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.
Answer:
5
Step-by-step explanation:
We know that a^2+b^2=c^2 from the Pythagorean theorem where a and b are the legs and c is the hypotenuse
4^2+3^2 = c^2
16+9 = c^2
25 = c^2
Taking the square root of each side
sqrt(25) = sqrt(c^2)
5 =c
Answer:
2408
Step-by-step explanation:
Answer:
THE ANSWER OM EDGE IS (C)
UR WELCOME
Step-by-step explanation:
have a great day :)
The correct answers are as follows:
11) 110 - alternate exterior angles
12) 84 - alternate interior angles
13) 80 - supplementary angles
14) 111 - alternate interior angles
15) 125 - alternate interior angles
16) 47 - alternate exterior angles
17) 53- corresponding angles
18) 113-alternate interior angles
Hope this answers the question.
