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Rudik [331]
3 years ago
7

How do you solve 10b-(-b)-9b-7a+10a+5

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
6 0

Answer: 2b-21a

Step-by-step explanation:

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Carmen is writing an article for a magazine. She will be paid a fee and also will be paid for each word that is published. The t
enot [183]

Answer:

The inverse function of f(x)=2.5*x+150 is f⁻¹(x)=\frac{2}{5}x-60

Step-by-step explanation:

An inverse or reciprocal function of f (x) is called another function f ⁻¹(x) that fulfills that:

If f(a)=b then f⁻¹(b)=a

That is, inverse functions are functions that do the "opposite" of each other. For example, if the function f (x) converts a to b, then the inverse must convert b to a.

To construct or calculate the inverse function of any function, you must  follow the steps below:

Since f (x) or y is a function that depends on x, the variable x is solved as a function of the variable y. And since inverse functions swap the input and output values ​​(that is, if f (x) = y then f⁻¹(y) = x), then the variables are swapped and write the inverse as a function.

You know that he function f(x) = 2.5*x + 150 or y=2.5*x +150

Solving for x:

2.5*x +150=y

2.5*x= y-150

x=\frac{y-150}{2.5}

x=\frac{y}{2.5}-\frac{150}{2.5}

x=0.4y-60=\frac{2}{5}y-60

Exchanging the variable, you obtain that <u><em>the inverse function of f(x)=2.5*x+150 is f⁻¹(x)=</em></u>\frac{2}{5}x-60<u><em></em></u>

7 0
3 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
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