Answer:
x^2-x-5
Step-by-step explanation:
1. expand the square
(+1)(+1)−3(+2)
2. distribute
(+1)+1(+1)−3(+2)
3. distribute
^2++1(+1)−3(+2)
4. multiply by 1
^2+++1−3(+2)
5. combine like terms
^2+2+1−3(+2)
6. distribute
^2+2+1−3−6
7. subtract the numbers
^2+2−5−3
8. combine like terms
^2−−5
9. gg you are done
^2−−5
Pakal triangle is hard to do with long things, but it is easy up to like 5 degree
we look at the row for 5th degree (6th row)
the sequence is
1,5,10,10,5,1
that is the coeficients
for (a+b)^5 that is
1a^5b^0+5a^4b^1+10a^3b^2+10a^2b^3+5a^1b^4+1a^0b^5
see the exponents each time add to 5
so
1x^5(-5)^0+5x^4(-5)^1+10x^3(-5)^2+10x^2(-5)^3+5x^1(-5)^4+1x^0(-5)^5=
x^5-25x^4+250x^3-1250
Hi there!

If cross sections were made perpendicular to the base, they would assume the shape of the lateral sides.
Thus, the cross sections would be rectangles. The correction answer is C.
You can make it with 10 because 10%times 5% is 50%
Hello there!
I hope you are having a good day!
Your answer would be c because it is adding by 3 each time, making it arithmetic.
Hope I helped!
Let me know if you need anything else!
~Zoe