There are 125 people and three door prizes. How many ways can three door prizes of $50 each be distributed? How many ways can do
1 answer:
We have been given that there are 125 people and three door prizes.
In the first part we need to figure out how many ways can three door prizes of $50 each be distributed?
Since there are total 125 people and there are three identical door prices, therefore, we need to use combinations for this part.
Hence, the required number of ways are:
In the next part, we need to figure out how many ways can door prizes of $5,000, $500 and $50 be distributed?
Since we have total 125 people and there are three prices of different values, therefore, the required number of ways can be figured out by using permutations.
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Answer:
D.)
Step-by-step explanation:
The zero's are referencing when y=0, note that when y=0 they are talking about the x-intercepts. You can graph the function and see when the graph crosses the x-axis or solve for the x-values. I will solve it via factoring and so:
Multiply the outer coefficients, in this case 1 and 6, and 1×6=6. Now let's think about all the factors of 6 we have: 6×1 and 2×3. Now is there a way that if we use any of these factors and add/subtract them they will return the middle term 5? Actually we can say 6-1=5 and 2+3=5. Let's try both.
First let's use 6 and -1 and so:
Notice how we have (x+6) and (x-6), these factors do not match so this is incorrect.
Now let's try 2 and 3 and so:
Notice how the factors (x+3) matched up so this is a factor and so is (x+2), now to solve for the zero's let's make f(x)=0 and solve each factor separately:
Case 1:
Case 2:
So your zero's are when x=-2 and x=-3.
D.) x=-3 and x=-2 because the graph crosses the x-axis at -3 and -2.
~~~Brainliest Appreciated~~~