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Vadim26 [7]
3 years ago
14

BRAINLIESTTT ASAP! PLEASE HELP ME :)

Mathematics
1 answer:
cupoosta [38]3 years ago
7 0

<em>See above information</em>

I am joyous to assist you anytime.

* Since you are in a hurry, check out this link for more information:

brainly.com/question/12725478?utm_source=android&utm_medium=share&utm_campaign=question

You might be interested in
The sum of twice a first number and five times a second number is 78. If the second number is subtracted from five times the fir
ss7ja [257]
The numbers are:  "9" and "12" .
___________________________________
Explanation:
___________________________________
Let:  "x" be the "first number" ; AND:

Let:  "y" be the "second number" .
___________________________________
From the question/problem, we are given:
___________________________________
     2x + 5y = 78 ;  → "the first equation" ; AND:

     5x − y = 33 ;  → "the second equation" .
____________________________________
From "the second equation" ; which is:

   " 5x − y = 33" ; 

→ Add "y" to EACH side of the equation; 

              5x − y + y = 33 + y ;

to get:  5x = 33 + y ; 

Now, subtract: "33" from each side of the equation; to isolate "y" on one side of the equation ; and to solve for "y" (in term of "x");

            5x − 33 = 33 + y − 33 ;

to get:   " 5x − 33 = y " ;  ↔  " y = 5x − 33 " .
_____________________________________________
Note:  We choose "the second equation"; because "the second equation"; that is;  "5x − y = 33" ;  already has a "y" value with no "coefficient" ; & it is easier to solve for one of our numbers (variables); that is, "x" or "y"; in terms of the other one; & then substitute that value into "the first equation".
____________________________________________________
Now, let us take "the first equation" ; which is:
  "  2x + 5y = 78 " ;
_______________________________________
We have our obtained value; " y = 5x − 33 " .
_______________________________________
We shall take our obtained value for "y" ; which is: "(5x− 33") ; and plug this value into the "y" value in the "first equation"; and solve for "x" ;
________________________________________________
Take the "first equation":
 ________________________________________________
      →   " 2x + 5y = 78 " ;  and write as:
________________________________________________ 
      →   " 2x + 5(5x − 33) = 78 " ;
________________________________________________
Note the "distributive property of multiplication" :
________________________________________________
     a(b + c) = ab + ac ; AND:

     a(b − c) = ab − ac .
________________________________________________
So; using the "distributive property of multiplication:

→   +5(5x − 33)  = (5*5x) − (5*33) =  +25x − 165 .
___________________________________________________
So we can rewrite our equation:

          →  " 2x + 5(5x − 33) = 78 " ;

by substituting the:  "+ 5(5x − 33) " ;  with:  "+25x − 165" ; as follows:
_____________________________________________________

          →  " 2x + 25x − 165 = 78 " ;
_____________________________________________________
→ Now, combine the "like terms" on the "left-hand side" of the equation:

              +2x + 25x = +27x ; 

Note:  There are no "like terms" on the "right-hand side" of the equation.
_____________________________________________________
    →  Rewrite the equation as:
_____________________________________________________
         →   " 27x − 165 = 78 " ;

      Now, add "165" to EACH SIDE of the equation; as follows:

         →    27x − 165 + 165 = 78 + 165 ;

        →  to get:      27x = 243  ;
_____________________________________________________
      Now, divide EACH SIDE of the equation by "27" ; to isolate "x" on one side of the equation ; and to solve for "x" ;
_____________________________________________________
               27x / 27  =  243 / 27 ; 

       →   to get:    x = 9 ; which is "the first number" .
_____________________________________________________
Now;    Let's go back to our "first equation" and "second equation" to solve for "y" (our "second number"):

     2x + 5y = 78 ; (first equation);
     
      5x − y = 33 ; (second equation); 
______________________________
Start with our "second equation"; to solve for "y"; plug in "9" for "x" ;

→ 5(9) − y = 33 ;  

    45 − y = 33;  
   
Add "y" to each side of the equation:
 
   45 − y + y = 33 + y ;  to get:

   45 = 33 + y ;  

↔ y + 33 = 45 ;  Subtract "33" from each side of the equation; to isolate "y" on one side of the equation ; & to solve for "y" ;  
 
 → y + 33 − 33  = 45 − 33 ;

to get:  y = 12 ;

