The number that is 1 less than the number whose digits have a sum of 12 and whose tens digit is greater than 4 but less than 6.
The tens digit is 5
56
thanks, Ms. Sue.
p(w)=4w^3-3w-4; find p(-4) and p(1/4)
To find the number that is 1 less than the number we need to determine first the number whose digits have a sum of 12 and whose tens digit is greater than 4 but less than 6.
Let's break down the problem step by step:
Step 1: Find the number whose digits have a sum of 12 and whose tens digit is greater than 4 but less than 6.
We know that the tens digit should be greater than 4 but less than 6. This means that the tens digit can only be 5. The units digit can be any number from 0 to 9 since there are no restrictions on it.
For the tens digit = 5, and sum of the digits = 12, we can set up an equation:
5 + x + y = 12
x represents the units digit, and y represents the hundreds digit.
Step 2: Solve the equation.
Since we have two variables, we need another equation to solve for both x and y. We can use the equation:
y = x + 1
Substituting this equation into the first equation:
5 + x + (x + 1) = 12
Combining like terms:
2x + 6 = 12
Subtract 6 from both sides:
2x = 6
Divide by 2:
x = 3
Substituting the value of x back into the equation y = x + 1:
y = 3 + 1
y = 4
So, the number whose digits have a sum of 12 and whose tens digit is greater than 4 but less than 6 is 543.
Step 3: Find the number that is 1 less than 543.
To find the number that is 1 less, we subtract 1 from 543:
543 - 1 = 542
Therefore, the number that is 1 less than the number whose digits have a sum of 12 and whose tens digit is greater than 4 but less than 6 is 542.