The sum of three consecutive odd integers is –15. What are the numbers?

1 answer:

3 0

Odd numbers are represented as 2n+1 where 2 is a natural number ( negative and possitive whole numbers)

consecutive odd numbers are 2 away from each other so 3 consecutive odd numbers would be
(2n+1),(2n+1+2),(2n+1+2+2)
the all add to -15 so they must all be negative
2n+1+2n+1+2+2n+1+2+2=-15
add like terms
6n+9=-15
subtract 9 from both sides
6n=-24
divide both sides by 6
n=-4
subsitute
first number=2n+1=2(-4)+1=-8+1=-7
second number=2n+1+2=2(-4)+3=-8+3=-5
third number=2n+1+2+2=-8+5=-3

the numbers are -7,-5, -3
the answe ris B

You might be interested in

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!