A system of equations with infinitely many solutions is a system where the two equations are identical. The lines coincide. Anything that is equal to
will work. You could try multiply the entire equation by some number, or moving terms around, or adding terms to both sides, or any combination of operations that you apply to the entire equation.
You could multiply the whole thing by 4.5 to get
. If you want, you could mix things up and write it in slope-intercept form:
. The point is, anything that is equivalent to the original equation will give infinitely many solutions x and y. You can test this by plugging in values x and y and seeing the answers!
The attached graph shows that four different equations are really the same.
Hello,
Using the theorem of Thalès,
PR/TP=QS/TQ==>QS=4*20/16=5
Answer A
So, let's first focus on the first year:
the population then would be: 23000* (100+2)%
100% is the old population the 2% is the incease, so in total it's 102%, or 1.02 - the same number written differently.
So it will be 23000*1.02
After two years it will be 23000*1.02*1.02, and so on
so after x years it will be:
y=23000*
Answer:
Step-by-step explanation:
First you would see how many times 3 would go into 22 and it would be 7 with a remainder of 1. Then you would do 7 times 3 which is 21. And 22 minus 21 is 1. Then your answer would be 7 with a remainder of 1. If you don't want a remainder, and you want a decimal then you would add a decimal point after the 22 and then add a 0. You would bring down the zero and then the remainder of one would be 10 then you would see how much 3 goes into 10, and it only goes 3 times. Your answer would be 7.3 with a remainder of 1, If you keep going you will see it becomes a repeating decimal. *Hope it helped*
Answer:
a) 
b) 
Step-by-step explanation:
Given Data:
Interest rate=
per year
No. of years=
Rate of continuous money flow is given by the function
a) to find the present value of money
Put f(t)=2000 and n=10 years and r=0.08
Now integrate
(b) to find the accumulated amount of money at t=10
Where P is the present worth already calculated in part a
