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Split the inequality......

x^4 + 4x^2  >  -4

x^4 + 4x^2 + 4  >  0     factor

(x^2  + 2) ( x^2 + 2) > 0

(x^2 + 2)^2  >  0         all  real numbers will work here

Also

x^4 + 4x^2  <  21

x^4 + 4x^2  - 21 <  0     factor

(x^2 + 7) ( x^2 - 3)  <  0

This  will be true   when       -sqrt (3)  < x < sqrt (3)

We  must  take  the most restrictive  interval....this is

x =  (-sqrt (3)  , sqrt (3) )

On the left, we  have  a 30 - 60 - 90  right triangle

Side  of  1  is opposite  thw  30 degree angle

So...in a 30-60-90   right triangle,  the side opposite  the 90 degree angle will be  twice  as long = 2

The  sum  of  the interior angles of a quadrilateral  =   360 degrees

So.....the missing angle , A,  at the top right  can  be  found as

90  + 90  + (60 + 90)  + A  =360

330 + A  = 360

A =  360  - 330   =  30°

And  the triangle on the  right  is also a 30-60-90 right triangle

The  side of 2 is opposite the 30 degree angle.....and x  is opposite  the 90 degree angle...so...as before....

x = 2 * 2   =    4 

The  missing side is  just the  height of  the  trapezoid

This  is   =     sqrt ( 7^2 -  (8 - 5)^2 )  =  sqrt [ 49 - 3^2 ]   = sqrt  [ 49 - 9 ]  = sqrt [ 40 ]  = sqrt (4) * sqrt (10) =

2sqrt (10)

Hint:

The little triangle is half of an equilateral triangle.

Use angle sum of a quadrilateral

then

trigonometry.

In a square  with a side of S.....the diagonal will always =  S*sqrt (2)

So

AC =    7sqrt(2)

log₁₀ x  = 3  + log ₁₀ y

Note  that   3  =  log ₁₀ 1000

So we actually have

log ₁₀ x   =  log ₁₀ 1000  +  log ₁₀ y       and  by  a log property  we  can write

log ₁₀ x  =   log ₁₀ ( 1000* y )

This implies that

x =  1000y       divide both sides  by  y

x / y   = 1000

Two coats  of  paint  will  cover 2 * 9 *10    =  180 sq  ft

And a 10 * 12  wall =   120sq ft

So

1 qt  /  180 sq ft  =    ?  qts  /  120 sq ft           cross-multiply

1 * 120  =  180 * ?          divide both sides  by 180

120  / 180  =  ?

2 / 3   qts  =  ? =    amt to cover  the  10 x 12   wall

-18  =     

-18 *  1    ⇒     t  =  -17

18 * - 1      ⇒   t  =  17

9 *  -2     ⇒      t  =     7

-9 * 2      ⇒      t  =   -7

6 * -3     ⇒       t   =  3

-6 * 3     ⇒       t =   -3

The second one is similar

b - 1  = sqrt ( 5b - 5)             (1)

(b - 1)^2  = 5b - 5

b^2  - 2b  +  1  =  5b  - 5

b^2 - 7b + 6  = 0

(b - 6)  ( b - 1)  =  0

b= 6     or  b  =1

Note  that BOTH solutions will  work, here.....check them in (1)  and see

a  =  3  +  sqrt ( 5a - 9)

a - 3  = sqrt (5a - 9)          square both sides

(a - 3)^2  = 5a  - 9

a^2  - 6a  +  9  =  5a  - 9             subtract 5a,  add 9   to both sides

a^2  - 11a + 18  =  0             factor

(a  -  9) ( a - 2)  =  0

Setting  both  factors   to  0   and solving for   a  gives  that

a = 9            or      a  = 2

Check  9                                                      Check 2

9  = 3  + sqrt ( 5*9 - 9)                               2  =  3  +  sqrt ( 5*2 - 9)

9  = 3  +  sqrt (36)                                      2  =  3  +  sqrt !

9  = 3 + 6                                                    2  =  3 + 1

good                                                           2  = 4

                                                                   no good

So....the only answer  is    a  =  9

g(f(x)=

2x - 4y  =  9                subtract  2x  from both sides

-4y =   - 2x  +  9           divide both sides by  -4

y  = (-2)/(-4)x  -9/ 4

y= (1/2)x  - 9/4

In the form y  = mx + b.....m is the  slope...so.....the slope of a parallel line  is  1/2

We  have  that

(x + 1)  ( x + 1 + 14)   x ( x + 21)    simplify

(x + 1) ( x + 15)  =  x^2 + 21x

x^2  + 16x  + 15  = x^2 + 21x        subtract x^2 from both sides

16x + 15  =  21x

15  + 16x - 21x  = 0

15 - 5x   =  0

15  = 5x

x =  3       you are correct  !!!

angle MFE  +  angle MFP   +  angle PFR  =180  .....so.....

(x + 6)  +  90  + ( 3x - 4)   =180     simplify

4x +  92  =  180

4x  = 180   - 92

4x =  88

x =  88 /  4   =  22

Here's  the  way  to do this

Take  (1/2)  of  22  = 11

Square it   =   11^2   =   121   =   what goes in  the  blank

Note  that   ( x + 11)^2  = x^2  +  22x  + 121

Let the centor of  the incircle  =  O

Then  angles  AEO  and  AFO  = 90°

And in quadrilateral AFOE....angle  FOE  can be found as

90 + 90 +78  + angle FOE  =  360

258  + angle FOE  =  360

angle FOE =  360 - 258   =   102° 

And angle EDF is an inscribed  angle in the  incircle that  subtends  the  same arc as angle FOE

So...angle  EDF  = (1/2) 102  =  51°

x^2  + y^2  - 10x  - 6y  + 18  = 0    complete the  square  on x, y

x^2  - 10y + 25  + y^2 - 6y + 9  =   -18  + 25 + 9      simplify

(x - 5)^2  +  ( y - 3)^2  =  16

The  center of this circle is   ( 5 , 3)    and the radius is  4

Then if  the  circles touch at   (9,3)  and they are identical....then  this  will be  the midpoint of  both centers

So.....let  the  center of the other circle  be (h, k)..so we have 

[ 5 + h) / 2 =   9                  (3 + k) / 2  =  3 

5 + h  =  9*2                       3 + k  =   3*2

5 + h = 18                           3 + k  = 6

h  =13                                     k = 3

The equation of  the second circle  is   ( x  -13)^2  + ( y - 3)^2  = 16

Here's a graph  :  https://www.desmos.com/calculator/efoxyb9720

Rounded to nearest  10, we  have

(20  /  10 ) + 20   = 

2 + 20  =  

22

Rounded to nearest whole number

( 15   / 15)  +   18  =

1  +  18  = 

19

22  -  19   =    3

I also have one more question:

Let n be a positive integer and a, b be invertible integers modulo n such that a = b^-1 (mod n). What is the remainder when a*b is divided by n?

Thank you in advance!

It sure is, CPhill!

No problem, Math Whiz.....easy to make errors on here  !!!!

Whoops, I made a mistake with saying that it was 40%, I forgot to subtract it from 100%. Sorry, those were the ones that are NOT correct.

Second one

Let  the  number  he needs   =  N

So

.90N  =  36

N = 36 /  .90  =  

40

First one.....number  of slices eaten = 50 -  20  =   30

30/50

. 6

60%

are all  correct