math homework..algebra 1 anyone can help?

Question

the sum of 4 consecutive odd numbers is less than or equal to 100. what are the largest possible values for the integers?

step-by-step explaining would help..
thx
=]

Answer 1

The answer is "C"

Answer 2

x = 1st number
x+2 = 2nd number
x+4 = 3rd number
x+6 = 4th number

adding an extra two each time because they are "odd" integers.

So, add up the numbers:

x+x+2+x+4+x+6 =

4x + 12 <= 100

4x <=88

x<=22

Therefore, x must be 21.

21, 23, 25, 27

Check the answer:

21 + 23 + 25+ 27 = 96.

If the number were any higher than 21, i.e. 23...the sum of all those odd integers would be more than 100.

Regards,

Mysstere

Answer 3

set up like an algebra problem, like this:

the consecutive odd integers are:

~ x
~ x + 2
~ x + 4
~ x + 6

their sum is 4x + 12

Now set up your inequality:

4x + 12 <= 100

divide by 4 to make things a little easier:

x + 3 <= 25

now subtract 3 from both sides to get:

x <= 22

Since x has to be odd, x=21 is the largest odd that satisfies this equation.

So your 4 consecutive odds are: 21, 23, 25, 27

HTH! :-)

Answer 4

first you need to set up your variables
let x = 1st consecutive odd # (con)
let x+2 = 2nd con
let x+4 = 3rd con
let x+6 = 4th con

each one is +2 more because odd numbers go up by 2 each time.

set up your inequality x + (x+2)+(x+4)+(x+6) <=100
now combine like terms and solve for x, once you have x you can find the rest of the numbers

check your answers to see if the values hold true for the problem

Answer 5

Let first number be '2n+1' (since it's odd, if it was even it would have been 2n)
The next three consecutive ODD numbers would be:
2n+3, 2n+5, 2n+7

Therefore, (2n+1) + (2n+3) + (2n+5) + (2n+7) <= 100
8n + 16 <= 100
8n <= 84
n <= 10.5

Therefore, the largest odd integer would be (2n+7)=2*10.5+7=28. Since it should be odd, next odd number is the odd number lesser than 28 which is: 27.

Therefore, numbers are: 27,25,23, and 21.

Answer 6

x + x+2 + x+4 + X+6 < or = 100


solve.

Answer 7

x+x+x+x<=100

21+25+27+29=100

Answer 8

You teach babies the fundamentals, then build upon their understand-how over the years. In my tenth grade math classification, the instructor rather advised us that a million+a million=3 for vast values of a million. Even that's purely too progressed to teach little ones in straightforward college. I won't even get into determination maths we studied in numeric strategies (as quickly as a million+a million led to 40 5).

Answer 9

21+23+25+27=96

Is that good enough? I busted out the calculator for you on that one. Good luck!

Answer 10

math is so easy all you have to do is if the top number is biger than botton number you divat
love [e-mail]

Answer 11

Sorry - beyond my algebra abilities!@