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Natali [406]
3 years ago
5

the trinomial x2 bx c factors to (x m)(x n). if b is negative and c is positive, what must be true about m and n?

Mathematics
2 answers:
tiny-mole [99]3 years ago
8 0

Answer:

Both m and n should be negative

Step-by-step explanation:

(x - m)(x - n) = x^{2} -(m + n)x + mn

Comparing this with

x^{2} -bx+c, we get,

b = m + n and c = mn

We have been given that c should be positive. So, we have two cases:

1. Both m and n should be positive and

2. Both m and n should be negative

<u>Case 1: Both m and n are positive</u>

If both m and n are positive, then so is mn = c.

But, note that m + n (= b) will also be positive. But, it is given that b should be negative. So, this case is not possible.

<u>Case 2: Both m and n are negative</u>

If both m and n are negative, then mn (= c) is positive.

Also, m + n (= b) is negative.

Hence, this is the correct case.

galina1969 [7]3 years ago
7 0

For c to be positive, and for b to be negative, m must be negative and n must be negative.


X^2 - bx + c = (x - m)(x - n).

c is the product of m and n. If both m and n are positive, c would be positive. However b is the sum of m and n, therefore to make b negative, both m and n must be negative to ensure that the product of m and n is positive


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german

Answer:

14

Step-by-step explanation:

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Vanyuwa [196]

Answer:

a + 20 = 7*12 (multiplied both sides by 12)

a + 20 = 84 (calculated 7*12)

a = 84 - 20 (subtracted 20 on both sides)

a = 64 (calculated 84-20)

4 0
2 years ago
8x - 3(5 + 9x what's the simplified algebraic expression p​
Mashcka [7]

=8x + (-3) (5) + (-3) (9x)

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4 0
3 years ago
In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, one at a time, until
Juliette [100K]

Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
  • n is the size of the sample.
  • k is the total number of desired outcomes.

In this problem:

  • There are 12 bulbs, hence N = 12.
  • 3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

  • Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.
  • The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.

Hence:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182

0.2182 x 1/8 = 0.0273.

0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

8 0
2 years ago
I need help someone please
statuscvo [17]

Answer: \frac{6}{2}

Step-by-step explanation:

1. Use this \frac{f(b)-f(a)}{b-a} to find the average rate of change

2. Average rate of change = \frac{f(8)-f(6)}{8-6}

3. f(8) = - 4 and f(6) = -10 (To find these you will need to look at the points on the graph. So the points (8,-4) and (6,-10))

4. Replace -4 to f(8) and -10 to f(6)

5. So it will look like\frac{-4-(-10)}{8-6}

6. Solve \frac{6}{2}

6 0
2 years ago
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