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AVprozaik [17]
3 years ago
9

HELP PLEASE???!!!!!!!

Mathematics
1 answer:
Fiesta28 [93]3 years ago
7 0

Answer:

<h2>A. (-4, -15)</h2>

Step-by-step explanation:

\left\{\begin{array}{ccc}y=\dfrac{3}{4}x-12&(1)\\y=-4x-31&(2)\end{array}\right\\\\\text{Substitute (1) to (2):}\\\\\dfrac{3}{4}x-12=-4x-31\qquad\text{multiply both sides by 4}\\\\3x-48=-16x-124\qquad\text{add 48 to both sides}\\\\3x=-16x-76\qquad\text{add}\ 16x\ \text{to both sides}\\\\19x=-76\qquad\text{divide both sides by 19}\\\\x=-4\\\\\text{Put it to (2):}\\\\y=-4(-4)-31\\y=16-31\\y=-15

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Answer:

a.Converge

b.Diverge

Step-by-step explanation:

We are given that

A_n=\frac{6n}{-4n+9}

\lim_{n\rightarrow \infty}A_n=\lim_{n\rightarrow \infty}\frac{6n}{-4n+9}=\frac{6n}{n(-4+\frac{9}{n})}

\lim_{n\rightarrow \infty}\frac{6}{-4+\frac{9}{n}}=\frac{6}{-4}=\frac{-3}{2}

Because \frac{9}{\infty}=0

Hence, sequence An converges to \frac{-3}{2} when n tends to infinity or - infinity.

\sum_{n=1}^{\infty}An=\sum_{n=1}^{\infty}\frac{6n}{-4n+9}=diverges

Necessary condition for a series to converge :

an\rightarrow 0 when n tends to infinity .

But A_n tends to \frac{-3}{2}when n tends to infinity.

Therefore, given series is divergent.

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 Write the equation of the circle with center (7, 3) and a radius of 2.
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A restaurant’s dining room measures 30 ft by 40 ft. If the fire code requires one square meter per guest, what is the seating ca
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S_A_V [24]
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One number is three more than twice another, the sum of the numbers is 12. find the two numbers
Alenkasestr [34]
The numbers are:  "3" and "9" .
_______________________________________________________
Explanation:
_______________________________________________________
Let "x" represent one of the two (2) numbers.

Let "y" represent the other one of the two (2) numbers.

x = 2y + 3  ;

x + y = 12 .
__________________________
Method 1)
__________________________
x = 12 <span>− y ; 

Plug this into "x" for "2y + 3 = x" ;

</span>→ 2y + 3 = 12 <span>− y ; 
</span>
Add "y" to each side of the equation; & subtract "3" from each side of the equation ;

→ 2y + 3 + y − 3 = 12 − y + y <span>− 3 ;
</span>
to get:  3y = 9 ;

Divide each side of the equation by "3" ; 
to isolate "y" on one side of the equation; & to solve for "y" ;

             3y / 3 = 9 / 3 ;

               y = 3 .
____________________________
Now:  x = 12 − y ;  Plug in "3" for "y" ; to solve for "x" ;

    →   x = 12 − 3 = 9
____________________________
So;  x = 9,  y = 3 .
____________________________
Method 2) 
____________________________
When we have:
____________________________
x = 2y + 3  ;

x + y = 12 .
____________________________
 →  y = 12 − x ;
_____________________________
Substitute "(12−x)" for "y" in the equation:

" x = 2y + 3 " ;

→ x = 2(12 − x) + 3 ; 
_____________________________________
Note the "distributive property of multiplication" :
_____________________________________
  a(b + c) =  ab + ac ;

  a(b − c) = ab − ac ;
_____________________________________
As such:
_____________________________________
     →  2(12 − x) = 2(12) − 2(x) = 24 − 2x ; 
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So; rewrite:
_____________________________________
        x = 2(12 − x) + 3 ;
as:
_____________________________________
  →  x = 24 − 2x + 3 ;

  →  x = 27 − 2x ;

Add "2x" to each side of the equation:

 →  x + 2x = 27 − 2x + 2x ;

 →  3x = 27 ;

Divide each side of the equation by "3" ;
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      3x / 3 = 27 / 3 ;

        x = 9 .
___________________________
Note:  "y = 12 − x" ;  Substitute "9" for "x" ; to solve for "y" ;

→  y = 12 − 9 = 3 ;

→  y = 3 .
__________________________
So,  x = 9 ; and y = 3.
____________________________
The numbers are:  "3" and "9" .
_____________________________
To check our answers:
Let us plug these numbers into the original equations;
   to see if the equations hold true ; (i.e. when, "x = 9" ; and "y = 3"
_____________________________
  → x + y = 12 ; 

   → 9 + 3 =? 12 ?  Yes!
_____________________________
 
   → x = 2y + 3 ;

   → 9 =? 2(3) + 3 ?? ; 

   → 9 =?  6 + 3 ?  Yes!!
_____________________________

3 0
3 years ago
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