The greatest common factor would be 5x because all the variables can be divided by 5x.
The answer itself already factored would be
5x(5x^2+2x+1)
No they’re not equivalent because for 3(2n+9) when you distribute the 3 you get 6n+27 & 5n+9+n equal to 6n+9 when you combine like terms (add the n’s).
we are given
we can check each options
option-A:
-1,1
we can plug x=-1 and x=1 and check whethet f(x)=0
At x=-1:
At x=1:
so, this is TRUE
option-B:
0,1
we can plug x=0 and x=1 and check whethet f(x)=0
At x=0:
At x=1:
so, this is FALSE
option-C:
-2,-1
we can plug x=-2 and x=-1 and check whethet f(x)=0
At x=-2:
At x=-1:
so, this is FALSE
option-D:
-1,0
we can plug x=-1 and x=0 and check whethet f(x)=0
At x=-1:
At x=0:
so, this is FALSE
Answer:
A.
Step-by-step explanation:
Answer:
the probability is 2/9
Step-by-step explanation:
Assuming the coins are randomly selected, the probability of pulling a dime first is the number of dimes (4) divided by the total number of coins (10).
p(dime first) = 4/10 = 2/5
Then, having drawn a dime, there are 9 coins left, of which 5 are nickels. The probability of randomly choosing a nickel is 5/9.
The joint probability of these two events occurring sequentially is the product of their probabilities:
p(dime then nickel) = (2/5)×(5/9) = 2/9
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<em>Alternate solution</em>
You can go at this another way. You can list all the pairs of coins that can be drawn. There are 90 of them: 10 first coins and, for each of those, 9 coins that can be chosen second. Of these 90 possibilities, there are 4 dimes that can be chosen first, and 5 nickels that can be chosen second, for a total of 20 possible dime-nickel choices out of the 90 total possible outcomes.
p(dime/nickel) = 20/90 = 2/9
