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
Answer:
wha?
Step-by-step explanation:
Answer:
option C.
Step-by-step explanation:
since the graph y = 6x is shrunk by a factor of ½
it becomes y = 3x
to find one coordinate of this graph we put x = 1 so that we get y = 3
so one point we have is (1, 3).
and the other point is origin
[ for if you keep x = 0 then 3 × 0 = 0 so y comes 0 too.]
now this graph is translated 9 units in the negative y direction
so the point that initially was (0, 0) now becomes (0, -9).
we see that the only graph passing thru (0, -9) is C.
so the answer is <u>option C.</u>

