Let x represent amount invested in the higher-yielding account.
We have been given that a man puts twice as much in the lower-yielding account because it is less risky. So amount invested in the lower-yielding account would be
.
We are also told that his annual interest is $6600 dollars. We know that annual interest for one year will be principal amount times interest rate.
, where,
I = Amount of interest,
P = Principal amount,
r = Annual interest rate in decimal form,
t = Time in years.
We are told that interest rates are 6% and 10%.


Amount of interest earned from lower-yielding account:
.
Amount of interest earned from higher-yielding account:
.

Let us solve for x.



Therefore, the man invested $30,000 at 10%.
Amount invested in the lower-yielding account would be
.
Therefore, the man invested $60,000 at 6%.
If your looking for the answer then it’s 34
Let the length of the diameter be d and radius be r, thus to solve for d we proceed as follows:
r²+31²=(r+15)²
solving for d we solve for r
r²+961=r²+30r+225
putting like terms together we obtain:
r²-r²+961-225=30r
736=30r
solving for r
r=24.53
thus
d=24.5333×2
d=49.067~49.1
Answer is D. 49.1
B. As the x-values increase, the y- valuebtend to decrease.
Answer:
None
Step-by-step explanation: