Answer:
So there are 8 five-point problems and 30 two-point problems
Answer:
C and E
Step-by-step explanation:
Recall that according to the complex conjugate theorem, if P(x) is a polynomial function with real coefficients, and a+bi is an imaginary root of P(x), then its conjugate (i.e a-bi) must also be a root.
Here we see that 2 of the given roots are imaginary : i and 2+i
by the theorem given above, the conjugates of these must also be roots
i.e -i and 2-i
Answer:
21 and 26 totaling 47sticks
Step-by-step explanation:
Th number of sticks form an arithmetic progression
1, 6, 11, 16...
The nth term of an arithmetic progression is expressed as;
Tn = a+(n-1)d
a is the first term = 1
d is the common difference
d = 6-1 = 11-6 = 5
n is the number of terms
The next two terms are the fifth and sixth term
T5 = 1+(5-1)5
T5 = 1+4(5)
T5 = 1+20
T5 = 21
T6 = 1+(6-1)5
T6 = 1+5(5)
T6 = 1+25
T6 = 26
Hence the number of sticks needed in the next two piles are 21 and 26 totaling 47sticks
Answer:
i d k lol im just tryna get points
Step-by-step explanation:
have a cookie and good luck tho
