How to become good at math

Mathematics

2 answers:

Colt1911 [192]4 years ago

8 0

Answer:

Step-by-step explanation

Do some practice problems or test online.

iren [92.7K]4 years ago

4 0

Answer:

Step-by-step explanation:

This may seem quite obvious but study! Math is said to be one of the hardest subjects, in middle, high school, and college. Thus, it is very important you are good at it; considering it will be essential in many cases in life. I'd suggest:

Doing your homework and assignments
Try some practice problems so you could get the gist of what is being asked of in the question
Try some online tests, quizzes, or lessons

Here's a tip:

Ask for help if you need it! It is important that you really understand everything, every little detail matters. Good luck!

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The points plotted below are on the graph of a polynomial. Some of the roots to this polynomial are integers. Which of the follo

Shkiper50 [21]

Answer:

I think its B but I have very little evidence. Don't even put B but I was just saying I think it is.

Step-by-step explanation:

3 0

3 years ago

7) PG & E have 12 linemen working Tuesdays in Placer County. They work in groups of 8. How many

BabaBlast [244]

Part A

Since order matters, we use the nPr permutation formula

We use n = 12 and r = 8

$_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\$

There are a little under 20 million different permutations.

<h3>Answer: 19,958,400</h3>

Side note: your teacher may not want you to type in the commas

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Part B

In this case, order doesn't matter. We could use the nCr combination formula like so.

$_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\$

We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.

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Another way we can compute the answer is to use the result from part A.

Recall that:

nCr = (nPr)/(r!)

If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320

So,

$_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\$