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[363  -  15] / 6  +  1 == 59 terms

If f(x)=(x+4)/(x^2+ax+b), and f(x) has two vertical asymptotes at x = 1 and x = -2, find the sum of a and b.  

You get an asymptote when the denominator is zero. 

The denominator is x² + ax + b     This is a quadratic equation, and a  

                                                      quadratic equation has two solutions. 

The problem itself tells us where the asymptotes are, thus it tells us what the solutions are.  

Since the asymptotes occur at  x = 1 and x = –2  then the factors of that  

quadratic equation are            (x – 1) and (x + 2)  which multiplies out to  x² + 1x – 2  

So a + b  which is  [ (+1) + (–2) ]  equals  –1  

_.

It's 69...

Hi Guest!

\(2x+y=-7 \hspace{0.65cm} (1) \\ x=4+2y \hspace{1.00cm} (2)\\ \)

Multiply equation (2) by -2:

Thus, our system of equations becomes:

\(2x+y=-7 \\ -2x=-8-4y \\\)

Now, add the above two equations:

\((2x-2x)+y=-7+(-8-4y) \\ \iff y=-15-4y \\ \iff 5y=-15 \\ y=-3\)

So, using either (2) or (1), we get (by (2)): \(x=4+2(-3)=4-6=-2\)

Therefore, \((-2,-3)\) is the solution to the system of the equations.
I hope this helps!

The force is inversely proportional to the distance squared, so:

A)  F2/F1 = 1/(2/5)^2 = (5/2)^2 = 25/4 = 6.25

B) F2/F1 = 1/7^2 = ...

tysm!!! the answer is -18 for anyone wondering :)

two numbers, x and y, each between 0 and 1, are multiplied. if the tenths digit of x is 1 and the tenghts digit of y is 2, what is the greatest possible value of the hundredths digit of the product?  

Not sure I understand the question, but I'll try. 

x = 0.1 and then some more 

y = 0.2 and then some more 

Let's try some numbers and see if we can see a pattern. 

0.11 • 0.21 = 0.0231  hundredths is 2 

0.12 • 0.22 = 0.0264  hundredths is 2 

0.13 • 0.23 = 0.0299  hundredths is 2 

0.14 • 0.24 = 0.0336  hundredths is 3 

Let's skip ahead and see what happens. 

0.18 • 0.28 = 0.0502  hundredths is 5  

0.19 • 0.29 = 0.0551  hundredths is 5 

0.1999999999999 • 0.2999999999999 = 0.5999999999995

I don't think it's going to break over to 6 unless you round it to fewer significant digits. 

So I'm going out on a limb here and say the largest number that can be in the hundredths place is 5. 

_.

There is something wrong with your numbers!

" The original ratio of cats to dogs before I change anything is 2:3"  I think this original ratio should read 3:2

Check your number carefully.

Let the original number==N

5N / [N + 7] ==2, solve for N

Cross multiply

5N ==2N + 14

5N - 2N ==14

3N ==14

N ==14 / 3 ==4 + 2/3 - the original number.

one number is chosen from the first three prime numbers, and a second number is chosen from the first three positive composite numbers. what is the probability that their sum is greater than or equal to 9? express your answer as a common fraction  

The first three prime numbers are 2, 3, & 5 

The first three composite numbers are 4, 6, & 8  

Make a grid – the numbers inside are the sums     

            4     6      8  

     2     6     8     10 

     3     7     9     11 

     5     9    11    13 

As you can see, there are 9 totals and 6 of them are greater than or equal to 9. 

Therefore the probability is 6/9 which reduces to 2/3  

_.

