Answer:
In the first account was invested 
at 3%
In the second account was invested 
at 5%
Step-by-step explanation:
we know that
The simple interest formula is equal to
where
I is the Interest Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
in this problem we have
First account
substitute in the formula above
Second account
substitute in the formula above
Remember that
The interest is equal to 
so
Adds the interest of both accounts
therefore
In the first account was invested at 3%
In the second account was invested 
at 5%
If X is increased by 7 to make 21
X + 7 = 21
then rearrange the equation to get X
X = 21-7
X = 14(b)
Answer:
512x−2+8=18
Step 1: Simplify both sides of the equation.
512x−2+8=18
512x+−2+8=18
(512x)+(−2+8)=18(Combine Like Terms)
512x+6=18
512x+6=18
Step 2: Subtract 6 from both sides.
512x+6−6=18−6
512x=12
Step 3: Divide both sides by 512.
512x512=12/512
x= 3/128
Answer:
x=3/128
The graph is misleading because the year’s interval is not constant.
The first year to the second year, the gap is 1 year; in the second to the
third year, the gap is 2; in the third to the fourth year is 4; and the fourth
to the fifth year is 6.
100. When rounding to the hundreds place, you must consider the value of the digit in the tens place. If the digit in the tens place is 5 or greater, the digit in the hundreds place increases by 1. If the digit in the tens place is 4 or less the digit in the hundreds place remains the same. In this case, the number in the hundreds place was originally zero.
