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I believe we don't need to use a table of values. 

The question is saying, when you plug a number, \(x\), into the two functions, the two results are the same. 

Let's call the number, \(n\). 

When we plug \(n\) into \(f(x)\), we get \(n^2 - 6n + 8.\)

When we plug \(n\) into \(g(x)\), we get \(n - 2\)

The problem states that these two values are equal, so we can write the equation: 

\(n^2-6n+8=n-2\\ n^2-7n+10=0\\ (n-5)(n-2)=0\\ n_1=5\ n_2=2\)

I hope this helped,

Gavin

Thanks Gavin and Melody,

Sorry, didn't know about it. I thought you were in a hurry, and left, so that's why I finished the rest of the solution.

From now on, I'll never do it again, therefore people can learn, and become smarter.

Yeah, I have realized. 

I was so caught up in writing the solution to the problem, 

I just made a careless error. 

Thanks for spotting it. 

smh 

(0.69, 0.87)

The following diagram should help. 

 \(\cos(73°)=c/b \\\sin(\theta)=c/b\\73°+\theta=?\)

It's algeberic expression 

17x5^2 = 17x 5 x 5

             = 17 x 25

             = 425

The answe is 16.99 (17) degrees

Thank you!

Answer for the 1st question is 12 (c)

Answer for the 2nd question is  C

Although I am not a computer expert, it is apparent to me why no one can access your link.

The "link" you provided is a file on your and only your computer. Unless I hacked you, I do not have access to these files. It is a PNG, so you should be able to upload the image. Click the items circled in red.

Try to do the following. I have created visual guidelines for you. 

You now have to find your file. Sift through your files and see if you can find the diagram.

I have answered this question where its link directs. 

If 15% of the apples picked one afternoon are rotten, then I would expect that the number of apples picked multiplied by 0.15 yields the approximate number of rotten apples. 

If I place 30 equally-sized pieces of paper in a hat, then I would expect that \(30*0.15\text{ or }4.5\) pieces of paper read "rotten."--not 3.

If I place 20 equally-sized pieces of paper in a hat, then I would expect that \(20*0.15\text{ or }3\) pieces of paper read "rotten." This is a match, and it is the correct answer.

If I place 15 equally-sized pieces of paper in a hat, then I would expect that \(15*0.15\text{ or }2.25\) pieces of paper read "rotten."--not 3.

If I place 12 equally-sized pieces of paper in a hat, then I would expect that \(12*0.15\text{ or }1.8\) pieces of paper read "rotten."--not 3.

SamJones   behave yourself!

If you put as much effort into your statistics as you do into cheating you might get an A

Hev123, do you and SamJones know each other? You guys are asking the same question.

You provided a link to a question SamJones posted. I answered it, but you are still asking for help. Is there something about my answer you do not understand? 

I already answered this question for SamJones.

That depends on what back to back means.

Why dont your draw it and think about the answer for yourself?

Thanks Gavin,

I know you meant well Tertre but I would like to encourage answerers NOT to give full answers.

If a person gives a part answer, in order to help the asker learn (just as Gavin has done) it is very frustrating when someone comes in and finishes it over the top of them. It diminishes their value as an educator.

We want to help the students learn not just spoon feed them the answers.

If the asker needs more help they should ask for it. 

Interactive learning is greatly encouraged!!

Hmm.

YOUR LINK DOES NOT WORK!!!!

There exist several positive integers x such that 1/(x^2+x) is a terminating decimal.

What is the second smallest such integer?

The terminating decimal is if and only if the denominator has powers of only 2 and/or 5.

\(\begin{array}{|r|r|l|} \hline x & x^2+x & factor \\ \hline 1 & 2 & = 2^1 \checkmark \\ 2 & 6 & = 2 \cdot 3^1 \\ 3 & 12 & = 2^2 \cdot 3^1 \\ 4 & 20 & = 2^2\cdot 5^1 \checkmark \\ \hline \end{array}\)

The second smallest  positive integer x is 4

Can someone help me?????  Do u really need a diagram?  You can copy and paste the diagram

center at 3,-5 and r= 5

from the formula of a circle (x-h)^2 + (y-k)^2 = r^2

with center at (h,k)

yes SSS by factor of 1/5

maybe AA cause it looks like two right triangles,

but usually we can't assume and have to use only what is given,

so to be safe we should not use AA

4

the vertical middle

the horizontal middle

the two diagonals