There are five consecutive odd integers. The sum of the three smallest is 3 more than the sum of the two largest. Find the integers. 15 points

11 13 15 17 19

Step-by-step explanation:

We will solve this word problem using 2k+1 which is one of the general forms of an odd integer.

Let 2k+1 be the first odd integer.

(2k+1) + (2k+3) + (2k +5) =  (2k+7) + (2k +9) + 3 (the +3 is to accommodate for the left side being 3 more than the right)

Solve for k

6k + 9 = 4k + 19

6k - 4k + 9 = 4k-4k + 19

2k + 9 = 19

2k + 9 - 9 = 19 - 9

2k = 10

2k/2 = 10/2

k = 5

To get the first digit in the series plug in 5 for k

2k+1

2(5) + 1

10 + 1 = 11

So our series is 11, 13, 15, 17, 19

Let's test it.

11 + 13 +15 = 39

17 + 19 = 36

So we are correct, the first 3 integers are 3 more that the sum of the last 2.

too many. (jk)

31 v bucks

step-by-step explanation:

since you bought 25 on friday and 6 on tuesday, you can just add them up.

25+6=31

check your work.

31-6=25

i hope this !

-yand3r3

the correct option is:

b. a value of the variable that makes the polynomial equal to zero.

the root or roots of a polynomial are the values of the variable that makes the polynomial equal to 0. we can find the roots of a polynomial making the expression equal to 0 and then, isolating the variable (it it has just one variable).

for example, given the following polynomial:

the root or zero of the polynomial, can be found by making the expression equal to 0, and then, isolating the variable, so, finding the root we have:

therefore, we have that the root of the polynomial is equal to 2.

we must remember that the number of roots that a polynomial can have, will depend on the degree of the polynomial, a first degree polynomial have will have one single root, a second degree polynomial (quadratic expression) may have two roots and so.

hence the correct option is:

b. a value of the variable that makes the polynomial equal to zero.

have a nice day!