Given that the number of years should be represented with x, the number of fish in the pond after x years should best be represented with f(x). The equation that would best show the given scenario in the problem above is,
f(x) = 500(2^x)
From the given, 500 is used as the initial population of the fish.
<h3><u>(1 + 5y)(1 - 5y) is the fully factored form of this polynomial.</u></h3>
This polynomial can be factored using difference of squares.
This polynomial is in the form of a^2 - b^2 = (a + b)(a - b)
Because 1 * 1 = 1, we can use this formula to simplify this polynomial.
1 - 25y^2 = (1 + 5y)(1 - 5y)
We can use FOIL to verify this.
(1 + 5y)(1 - 5y)
1 - 5y + 5y - 25y^2
1 - 25y^2
Step-by-step explanation:
First, replace f(x) with y . ...
Replace every x with a y and replace every y with an x .
Solve the equation from Step 2 for y . ...
Replace y with f−1(x) f − 1 ( x ) . ...
Verify your work by checking that (f∘f−1)(x)=x ( f ∘ f − 1 ) ( x ) = x and (f−1∘f)(x)=x ( f − 1 ∘ f ) ( x ) = x are both true.
Answer: for 9 attendees it would cost $18
Step-by-step explanation: First you have to find the unit rate. So for every 7 attendees it costs $14, divide them both by the GCF which is 7. 14÷7=2
7÷7=1
So for every 1 attendee it is $2.
Now to figure out how much it would cost for 9 attendees, figure out what you have to do to 1 to get 9. Multiply it by 9.
And whatever you do to one number you have to do for the other. So $2 • 9 = $18
So for every 9 attendees it costs $18.
50*8=400
400+175=575
it subtracted 575 on the last round
