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Mama L [17]
3 years ago
6

Sofia is making a copy of a ticket from the school play to put in her memory album. the original ticket is 7 inches long and 5 i

nches wide her copy is 2 inches long. what is the scale of the copy of the ticket?
Mathematics
1 answer:
kotykmax [81]3 years ago
7 0

Answer:

The scale of the copy of the ticket is \frac{2}{7}

Step-by-step explanation:

we know that

To find the scale divide the long of the copy of a ticket by the long of the original of a ticket

so

\frac{2}{7}

You might be interested in
How do you Simplify :(2p^-3)^5
iragen [17]

Answer:

The final simplification is (32p^-15).

Step-by-step explanation:

Given:

(2p^-3)^5 we have to simplify.

Property to be used:(Power rule)

Power rule states that: (a^x)^y=a^x^y ...the exponents were multiplied.

Using power rule.

We have,

⇒(2p^-3)^5          

⇒(2^5)(p^-3^*5)     ...taking exponents individually.

⇒32(p^-^1^5)         ...2^5=2*2*2*2*2=32

⇒32p^-^1^5

So our final values are 32p^-15

3 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
Solve the inequality: (x+3)(x−4)&lt;−12
melamori03 [73]

Answer:

0<x<1

Step-by-step explanation:

(x+3)(x−4)<−12

FOIL

x^2 -4x+3x -12 <-12

x^2 -x -12 <-12

Add 12 to each side

x^2 -x -12+12 <-12+12

x^2 -x <0

Factor out an x

x(x-1) <0

Using the zero product property

x(x-1) =0

x=0  x-1=0

x=0  x=1

Check the ranges

x<0

x(x-1) <0        

- * - >0

False

0<x<1

x(x-1) <0        

+ * - <0

True

x>1

x(x-1) <0        

+ * + >0

False

3 0
3 years ago
7800 nm + 95 pm in scientific notation
azamat
7800 nm+95pm=7800* 10^{-9} +95* 10^{-12}
=7800000* x^{-12} +95* 10^{-12} =7800095* 10^{-12} =7.800095* 10^{-6}
3 0
3 years ago
At the market 8 batteries cost 10 dollars how much will 6 cost
NemiM [27]
Answer: 4.8
Why: 8/10=.8
.8x6=4.8
4 0
3 years ago
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