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y=8x^2+4x-1 to form y=a(x-h)^2+k
y=8(x^2+x/2)-1
now must complete square of bracket terms
y=8(x^2 +x/2 +1/16) -1 -8(1/16)
simplify
y=8(x+1/4)^2-1.5
this fits form with
a=(8)
h=(-1/4)
k=(-1.5)
checking
let x=3 solve for y
initial eqn y=8x^2+4x-1= 8(3^2) +4(3)-1=72+12-1=83
final form y=8(x+1/4)^2-1.5=8(3.25)^2-1.5= 8(10.5625)-1.5=84.5-1.5=83 ok
to complete the square we take coefficient of middle term, divide by two, and square it
ie (1/2) , (1/2)/2=1/4, (1/4)^2=(1/16)
this becomes constant term of "square"
since we added this to complete square terms we must subst from constant term to balance eqn. In this case we added 8(1/16) in total therefore we substract (8/16) from constant

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basic form is log base a of b = ( log b)/(loga)
log base 5 of sq rt 125 - log base 2 of sq rt 2
log of neg number is undefined therefore only use positive sq rts
(log sq rt 125 / log 5 ) -(log sq rt 2/ log 2)
(log 11.18/log5) - (log 1.414/log2)
1.5-.5
1.0

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3+sq rt (5x-1) = x
subt 3 each side
sq rt(5x-1) = x-3
sq both sides
+-(5x-1) = (X-3)^2
USING ONLY + answer ( - answer gives imaginary answers)
+(5x-1) = (x-3)(x-3)= x^2-6x+9
subt 5x each side
-1 = x^2 -6x +9 -5x
add 1 to each side
0= X^2 -6x +9 -5x +1
simplifying
0=x^2 -11x +10
factoring
0=(x-10)(x-1)
x=10 and x=1
checking
3 +sq rt (5x-1) = x
3 + sq rt (5(1)-1) = (1)
3 + sqrt 4=1
3+-2 =1 ok with -2
for x=10
3+sqrt (5(10)-1) = (10)
3+ sqrt 49 =10
3+-7=10 ok with +7

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if a drawing were made showing a line from a point on the ground thru top of tower and then thru top of bldg, we could see two similar triangles. Similar triangles have sides proportional.
in our case h/50.5=2.9/1.58
cross multiplying
1.58 *h=50.5 * 2.9
1.58 *h = 146.45
divide by 1.58
h=92.68987= 92.69

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assume the 3:15 is 3 hrs and 15 minutes from zero. AM or PM is not given.
The hr hand has rotated once for every hour plus 90 degrees for 15 minutes or
hr = 360 *3 +90 = 1170 degrees.
the minute hand has rotated 60 times for each hour plus 90 degrees for 15 minutes or
minute hand = 3 *60*360 +90= 64,890 degrees
The difference is 63,720

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the definition of a square defines each side as equal
per = 2l +2w for rectangle, but square is l=w, or per =4s
per=4s
144 = 4s
divide by 4
36=s (ft)

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basic formula for a circle is
(x-h)^2 + (y-k)^2 = r^2, where h is x center, k is y center, r is radius
to fit this form the given problem must "complete the square"
(x^2 +4x +4) + (y^2 +6y+9) = 2 + (+4)+ (+9)
reviewing take coefficient of the middle term, 4x, divide by 2 and then square, (4/2)^2 =4 and add to both sides of eqn
for y terms, ( 6/2)^2 =9 and add to both sides
Now factor x and y sets
(x+2)^2 + (y+3)^2 = 15
comparing to basic form we find
x center is (-2)
y center is (-3)
rad^2 = 15, rad =sq rt 15 = 3.87

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volume = length * width * height
or V=l*W*H
for given info
20,000= (100) (20) *H
20,000=2,000*H
10=H

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Opposites are just as you stated. Just opposite sign. The two add to zero.
Normally with polynomials, we talk about inverses. This is the reflextion over y=x. We usually find these by setting x=y, and y=x.
For example if y=2x, inverse is x=2y or y=x/2.

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This is a physics problem under "projectile motion"
Sometimes it is found in algebra under quadratic topics as a actual application example.
For today, we are asked how high the projectile will go, in a given time, with an initial velocity. The velocity provides the upward energy, and gravity provides the downward energy.

