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Leto [7]
3 years ago
7

In a set of five consecutive integers, the smallest integer is more than 2/3 the largest. What is the smallest possible value of

the sum of the five integers?
Mathematics
2 answers:
astraxan [27]3 years ago
4 0
The answer is: "55".
________________
"55" is the smallest possible value of the sum of the five integers.
________________________________________
We are given five (5) consecutive integers.
___________________________________
Let us represent the first integer as: "x"; 
___________________________________
The second integer as: "x + 1" ; 
___________________________________
The third integer as: "x + 2" ; 
___________________________________
The fourth integer as: "x + 3" ; 
___________________________________
The fifth integer as: "x + 4" ; 
___________________________________
We are given: The smallest integer is MORE THAN (⅔ of the largest integer.
____________________________________
The smallest integer, "x" is greater than: "(⅔) (x + 4)"
________________________________________________
Note the distributive property of multiplication:
__________________________________
→  a*(b + c) = ab + ac ;
__________________________________
→ As such; (⅔) (x + 4) = [(⅔)*(x)] + [⅔)*(4); 
________________________________________
→ (⅔) (x + 4) = [(⅔)*(x)] + [⅔)*(\frac{4}{1})
______________________________________________
{Note: [⅔)*(\frac{4}{1})=\frac{2*4}{3*1}=\frac{8}{3}}; 
_______________________________________________
→ Rewrite the equation: → ⅔ of the largest integer =
______________________
→ (⅔) *(x + 4) = (⅔)x + \frac{8}{3} ;
_______________________________
The smallest, "x", is GREATER than: (⅔)x + \frac{8}{3} ; 
________________________________
→ Write as:  x > (⅔)x + \frac{8}{3}
______________________________________
→ Multiply the ENTIRE "inequality" (BOTH SIDES) by "3", to get rid of the "fractions":
_______________________________
3*{  x > (⅔)x + \frac{8}{3} } ; to get: 
________________________________________
→ 3x > 2x + 8 ; → Now, subtract "2x" from EACH SIDE of the inequality;
_________________________
 → 3x − 2x > 2x + 8 − 2x ;  
______________________
→ to get:  →  x > 8
____________________
Now, we want to to know the "smallest possible value of the sum five          integers".
___________________________________
→That is, the smallest possible value of the sum of :
___________________________________________
→x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = 5x + 10.
__________________
So, we want to know the smallest value of "(5x + 10)" ; x > 8, 
_______________________________________________
→ Let us solve for "5x + 10" ; when x = 10; by substituting "8" for "x"
________________
→ 5x + 10 = (5*8) + 10 = 40 + 10 = 50
___________________________________
So, if "x > 8"; then "x" must be greater than "8"; 
_____________________________
When "x = 8", the sum of the 5 (five) integers in our problem = 50.
___________________________
Since "x" in the first of the consecutive integers in the problem; we know that "x > 8"; then the smallest possible value for "x" would be "9"; since "x" has to be an integer.
______________________________
So, we know that:
The smallest possible value of sum of the value integers =
 5x + 10; when x = 9; 
_________________
→ So, we plug "9" for the value of "x" and solve:
______________________________
→ 5(9) + 10 = 45 + 10 = 55 ; → which is our answer.
_____________________________________
Let us check our answer:
_______________________
→ x + (x + 1) + (x + 2) + (x + 3) + (x + 4) = ? 55?
                                                           (when "x = 9" ?)?? ;
___________________________________________________
→  9 + (9 + 1) + (9 + 2) + (9 + 3) + (9 + 4) = ? 55? ;
______________________________________
→  9  + 10 +  11 + 12 + 13 = ? 55 ?? ;  Yes!.
__________________________________________
Also, is the answer: 55 , Reasonable? Yes, since it is an integer, and 5 consecutive integers added together would "add up" to an integer value.
__________________________________________
Hope this answer and lengthy explanation is helpful.
Best of luck!
_______________________
Natali [406]3 years ago
3 0

Answer:

55

Step-by-step explanation:

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Step-by-step explanation:

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(2)  (\sqrt{2})× (\sqrt{3}) =

   Also applying the the law of multiplication of surd

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A $33$-gon $P_1$ is drawn in the Cartesian plane. The sum of the $x$-coordinates of the $33$ vertices equals $99$. The midpoints
lisabon 2012 [21]

1.<span>
The midpoint </span>MPQ of PQ is given by  (a + c / 2, b + d / 2)<span>

2.
Let the x coordinates of the vertices of P_1 be : 

x1, x2, x3,…x33

the x coordinates of P_2 be :

</span>z1, x2, x3,…z33<span>

and the x coordinates of P_3 be:


w1, w2, w3,…w33</span>

<span>
3.
We are given with: 


</span>

X1 + x2 + x3… + x33 = 99

We also want to find the value of w1 + w2 + w3… + w33.<span>

4.

Now, based from the midpoint formula:</span>

 

Z1 = (x1 + x2) / 2

Z2 = (x2 + x3) / 2

Z3 = (x3 + x4) / 2

Z33 = (x33 + x1) / 2<span>

and 

</span>

<span>W1 = (z1 + z1) / 2


W2 = (z2 + z3) / 2</span>

<span>W3 = (z3 + z4) / 2

W13 = (z33 + z1) / 2

.
.

5.</span>

<span>W1 + w1 + w3… + w33 = (z1 + z1) / 2 +  (z2 + z3) / 2 + (z33 + z1) / 2 = 2 (z1 + z2 + z3… + z33) / 2</span>

<span>Z1 + z1 + z3… + z33 = (x1 + x2) / 2 + (x2 + x3) / 2 + (x33 + x1) / 2

</span>2 (x1 + x2 + x3… + x33) / 2 = (x1 + x2 + x3… + x33 = 99<span>


<span>Answer: 99</span></span>

8 0
3 years ago
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