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Gnesinka [82]
2 years ago
13

F(x) = 7 - What is the inverse

Mathematics
2 answers:
slava [35]2 years ago
8 0

Answer:

Add x x and 0 0 . Since g(f(x))=x g ( f ( x ) ) = x , f−1(x)=x+7 f - 1 ( x ) = x + 7 is the inverse of f(x)=x−7 f ( x ) = x - 7

Step-by-step explanation:

luda_lava [24]2 years ago
8 0
The inverse would be f(x) = x-7
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Divide (x^3+2x^2-x-2) by (x-1)
Ymorist [56]
\frac{x^3+2x^2-x-2}{(x-1)} =x^2+3x+2 \\  \\ x^3+2x^2-x-2= \\= x^2(x+2)-(x+2) \\ =(x+2)(x^2-1)=(x+2)(x-1)(x+1) \\  \\  \frac{x^3+2x^2-x-2}{(x-1)}  = \frac{(x+2)(x-1)(x+1)}{x-1} =(x+2)(x+1)=x^2+3x+2
7 0
3 years ago
A cube with 40-cm-long sides is sitting on the bottom of an aquarium in which the water is one meter deep. (Round your answers t
Molodets [167]

Answer:

940.8 N

1254.4 N

Step-by-step explanation:

I would think the questions would be to calculate the forces at the top of the cube and at the sides. Thus:

On the top:

F = pressure * area

P = density * gravity * height

the height would be:

1m - 0.4m = 0.6m

replacing:

P = 1000 * 9.8 * 0.6 = 5880

A = (0.4) ^ 2 = 0.16

F = 5880 * 0.16

F = 940.8 N

On the sides:

dF = d * g * h * dA

dA = 0.4 * dh replacing

dF = 1000 * 9.8 * h * 0.4 * dh

dF = 3920 * h * dh

We integrate both sides and we have:

F = 3920 * (h ^ 2/2), h = 0.6 up to h = 1

F = (3920/2) * (1 ^ 2 - 0.6 ^ 2)

F = 1254.4 N

3 0
3 years ago
A lunch box costs $4.50 plus sales tax. After sales tax, the lunch box is $4.86. What is the sales tax rate?
Alborosie
8 percent (C)

To find 1% you divide 4.50 by 100, which is 0.045.

Then you have to find what is added on.
4.86-4.50 is .36

Now you divide .36 by 0.045. An easier way to do this is 360 divided by 45. This equals 8.

Hope this helped
3 0
2 years ago
Simplify u^2+3u/u^2-9<br> A.u/u-3, =/ -3, and u=/3<br> B. u/u-3, u=/-3
VashaNatasha [74]
  The correct answer is:  Answer choice:  [A]:
__________________________________________________________
→  "\frac{u}{u-3} " ;  " { u \neq ± 3 } " ; 

          →  or, write as:  " u / (u − 3) " ;  {" u ≠ 3 "}  AND:  {" u ≠ -3 "} ; 
__________________________________________________________
Explanation:
__________________________________________________________
 We are asked to simplify:
  
  \frac{(u^2+3u)}{(u^2-9)} ;  


Note that the "numerator" —which is:  "(u² + 3u)" — can be factored into:
                                                      →  " u(u + 3) " ;

And that the "denominator" —which is:  "(u² − 9)" — can be factored into:
                                                      →   "(u − 3) (u + 3)" ;
___________________________________________________________
Let us rewrite as:
___________________________________________________________

→    \frac{u(u+3)}{(u-3)(u+3)}  ;

___________________________________________________________

→  We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ;  since:

" \frac{(u+3)}{(u+3)} = 1 "  ;

→  And we have:
_________________________________________________________

→  " \frac{u}{u-3} " ;   that is:  " u / (u − 3) " ;  { u\neq 3 } .
                                                                                and:  { u\neq-3 } .

→ which is:  "Answer choice:  [A] " .
_________________________________________________________

NOTE:  The "denominator" cannot equal "0" ; since one cannot "divide by "0" ; 

and if the denominator is "(u − 3)" ;  the denominator equals "0" when "u = -3" ;  as such:

"u\neq3" ; 

→ Note:  To solve:  "u + 3 = 0" ; 

 Subtract "3" from each side of the equation; 

                       →  " u + 3 − 3 = 0 − 3 " ; 

                       → u =  -3 (when the "denominator" equals "0") ; 
 
                       → As such:  " u \neq -3 " ; 

Furthermore, consider the initial (unsimplified) given expression:

→  \frac{(u^2+3u)}{(u^2-9)} ;  

Note:  The denominator is:  "(u²  − 9)" . 

The "denominator" cannot be "0" ; because one cannot "divide" by "0" ; 

As such, solve for values of "u" when the "denominator" equals "0" ; that is:
_______________________________________________________ 

→  " u² − 9 = 0 " ; 

 →  Add "9" to each side of the equation ; 

 →  u² − 9 + 9 = 0 + 9 ; 

 →  u² = 9 ; 

Take the square root of each side of the equation; 
 to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ; 

→ √(u²) = √9 ; 

→ | u | = 3 ; 

→  " u = 3" ; AND;  "u = -3 " ; 

We already have:  "u = -3" (a value at which the "denominator equals "0") ; 

We now have "u = 3" ; as a value at which the "denominator equals "0"); 

→ As such: " u\neq 3" ; "u \neq -3 " ;  

or, write as:  " { u \neq ± 3 } " .

_________________________________________________________
6 0
3 years ago
A computer online service charges one hourly rate for regular use but a higher hourly rate for designated "premium" areas. One c
N76 [4]

Answer: the service charge per hour for premium services is $5.5

the service charge per hour for regular services is $3

Step-by-step explanation:

Let x represent the service charge per hour for premium services.

Let y represent the service charge per hour for regular services.

One customer was charged $38 after spending 2 h in premium areas and 9 regular hours. It means that

2x + 9y = 38- - - - - - - - - - - 1

Another customer spent 3 h in premium areas and 6 regular hours and was charged $34.50. It means that

3x + 6y = 34.5- - - - - - - - - - -2

We would eliminate x by multiplying equation 1 by 3 and equation 2 by 2. It becomes

6x + 27y = 114

6x + 12y = 69

Subtracting, it becomes

15y = 45

y = 45/15

y = 3

Substituting y = 3 into equation 1, it becomes

2x + 9 × 3 = 38

2x + 27 = 38

2x = 38 - 27 = 11

x = 11/2 = 5.5

8 0
3 years ago
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