Usually, the angle of the initial velocity is important, but for today we will assume straight up.
If not, let me know.
The basic eqn is:
delta y = v*t - (1/2) (32.2) t^2
where :
delta y is change in height. Note as time progresses, the projectile goes up, reaches a peak, and then comes down.
v is initial velocity. units in this case are ft per sec or ft/sec
t is elapsed time. units are sec
32.2 is gravity. we are using English today. be careful of metric. english units are 32.2 ft per sec squared, or 32.2 ft/sec^2 (metric is 9.8 m/sec^2)
1/2 is a constant to get average acceleration

1st problem
y= vt-1/2 (32.2) t^2
= (45) (3) - 1/2 (32.2) (3^2)
= 135 - 144.9
= (-9.9) ft
Note we started at 10 ft above ground, therefore 10-9.9 = (.1) ft above ground
This is not so nice a first problem. The projectile has been in the air long enough to go fully up, stop, and now decend to just above ground height.
Additional calc shows that the projectile goes up to a peak of about 31.4 ft in 1.4 sec before starting down.
2nd problem
y=vt-16.1t^2 = 95*2 -16.1 * 2^2= 190-64.4=125.6 ft
Again, initial height was 50 ft, therefore final position is 50+125.6 =175.6 ft
additional calc shows peak at about 3 sec of about 140 ft

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1) x-y=0
2) 12x-5y=-21
In substitution we use one eqn to establish a x to y relationship.
Then we substitute this relationship in the other eqn
for example from (1) we know x-y=0, therefore x=y
substituting into (2)
12 (y) -5y = -21
simplify
7y=-21
divide by 7
y=(-3)
substitute this in either (1) or (2) to find x
using (1) x-y=0
x-(-3) =0
y=(-3)
checking
(1) (-3)-(-3)=0 ok
(2) 12(-3) -5(-3) =-21
-36 +15 = -21 ok

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5ln(1-x) =8
ln(1-x)^5 =8
e^8=(1-x)^5
2980.958=(1-x)^5
taking 5th root
4.95303 = (1-x)
subt 1 both sides
3.95303 =-x
x=(-3.95303)
checking
5 ln (1-(-3.95303) )=8
5ln 4.95303 =8
5(1.5999) =8
8=8 ok
to 2 dec places x= (-3.95)

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5x /(x+1)
vertical asymptote at den =0, x+1=0, x=-1
as degrees num and den equal, hor asymptote at 5/1 = +5
sketch vertical line at x=-1, and horizontal line at y=+5
rough sketch shows function starting low at x=-1, rising to rt thru (0,0) to just under asy y=5
starting high at x=-1, function decays to just above hor asy at y=5 as it goes to left
end behavior: as x approaches -infinity, y approaches +5 from +side
as x approaches + infinity, y approaches +5 from - side

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Draw a rough sketch of two similar triangles
label sides carefully
as triangles are similar, sides are proportional
per of given triangle abc = 9 + 18 + 21 = 48.
per of new triangle is 112
as perimeters vary as scale factor to 1st power
sf = 112/48 =2 1/3 =7/3
xy = (7/3)ab =7/3 * 9 = 21
yz = (7/3) bc = 7/3*18 = 42
xz = (7/3) ac = 7/3*21 = 49
checking per 2nd triangle = 21+42+49= 112 ok

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z= k/t
12 =k/10
cross multiply
k=120
now proportion is
z=120/t
for t=6
z=120/6 = 20

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1) 5x +2y = 10
2) 3x +2y =6
subt (1) - (2)
2x =4
divide by 2
x=2
subst in (1)
5(2) +2y =10
10 +2y =10
subt 10 both sides
2y =0
y=0
answer (2,0)
checking
1) 5(2) +2(0) =10 ok
2) 3(2) = 2(0) =6 ok

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(y-1) -y =y+1
simplify by combining like terms
-1 =y+1
subt 1 from both sides
-2=y
checking
(y-1)-y=y+1
( (-2) -1) -(-2) = (-2) +1
-3 +2 = -2 +1
-1 =-1 ok

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(5/2)x +(1/4)x =(7/4) +x
Multiply by 4
10x +x=7+4x
7x=7
x=1
check
5/2 +1/4 =7/4 +1
10/4 +1/4 =7/4 +4/4
11/4 = 11/4 ok
if you really did mean
5/(2x) +1/(4x) =7/4 +x
mult by 4x
10 +1 = 7x +4x^2
0= 4x^2 -7x -11
0= (4x+11) (x-1)
x=(1) and x=(-11/4)
checking x=1
5/2 +1/4 =7/4 +1 ok
checking x=( -11/4)
5/-22/4 +1/-44/4 = 7/4 -11/4
-20/22 -2/22 = -4/4
-22/22 =-4/4 ok

