Dilbert invests a total of $22,500 in two accounts paying 6% and 14% annual interest, respectively. how much was invested in each account if, after one year, the total interest was $2,510.00.

$8000 was invested into the account paying 6% interest. $14,500 was invested into the account paying 14% interest.

Step-by-step explanation:

Let "a" be the money invested in the first account

let "b" be the money invested in the second account

The equation for the total interest is:

0.06a + 0.14b = 2510

I converted 6% and 14% to decimal form by dividing by 100.

The equation for total investment is:

a + b = 22500

Solve the system between these two equations:

Rearrange a + b = 22500 to isolate one of the variables.

a + b = 22500

b = 22500 - a

Substitute b = 22500 - a into the other equation

0.06a + 0.14b = 2510

0.06a + 0.14(22500-a) = 2510     Use distributive property over brackets

0.06a + 3150 - 0.14a = 2510     Combine the like terms with variable "a"

3150 - 0.08a = 2510   Subtract 3150 from both sides

-0.08a = -640    Divide both sides by -0.08

a = 8000

Substitute a = 8000 into the simpler equation to find "b"

b = 22500 - a

b = 22500 - 8000

b = 14500

Therefore $8000 was invested into the account paying 6% interest. $14,500 was invested into the account paying 14% interest.

the 2 figures are congruent bc if you reflect abcd over the y axis then over the d axis the figures overlap

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)