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dezoksy [38]
3 years ago
11

Mr. Hawkins is covering a wall with wallpaper. The wall is a rectangle and measures 14 feet by 12 feet. Each square foot of wall

paper costs $2.90. What is the cost of adding wallpaper to the room? HINT: Find the area of the wall, then use that to determine the cost.
(Please explain your answer)
Mathematics
2 answers:
Alexandra [31]3 years ago
8 0
Answer:

$487.2

Step-by-Step explanation:

You multiply 14 x 12 and you get 168. Then after you multiply 14 and 12, multiply the 168 with $2.90. Then you get the answer $487.2

ELEN [110]3 years ago
3 0

Answer:

$487.2

Step-by-step explanation:

you multiply 14 and 12 then you get the answer of 168 and multiply that with $2.90 then you get the answer of $487.2.

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At the beginning of the day the stock market goes down 4014 40 1 4 points and stays at this level for most of the day. At the en
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Answer:

<h2>124.51%</h2>

Step-by-step explanation:  

In this problem, we are expected to solve for the percentage change, given the opening  point and the closing point

given that the low is 4014

and the high is  9012

the formula to calculate the percentage change is given as

%change= (high-low)/low*100

substituting our given data we have

%change= (9012-4014)/4014*100

%change= (4998)/4014*100

%change= (1.2451)*100

%change= 124.51%

The total change in the stock market from the beginning of the day to the end of the day is 124.51%

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A triangular pyramid has a right triangle for its base. The perpendicular sides of the triangle are 10 inches and 18 inches long
igor_vitrenko [27]
The formula for a triangular pyramid is 1/3AH. A is the area of the base. H is the height of the pyramid. So to find the area of a right triangle, you would use the formula (a x b)/2. So (10 x18)/2 is 90 inches^2. Now you can fill in the first formula. 1/3(90 x 14), which equals 420 inches^3. I hope you find this helpful (if correct.)
6 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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