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6-2x=5x-9x+2
6-2x = -4x +2
add(2x) and subt (2)
4=-2x
divide by (-2)
x=(-2)
check
6-2(-2)=5(-2) -9(-2) +2
6+4=-10 +18 +2
10=10 ok

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-14w +2w^2 +wz-7z
reorder
2w^2 -14w + wz-7z
factor by grouping
2w(w-7) + z(w-7)
(2w+z) (w-7)
check by FOIL
2w^2 -14w +zw -7z

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an imaginary number (i) is = sq rt (-1)
x^2 +25 =0
x^2 = -25
x = +- 5 sq rt (-1) = +- 5i

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(2x-3)^2=24
take sq rt of both sides
2x-3 = sq rt 24 = +- (4.899)
2x-3 =+4.899
2x = 7.899
x= 3.95
or 2x-3= -4.899
2x = -1.899
x=-.95
checking
(2(3.95) -3 )^2= 24 ok
( 2(-.95) -3)^2 =24 ok

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Let x = cost of 17 hrs labor
300/1=x/17
cross mult
300*17=$ 5100 =x

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(5x+1)/ (x^2+5x-14)
(5x+1) / ( (x+7) (x-2) )
therefore Domain is all Real numbers except, x=(-7) and x=(2)

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good work
an easier approach might be
y=-2(x-8)/(8-x)
simplify to
y= 2(x-8) / (x-8)
cancelling (x-8) 's gives us y=+2 but with a hole at x=8
no vertical asymptotes ( ie den =0)
no real horizontal asymptotes except y does always =2
review of horiz asymptotes, y = a(n)x^n +..... / b(m) x^m +.....
if n less than m, hor asy at y=0
if n=m, hor asy at y=a(n)/b(m)
and if n>m, no hor asy

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let smaller +x
let larger = x+2
The problem only makes sense if " is three than" reads " is 3 smaller than"
to solve this problem
l^2 = 12 S-3
(x+2)^2 =12(x) -3
x^2 +4x+4 =12x -3
subt 12 x both sides
x^2 +4x -12x +4 =-3
now add 3 to both sides
x^2 +4x -12x +4 +3 =0
x^2 -8x +7 =0
factoring
(x-7) (x-1) =0
setting both =0
x-7=0, x=7 (smaller) & larger =9
or x-1=0, x=1 (smaller) & larger is 3
checking (l^2=12*s-3)
3^2=12(1)-3 ok
9^2 = 12(7)-3 or 81=84-3 ok
Therefore both sets are correct, 1 &3, and 7&9

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GOOD JOB
END BEHAVIOR is fancy for what happens to function for very big or small numbers
for the ODD functions they in general slant up for positive leading coefficients
or slant down for negative leading coefficients.
This is a positive therefore it will slant up
the end on the left side will be y approaches (-infinity) as x approaches (- infinity)
the end on the right side wiil be y approaches (+ infinity) as x approaches ( + infinity)
as it is 5th degree it should have 4 turns as it goes from lower left to upper right on graph

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sq rt (3*2) +sqrt (50)
sqrt 3 *sqrt2 + sqrt 25*sqrt2
(sqrt3 +5) * sqrt (2)
(5 +sqrt3 ) *sqrt2
check sqrt 6 +sqrt50 =9.52
(5+sqrt3) * sqrt2 = 9.52

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review original problem please
IF ( (-6) +(11-5) ) / 4^2 - (-9) or thereabouts
numerator will be zero and problem will be zero
STATED PROBLEM
(-6) + (-11+5)/4^2 -(-9)
-6+ (-6)/16 +9
3 -(3/8)
2 5/8
2.625

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f(x) =y=(5x-1)/(x^2-16)
Normally in Domain problems we seek when denominator =0
this becomes easier to see if we factor denominator
y= (5x-1) / ( x+4) (x-4)
x+4 =0, x=-4
x-4 =0, x=4
Therefore domain is all Real numbers except x= +4, or =-4

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INVERSE of a number is reciprocal, ie 2, inv 2 = 1/2
however inverse of FUNCTION indicates a function that is reflected over y=x
Normally this is solved by substituting x for y ,,,, and y for x
f(x) =(5x-3) / (x-2)
but f(x) =y
y = (5x-3) / (x-2) DOMAIN is all REAL numbers except x=2
inverting
x = (5y-3) / (y-2)
or (y-2) = (5y-3) /x Domain is all REAL numbers except x=0