Answer:
The numbers are 56, 280 and 156
Step-by-step explanation:
Let
x -----> the first number
y ----> the second number
z ----> the third number
we know that
y=5x -----> equation A
z=x+100 ----> equation B
x+y+z=492 ----> equation C
Solve the system of equations by substitution
substitute equation B and equation A in equation C and solve for x
x+5x+x+100=492
7x=492-100
7x=392
x=56
<u><em>Find the value of y</em></u>
y=5x -----> y=5(56)=280
<u><em>Find the value of z</em></u>
z=x+100 ----> z=56+100=156
therefore
The numbers are 56, 280 and 156
If I am reading this right, it looks like the 10, 3, 2, 1 are Adjustments and the Adjusted TB should equal the difference. Make sure you know how to add and subtract the debit and credit adjustments correctly.
TB +/- Adj = ATB
If you would like to simplify <span>(5a^2 * b^3)^0, you can do this using the following steps:
</span>
<span>(5a^2 * b^3)^0</span> = 5^0 * (a^2)^0 * (b^3)^0 = 1 * 1 * 1 = 1
The correct result would be 1.
