A linear equation has maximum power of a variable(variables) 1.
In order to find the linear equations we need to check the equations with variable with power one.
Also the variable should be in numerator only.
Let us check option one by one:
a-y=√x + 1 we have a radical on x.
Therefore, it's not a linear equation.
b-y=1/x
We have x in the denominator of 1.
Therefore, it's not a linear equation.
c-y=x^2 +4
We have power 2 of x. Therefore, it's not a linear equation.
d-y = -2/3x + 4/3.
We have x and y powers only 1.
Therefore, it's the only linear equation.
<h3>Therefore, correct option is D option. d-y = -2/3x + 4/3</h3>
The answer to your question is (1,3) because I graphed it that is where they intersected at. I hope this helps.
We have been given that Willy has compounded monthly to invest his summer earnings of $4259 in the Rock Solid Bank. The bank is offering 6%. We are asked to find the amount of money will be after 5 years.
We will use compound interest formula to solve our given problem.
, where,
A = Final amount after t years,
P = Principal amount,
r = Annual interest rate in decimal form,
n = Number of times interest is compounded per year.
t = Time in years.

Since interest is compounded monthly, so
and
.





Therefore, Will will have approximately
in 5 years.
Well. If you get 29% of 5 cars that is 145%. This was a little confusing but I am defiantly sure it is correct. I like to use simpler numbers to see if I am doing the work right. So I said if he has a 50 % likely hood to find a car that was expired and had 1 car. It would be 50 percent. Now if he had 2nt got it the first time it would be a 100 % chance to find the car expired . Hope I didn’t co fuse you more