Answer:
31, 33, and 35
Step-by-step explanation:
31, 33, and 35 are consecutive odd integers and they add up to 99. So there isn't any other three consecutive odd intergers add up to greater than 99 and less than 100.
To find the greatest possible values of three consecutive odd integers whose sum is less than 100, we can set up an equation and solve it for the variable representing the largest odd integer.
To find the greatest possible values of three consecutive odd integers whose sum is less than 100, we can start by considering the largest odd integer that is less than 100, which is 99. Let's represent this as x.
The next two consecutive odd integers would be (x+2) and (x+4). The sum of these three integers can now be expressed as:
x + (x+2) + (x+4) = 3x + 6
Since the sum must be less than 100, we can set up the following inequality:
3x + 6 < 100
Solving this inequality, we get:
3x < 94
x < 31.333
Therefore, the greatest possible value for x would be 31. This means the three consecutive odd integers would be 31, 33, and 35.
#SPJ2
13,15,17,x,20,21; The median is 18.
Answer:
1
Step-by-step explanation:
13 15 17 x 20 21
Cross out all outer numbers
We are left with 17 and x
With double medians, you add those two together and then divide them by two.
With the median (total) being 18, we will first add 17+x then divide Y/2 = 18
The value of x is 1
What is the solution to the following inequality?
Answer:
3x - 2x -1 < 4
3x -2x <4+1
x < 4+1
x < 5
The area of this triangle is______ m2
Answer:120 m 2
Step-by-step explanation: got it right (OW)
Answer:
120 m 2
Step-by-step explanation:
37.5 * 6.4 / 2 = 120 m 2