So;  x = 9 ; and y = 12 .  The numbers are:  "9" and "12" .
____________________________________________
 To check our work:
_______________________
1)  Let us plug these values into the original "second equation" ; to see if the equation holds true (with "x = 9" ; and "y = 12") ; 

→ 5x − y = 33 ;  → 5(9) − 12 =? 33 ?? ;  → 45 − 12 =? 33 ?? ;  Yes!
________________________
2)  Let us plug these values into the original "second equation" ; to see if the equation holds true (with "x = 9" ; and "y = 12") ;

→ 2x + 5y = 78 ; → 2(9) + 5(12) =? 78?? ; → 18 + 60 =? 78?? ; Yes!
_____________________________________
So, these answers do make sense!
______________________________________
3 0
3 years ago
Helppp meeeee with question 1 or can the all pleaseeee !!!
Liula [17]

Answer:

1. 2c+9=m

Step-by-step explanation:

6 0
3 years ago
The rectangle has a length of (x+3) and the width (x+1)
Tamiku [17]
The answer is b
Because you have to add da x and dat is 2x den add da numbers which is 4
7 0
3 years ago
Angie divides the polynomial function f(x)=x3+3x2−7x−21 by (x + 3). What can she conclude from the fact that the remainder is eq
iren [92.7K]

Answer:

See below

Step-by-step explanation:

<u>Given function</u>

  • f(x)=x³ + 3x²- 7x - 21

<u>We can find its factors:</u>

  • x³ + 3x² - 7x - 21 =
  • x²(x + 3) - 7(x + 3) =
  • (x² - 7)(x + 3)

We can conclude that the given polynomial function is fully divisible by (x + 3) and the quotient is (x² - 7)

Also one of zeros of the function is -3

8 0
3 years ago
A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the number of cakes sold per
HACTEHA [7]

Answer:

A) Revenue function = R(x) = (580x - 10x²)

Marginal Revenue function = (580 - 20x)

B) Fixed Cost = 900

Marginal Cost function = (300 + 50x)

C) Profit function = P(x) = (-35x² + 280x - 900)

D) The quantity that maximizes profit = 4

Step-by-step explanation:

Given,

The Price function for the cake = p = 580 - 10x

where x = number of cakes sold per day.

The total cost function is given as

C = (30 + 5x)² = (900 + 300x + 25x²)

where x = number of cakes sold per day.

Please note that all the calculations and functions obtained are done on a per day basis.

A) Find the revenue and marginal revenue functions [Hint: revenue is price multiplied by quantity i.e. revenue = price × quantity]

Revenue = R(x) = price × quantity = p × x

= (580 - 10x) × x = (580x - 10x²)

Marginal Revenue = (dR/dx)

= (d/dx) (580x - 10x²)

= (580 - 20x)

B) Find the fixed cost and marginal cost function [Hint: fixed cost does not change with quantity produced]

The total cost function is given as

C = (30 + 5x)² = (900 + 300x + 25x²)

The total cost function is a sum of the fixed cost and the variable cost.

The fixed cost is the unchanging part of the total cost function with changing levels of production (quantity produced), which is the term independent of x.

C(x) = 900 + 300x + 25x²

The only term independent of x is 900.

Hence, the fixed cost = 900

Marginal Cost function = (dC/dx)

= (d/dx) (900 + 300x + 25x²)

= (300 + 50x)

C) Find the profit function [Hint: profit is revenue minus total cost]

Profit = Revenue - Total Cost

Revenue = (580x - 10x²)

Total Cost = (900 + 300x + 25x²)

Profit = P(x)

= (580x - 10x²) - (900 + 300x + 25x²)

= 580x - 10x² - 900 - 300x - 25x²

= 280x - 35x² - 900

= (-35x² + 280x - 900)

D) Find the quantity that maximizes profit

To obtain this, we use differentiation analysis to obtain the maximum point of the Profit function.

At maximum point, (dP/dx) = 0 and (d²P/dx²) < 0

P(x) = (-35x² + 280x - 900)

(dP/dx) = -70x + 280 = 0

70x = 280

x = (280/70) = 4

(d²P/dx²) = -70 < 0

Hence, the point obtained truly corresponds to a maximum point of the profit function, P(x).

This quantity demanded obtained, is the quantity demanded that maximises the Profit function.

Hope this Helps!!!

8 0
3 years ago
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