Answer:

Step-by-step explanation:
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:
h= 7.5
Step-by-step explanation:
8h=60
One solution was found :
h = 15/2 = 7.500
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
8*h-(60)=0
Step by step solution :
Step 1 :
Pulling out like terms :
1.1 Pull out like factors :
8h - 60 = 4 • (2h - 15)
Equation at the end of step 1 :
Step 2 :
Equations which are never true :
2.1 Solve : 4 = 0
This equation has no solution.
A a non-zero constant never equals zero.
Solving a Single Variable Equation :
2.2 Solve : 2h-15 = 0
Add 15 to both sides of the equation :
2h = 15
Divide both sides of the equation by 2:
h = 15/2 = 7.500
One solution was found :
h = 15/2 = 7.500
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