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<img src="https://tex.z-dn.net/?f=3%5En%5E%2B%5E1%2B9%2F3%5En%5E-%5E1%2B1" id="TexFormula1" title="3^n^+^1+9/3^n^-^1+1" alt="3^n

abruzzese [7]

Answer:

Hello,

Step-by-step explanation:

$\dfrac{3^{n+1}+9}{3^{n-1}+1} \\\\=\dfrac{9*(3^{n-1}+1)}{3^{n-1}+1}\\\\=9\\$

4 0

3 years ago

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

4 years ago

A STUDENT FINISHES THEIR FISR HALF OF AN EXAM IN 2/3 THE TIME IT TAKES HIM TO FINISH THE SECOND HALF. IF THE ENTIRE EXAM TAKES H

shepuryov [24]

Answer: First half was 24 minutes

Step-by-step explanation:

Let the time taken to finish the second half be y.

Since the student used 2/3 of the second half time to finish the first half, first half = 2/3 × y = 2y/3

The entire exam is an hour which equals 60 minutes

First half + Second half = 60minutes

Note that first half is denoted as 2y/3 and second half is denoted by y.

2y/3 + y = 60

5y/3 = 60

Cross multiply

5y = 60 × 3

5y = 180

y = 180/5

y = 36

Second half took 36 minutes

Since first half is 2y/3, it will be:

(2×36) / 3

= 72/3

= 24minutes

First half took 24 minutes.

8 0

3 years ago

PLEASE HURRY I WILL DO ANYTHING (not anything but give brain list)

Sergeeva-Olga [200]

The cost of the sheets is less for the second example, the second example is the better deal.

Consider the first example:

Here, A brand of sheets is on sale for “Buy one get one at 50% off”. If each sheet costs $18.00.

So, the value of the first sheet will be $18,

And since the other sheet is at 50% off .

That is it will be half of the original value

= 50/100 * 18

=9

Therefore, the cost of two sheets

=$18 + $9

= $27.

So, Six sheets will cost

=$27+$27+$27

= $81

Cost of six sheets will be $81

Now, consider the second example

In the second example sheets are 30% off.

So, 30% of 18

= $6.

Subtract that value from 18 and you will get $12.

So, cost of one sheet = $12.

cost of 6 sheet

= $12×6

= $72

In the first example, Six sheets costs $81, while in the second example six sheets cost $72.

Since the cost is less for the second example. So, the second example is the better deal.

4 0

2 years ago

Determine dois números ímpares consecutivo cujo produto seja 195

miv72 [106K]

Let 1st integer = xLet 2nd integer = x + 1 We set up an equation. x(x + 1) = 195 x2 + x = 195 x2 + x - 195 = 0

We will use the quadratic formula: x = (-b ± √(b2 - 4ac) / (2a) x = (-1 ± √(1 - 4(-195))) / 2 x = (-1 ± √(781)) / 2 x = (-1 ± 27.95) / 2 x = 13.48x = -14.78
We determine which value of x when substituted gives us a product of 195. 13.48(14.48) = 195.19-14.48(-13.48) = 195.19 The solution is 2 sets of two consecutive number Set 1 The 1st consecutive integer is 13.48The 2nd consecutive integer is 14.48
Set 2 The 1st consecutive integer is -14.48The 2nd consecutive integer is -13.48Hopefully this helped, hard work lol :)

8 0

3 years ago

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