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ohaa [14]
4 years ago
15

Which polynomial represents the sum below? (14x2 – 14) + (–10x2 – 10x + 10) A. 4x2 – 10x + 24 B. 4x2 – 10x – 4 C. 24x2 – 24x – 1

0 D. 14x3 + 4x2 + 10x – 4
Mathematics
2 answers:
bekas [8.4K]4 years ago
8 0

Answer:

The answer is B.4x^2-10x-4

Step-by-step explanation:

In order to solve the problem, we have to understand the addition of similar term. In general, any term has a coefficient and a variable. We have to find every different variable in the addition and then we have to add the coefficients of every term which has the same variable.

In this case:

14x^2-14-10x^2-10x+10

There are 3 terms: x^2, x, non-variable

So,

(14x^2-10x^2)+(-10x)+(-14+10)\\4x^2-10x-4

Finally, the answer is letter B.4x^2-10x-4

Gennadij [26K]4 years ago
3 0

Answer:

answer : B

Step-by-step explanation:

(14x2 – 14) + (–10x2 – 10x + 10) = (14x² - 10x²) -10x + (-14+10)

(14x2 – 14) + (–10x2 – 10x + 10) = = 4x² - 10x - 4

answer : B

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3 years ago
A customer at Marty’s Fruit Stand picks a sample of 3 oranges at random from a crate containing 60 oranges, of which 4 are rotte
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Answer:

340 ways

Step-by-step explanation:

Given:

Total number of oranges = 60

Number of rotten oranges = 4

Number of oranges picked = 3

Now, number of good oranges = Total number - Rotten oranges

                                                    = 60 - 4 = 56

Now, we need to pick at least two rotten oranges.

So, the possible outcomes can be as follows:

  1. 2 rotten oranges + 1 good orange = 3 oranges
  2. 3 rotten oranges + 0 good orange = 3 oranges

Now, number of ways of picking 'r' distinct objects from a total of 'n' objects is given  as:

^nCr=\frac{n!}{r!(n-r)!}

Now, picking 2 rotten oranges from a total of 4 rotten oranges is:

^4C_2=\frac{4!}{2!2!}=\frac{4\times 3\times 2}{4}=6

Similarly, picking 3 rotten oranges from a total of 4 rotten oranges is:

^4C_3 =\frac{4!}{3!\times1!}=\frac{4\times 3!}{3!}=4

Now, picking 1 good orange from a total of 56 good oranges is:

^{56}C_1=56

Picking 0 good oranges means picking no good oranges.

Therefore, the total number of ways of picking at least 2 rotten oranges is the sum of the above two possibilities and is given as:

At least 2 rotten out of 3 picked = (2 rotten and 1 good) or 3 rotten

                                                       = 6 × 56 + 4

                                                       = 336 + 4 = 340 ways

Therefore, there are 340 ways of picking at least 2 rotten oranges when 3 oranges are picked from a total of 60 oranges.

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