Answer:
Hence, kaitlin will need 2 albums if she fills each page with her coins.
Step-by-step explanation:
The given information is:
Kaitlin has 46 coins in her coin collection.
She has albums that can hold 5 coins per page.
The albums have 6 pages each.
Now we are asked to find how many albums will be needed to fill each page with her coins.
As we know that 1 album has 6 pages.
1 page can hold=5 coins.
So, 1 album can old=6×5=30 coins.
Now 46-30=16 coins are left
So we will need one ore album so that these 16 coins could be placed in it.
Hence, kaitlin will need 2 albums if she fills each page with her coins.
Answer:
![a_n=9(4^{n-1})](https://tex.z-dn.net/?f=a_n%3D9%284%5E%7Bn-1%7D%29)
Step-by-step explanation:
we know that
In a <u>Geometric Sequence</u> each term is found by multiplying the previous term by a constant, called the common ratio (r)
In this problem we have
![a_2=36\\ a_5=2,304](https://tex.z-dn.net/?f=a_2%3D36%5C%5C%20a_5%3D2%2C304)
Remember that
----->
-----> equation A
Substitute the values of a_5 and a_2 and solve for r
Find the value of a_1 in equation A
![36=a_1(4)](https://tex.z-dn.net/?f=36%3Da_1%284%29)
![a_1=9](https://tex.z-dn.net/?f=a_1%3D9)
therefore
The explicit rule for the nth term is
![a_n=a_1(r^{n-1})](https://tex.z-dn.net/?f=a_n%3Da_1%28r%5E%7Bn-1%7D%29)
substitute
![a_n=9(4^{n-1})](https://tex.z-dn.net/?f=a_n%3D9%284%5E%7Bn-1%7D%29)
(x-p)(x-q)
y(x-q) .... let y = x-p
yx - yq ... distribute
x(y) - q(y)
x(x-p) - q(x-p) ... replace y with x-p
x^2 - px - qx + pq .... distribute
x^2 + (-p-q)x + pq
The last expression is in the form ax^2+bx+c with a = 1, b = -p-q and c = pq
The constant term is the term without any variable x attached to it (either x or x^2), so the constant term is pq
The answer i got is the last one :)))))
Answer:
pretty sure its 5
Step-by-step explanation: