Answer:
In the pic
Step-by-step explanation:
If you have any questions about the way I solved it, don't hesitate to ask me in the comments below ÷)
It would be 30 because 10^4 is 10000 and 10000 times 12 is 120000 and 10^3 is 1000 and 1000 times 4 is 4000 so 120000 divided by 4000 is 30
Answer:
The solution to the system of equations are;
x = -4/3
y = 5/3
Step-by-step explanation:
To find the Solution, we would carry the Operation simultaneously.
4x + 2 = -2y .........(i)
6y - 18 = 6x ..........(ii)
First let's rearrange the equations, to make the journey smoother
2y + 4x = -2 ...........(iii)
6y - 6x = 18 ...........(iv)
Let's Multiply equation (III) by 3 so as to have a uniform spot to begin elimination.
3.2y + 3.4x = -2 . 3
6y + 12x = -6............... (v)
Let's subtract equation (v) from equation (iv)
= 0y - 18x = 24
-18x = 24
x = - 24 / 18
x = -4/3
Let's substitute (x = -4/3) in equation (ii), so that we can solve for the value of y:
6y - 18 = 6x
6y - 18 = 6 (-4/3)
6y - 18 = -8
6y = -8 + 18
6y = 10.
y = 10 / 6
y = 5/3
The solution to the system of equations are;
x = -4/3
y = 5/3
Answer:
800 months
Step-by-step explanation:
The formula for interest is:
(Capital * saving account * time) / 100
So:
(10000 * 0.06 * x) / (100) = 4800
We clear x:
(10000 * 0.06 * x) = (100) * 4800
x = 480,000 / (10000 * 0.06)
x = 800 months (66.67 years)
<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>
Step-by-step explanation:
Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.
From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by 
= 
.
So, the population in the year t can be given by 
Population in the year 2000 = 
=
Population in year 2000 = 3,762,979
Let us assume population doubles by year 
.
≈
∴ By 2033, the population doubles.