The probability of landing on a 6 == 1/6

Then, the probability of landing on any number less than 6 is:

1  -  1/6 ==5 / 6

a)  P = A (1 + i)^(-n)

P ==1000(1.10)^-10

P ==1000  x  0.3855432

P ==$385.54 - initial investment.

b)  i = (A/P)^(1/n) - 1

i ==(1200 / 1000)^(1/2) - 1

i ==Sqrt(1.2) - 1

i ==1.095445 - 1  x 100 ==9.5445% - interest rate.

c) Use "Rule of 72".  $100 at 10% will double to $200 in: 72 / 10 [interest rate] ==~7.2 years.

d)  5225 ==2100 x [1 + 0.20]^N.  Note: We don't have to "estimate" but will solve it exactly!!!

5225 / 2100 ==1.20^N

2.488095238 ==1.20^N

Take the log of both sides

N ==log(2.488095238) / log(1.20)

N ==5 years for 2100 to grow to 5225 at 20% comp. annually.

A ==32  +  6t

B ==20  + 9t, solve for t

A == B after t years

32  +  6t ==20  +  9t, solve for t

32  -  20 ==9t  -  6t

12 == 3t

t ==12 / 3 ==4 years - when the 2 trees will be the same height.

Thanks Ginger,

When i was teaching in a classroom:

 I think the biggest problem with teaching mathematics to older children,  especially older children who have had very limited success in the past, is that they have too many holes in their knowledge and understanding. 

An analogy: You cannot be taught to spell if you do not know the names of the letters. 

On top of their lack of knowledge they lack confidence and interest in trying to learn.  

Put 30 of these children in a classroom and a syllabus that is unsuitable, and chaos reults.

Now days:

I suspect the new problem to teaching maths, perhaps more relevant to the more mathematically capable students, is the climate of  instant gratification.

Sometimes, when I was a kid, I would chew on a problem for days.  These days people, almost immediately consult the internet.  This is not always bad but it has to cut down on an individuals ability to nut a problem out by themselves.

For the benefit of others:   

This is why sites like AoPS   (Art of problem solving)   dislike sites like this one so much.  

How can their students master problem solving if they just get instant internet answers?

P( no zero, no 1) = (10/12)^4

P(1 zero, 1 one, 2 others  different ones)  =   1/12 *  1/12  *  (10/12) *  (9/12) * 4!  

                                                                  =    (10*9*24) / (12*12*12*12)  =  180/  (12^3) = 15 / 144 

P(1 zero, 1 one, 2 others  that are the same as each other)  =   1/12 *  1/12  *  (10/12)  *1  *   4!/2!

                                                                                                =       10/(12^3) * 12

                                                                                                =      10/144

P(2zeros and 2 ones) =  (1/12)* (1/12) *   4!/(2!2!)  =   ( 1/144)  *6 =   6/144

Total  =   (15+10+6) / 144    =     31/144

You need to decide whether on not to accept this answer. 

ok, I'm not a mind reader.  

Actually, you are ...in a metaphorical sense. You’ve demonstrated such skills on more than one occasion. I think most teachers need above average skills in mind reading to be a good teacher. Beyond basic mind-reading is the quintessential skill of filling in the thoughts for thought processes that are missing essential information in the student’s mind.

For teaching younger students, this is a common and relatively easy requirement because younger students are usually just missing information. The difficulty is in persuading the student to use his/her mind to its full capacity.

For the older student, however, the requirements are more arduous because wrong and errant information may be firmly entrenched. The student needs to be willing to excise errant thought process to make room for the correct processes.     

There is a valid argument that to read a mind, there needs to be a mind to read. Mindlessness is a real phenomenon. It’s exceptionally difficult to comprehend and hence exceptionally difficult to treat. 

GA

--. .- 

The following graphs should help:

By inspection:

\(\frac{1^3+6}{1}=7\)

\(\frac{2^3+6}{2}=7\)

Unfortunately  \(\frac{3^3+6}{3}\ne7\) 

However \(\frac{(-3)^3+6}{-3}=7\)

I'll leave you to finish.

ok, I'm not a mind reader.   

What is the difference in elevation between the bottom of the canyon and the​ bird's nest?

Hello Guest!

\(\Delta H=(56+\frac{3}{5}+584+\frac{3}{5})ft=\color{blue}641\frac{1}{5}ft\)

  !