Answer:
6
come to z o o m
z o o m id 85675494491 pwd eu3nFG
The answer is for your paper is 21
The answer can be readily calculated using a single variable, x:
Let x = the amount being invested at an annual rate of 10%
Let (8000 - x) = the amount being invested at an annual rate of 12%
The problem is then stated as:
(x * 0.10) + ((8000 - x) * 0.12) = 900
0.10(x) + ((8000 * 0.12) - 0.12(x)) = 900
0.10(x) + 960 - 0.12(x) = 900
0.10(x) - 0.12(x) = 900 - 960
-0.02(x) = -60
-0.02(x) * -100/2 = -60 * -100/2
x = 6000 / 2
x = 3000
Thus, $3,000 is invested at 10% = $300 annually; and $8,000 - $3,000 = $5,000 invested at 12% = $600 annually, which sum to $900 annual investment.
Answer:
The amount of nuclear fuel required is 1.24 kg.
Step-by-step explanation:
From the principle of mass energy equivalence we know that energy generated by mass 'm' in an nuclear plant is

where
'c' is the speed of light in free space
Since the power plant operates at 1200 MW thus the total energy produced in 1 year equals

Thus using the energy produced in the energy equivalence we get

Now since the efficiency of conversion is 34% thus the fuel required equals

Answer: 13% increase?
Step-by-step explanation: 19-6=13 which means it would be 1.13 the original amount?