The only way 3 digits can have product 24 is
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4
To be divisible by 3 the sum of the digits must be divisible by 3.
1+ 3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3.
So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.
Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.
The others must end in 3.
They must be greater than 152 which is 225. So the
First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.
The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
If you would like to find the function that gives the profit Betty makes by selling a number of glasses of lemonade, you can find this using the following steps:
p ... profit
g ... glasses of lemonade
f(g) = p = g * $1 - $50
The correct result would be f(g)<span> = g * $1 - $50.</span>
Pairs, in this case, relates to a group of 2 or more. We have 6 friends. Let's call them A,B,C,D,E,F. This will allow us to make a [some sort of] combination tree:
1. ABC against DEF
2. ABD against CEF
3. ABE against CDF
4. ABF against CDE
5. ACD against BFE
6. ACE against BDF
7. ACF against BDE
8. ADE against BCF
9. ADF against BCE
10. AEF against BCD
I believe there are 12 combinations... I just can't think of the last 2 though.
Answer:
Step-by-step explanation:
You can use Pythagoras Theorem,

∴
Hope that helps!!
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Square units are the units of measuring area. Therefore, this question asks for the area of the room.
The room has the shape of a rectangle with:
length = 13 units and width = 9 units
The are of the rectangle can be calculated using the following rule:
area of rectangle = length * width
area of the room = 13 * 9 = 117 square units