

\(d=\sqrt{100}\)

\(d=10\)

so the radius is 5

The midpoint of the line PQ is the center of the circle, so plug P and Q into the midpoint formula

\(m=(\frac{-1+7}{2},\frac{8+2}{2})\)

\(m=(\frac{6}{2},\frac{10}{2})\)

m=(3,5)

Plugging the midpoint and the radius into the equation for the circle gets:

25=(x-3)²+(y-5)²