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3 years ago

(2x-1) (4x+1) solve equation

2 answers:

8 0

(2x-1)(4x+1) just multiply the terms in the parentesis

Do the following:

2x*4x
2x*1
-1*4x
-1*1

You get:

8x^2+2x-4x-1

Final result:

8x^2-2x-1

Ymorist [56]3 years ago

4 0

(6x-1) is the answer to (2x-1) (4x+1)

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The sum of two numbers is equal to 495. The last digit of one of them is zero. If you cross the zero off the first number you wi

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You will get 4095 hope this is helpful

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3 years ago

10 points and brainliest!

Feliz [49]

Answer:

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Step-by-step explanation:

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3 years ago

Suppose that 60 batteries are shipped to an auto parts store, and that 5 of those are defective. A fleet manager then buys 6 of

PIT_PIT [208]

Answer:

In 8910 ways can at least 4 defective batteries be included in the purchase

Step-by-step explanation:

The order in which the batteries are there is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

In this question:

4 defective:

4 defective from a set of 5 and 2 non-defective from a set of 60 - 5 = 55. So

$C_{5,4}C_{60,2} = \frac{5!}{1!4!} \times \frac{60!}{2!58!} = 8850$

5 defective:

5 defective from a set of 5 and 1 non-defective from a set of 60 - 5 = 55. So

$C_{5,5}C_{60,5} = \frac{5!}{0!5!} \times \frac{60!}{1!59!} = 60$

In how many ways can at least 4 defective batteries be included in the purchase?

$T = 8850 + 60 = 8910$

In 8910 ways can at least 4 defective batteries be included in the purchase

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3 years ago

For a regular 30-sided polygon, what is the degree of rotation?

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4 years ago

Read 2 more answers