No matter what I do, I do, Mr. Bradford
coins
always have eight times
as many coin as i The first year, I had 4 coins, the second
Year, I had 6 the third year I had 8 At this rate, how many
Coins will Bradford have the fifth years

96

Step-by-step explanation:

Basically you're multiplying how many coins you have that year by eight.

The fist year you had 4 coins. 4 multiplied by 8 equals 32. Mrs Bradford had 32 coins.

The second year you had 6 coins. 6 multiplied by 8 equals 48. Mrs Bradford had 48 coins.

The third year you had 8 coins. 8 multiplied by 8 equals 64. Mrs Bradford had 64 coins

For the fourth and fifth year you have to add 2 for each year.

So for the 4th year you had 10 coins and 10 multiplied by 8 equals 80. Mr.Bradford had 80 coins.

And Finally for year five you had 12 coins. 12 multiplied by 8 equals 96. Mr.Bradford had 96 coins in year five.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

1.) 21

2.) 0.453

3.) 1.5

4.) 48

5.) $ 22

step-by-step explanation: