hectictar

Wow!! 👍

Oh no!!! I said I was gonna do better about always waiting till the last minute!!! Haha

1.

f(x) + g(x)

                                                   Since  f(x)  =  -5x - 4, we can substitute  -5x - 4  in for  f(x)

f(x) + g(x)  =  (-5x - 4) + g(x)

                                                   Since  g(x)  =  6x - 7, we can substitute  6x - 7  in for  g(x)

f(x) + g(x)  =  (-5x - 4) + (6x - 7)

f(x) + g(x)  =  -5x - 4 + 6x - 7

                                                   Combine like terms.

f(x) + g(x)  =  x - 11

2.

f(x) - g(x)

                                                   Since  f(x)  =  3x + 2, we can substitute  3x + 2  in for  f(x)

f(x) - g(x)  =  (3x + 2) - g(x)

                                                   Since  g(x)  =  6x - 7, we can substitute  6x - 7  in for  g(x)

f(x) - g(x)  =  (3x + 2) - (6x - 7)

                                                   Distribute the  -1  to the terms in parenthesees.

f(x) - g(x)  =  3x + 2 - 6x + 7

                                                   Combine like terms.

f(x) - g(x)  =  -3x + 9

4.

\(\frac{f}{g}\,=\,\frac{f(x)}{g(x)} \\~\\ \frac{f}{g}\,=\,\frac{x^2-16}{x+4}\)

Before simplifying, let's get the domain.  x + 4  is in a denominator, so...

x + 4 ≠ 0

x  ≠  -4     The domain is all real numbers with the exception that  x ≠ -4

\(\frac{f}{g}\,=\,\frac{x^2-16}{x+4}\)

                                   Factor the numerator as a difference of squares.
\(\frac{f}{g}\,=\,\frac{(x-4)(x+4)}{x+4}\)

                                   Reduce the fraction by  (x + 4) .

\(\frac{f}{g}\,=\,\frac{(x-4)}{1} \\~\\ \frac{f}{g}\,=\,x-4\)

3.

f ⋅ g   =   f(x) ⋅ g(x)

                                                Since  f(x)  =  -5x + 3  , we can replace  f(x)  with  -5x + 3

f ⋅ g   =   (-5x + 3) ⋅ g(x)

                                                Since  g(x)  =  6x - 2  , we can replace  g(x)  with  6x - 2

f ⋅ g   =   (-5x + 3) ⋅ (6x - 2)

Before we simplify, let's get the domain.

There are no denominators or anything, so the domain is all real numbers.

f ⋅ g   =   (-5x)(6x) + (-5x)(-2) + (3)(6x) + (3)(-2)

f ⋅ g   =  -30x²  +  10x  +  18x  +  -6           Combine like terms...

f ⋅ g   =  -30x²  +  28x  -  6

Let the number of candy bar packages be  n .

$18  -  $6.50  -  $1.15n   ≥   $5.75

Now we can drop the dollar signs from every number and simplify.

18  -  6.5  -  1.15n   ≥   5.75

11.5  -  1.15n   ≥   5.75

                                            Subtract  11.5  from both sides of the inequality.

-1.15n  ≥  -5.75

                                            Divide both sides by  -1.15  and flip the inequality sign.

n   ≤  5

The number candy bar packages must be less than or equal to  5 . The maximum number is  5  .

Looking at triangle  PQR....

RP  =  PQ    so    ∠QRP  =  ∠PQR  =  x°

x + x + 3y   =   180

2x + 3y   =   180

Looking at triangle SPQ....

∠SPQ  =  180 - 3y

x + 2y + (180 - 3y)  =  180

x - y + 180   =   180

x - y   =   0

x   =   y

Substitute  y  for  x  in the equation  2x + 3y  =  180

2y + 3y  =  180

5y  =  180

y  =  36

3y  =  3 * 36

3y  =  108

A.

60 cm  -  2 cm/min *  5 min   =   60 cm - 10 cm   =   50 cm

8 cm  +  1.25 cm/min * 5 min   =   8 cm  +  6.25 cm   =   14.25 cm

B.

water level of old aquarium  =  W  =  60 - 2T

water level of new aquarium  =  W  =  8 + 1.25T

C.

water level of new aquarium  >  water level of old aquarium

8 + 1.25T  >  60 - 2T

D.

8 + 1.25T  >  60 - 2T

3.25T  > 52

T > 16

E.

The new aquarium's water level will be greater than the old aquarium's water level after  16  mins.

F.

Find the time it takes for the water level of the old aquarium to be  0  .

0  =  60  - 2T

2T  =  60

T  =  30

Find the water level of the new aquarium at this time.

W  =  8 + 1.25(30)

W  =  45.5

*edit again* I just thought of this....

area of small circle  +  area of middle ring + area of outer ring   =   area of big circle

area of small circle + area of small circle + area of outer ring   =   area of big circle

area of small circle + area of small circle + area of small circle   =   area of big circle

3(area of small circle)   =   area of big circle

3π r²  =  π * 12²      ,  where  r  is the radius of the small circle.

3π r²  =  144π

                               Divide both sides by  3π .

r²  =  48

                               Take the positive square root of both sides.

r  =  4√3

5.   9^1/3  *  81^1/3   =   ( 9 * 81 )^1/3   =   ( 9 * 9 * 9 )^1/3   =   (9³)^1/3   =   9^3/3   =   9¹   =   9

6.   \(\frac{8}{\sqrt[7]{x^{15}}}\,=\,\frac{8}{x^{\frac{15}{7}}}\,=\,8x^{-\frac{15}{7}}\)