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Sorry Alan,

I was working on it at the same time as you.

Patrick bought some dictionaries and files for $1 474.

The number of dictionaries he bought was 2/3 the number of files he bought.

The cost of a dictionary was $12 more than the cost of a file.

He spent $22 more on dictionaries than on files.

How much did a dictionary cost?     [Find n]

d dictionaries are bought for n  dollars each    So they will cost a total of   nd dollars

f files are bought for  m dollars   each                so they will cost a total of    mf   dollars

dn + fm = 1474

d= (2/3)f      and   n=m+12

substituting gives us

\(\frac{2}{3}*f (m+12)+fm=1474\\ \frac{2f}{3}*(m+12)+\frac{3fm}{3}=1474\\ 5fm+24f=1474*3\\ f(5m+24)=1474*3\qquad (1)\)

-----------------

\(dn=fm+22\\ \frac{2f}{3}(m+12)=fm+22\\ 2fm+24f=3fm+66\\ 24f-fm=66\\ f(24-m)=66\qquad (2)\)

---------------------------

(1) divided by (2)

\(\frac{f(5m+24)}{f(24-m)}=\frac{3*1474}{66}\\ \frac{(5m+24)}{(24-m)}=67\\ 5m+24=67*24-67m\\ 72m=66*24\\ m=22\\~\\ n=22+12=34\)

the dictionaries cost   $34 each

See https://web2.0calc.com/questions/help-pls_85792#r1

See the following:

You now hve four independent equations in four unknowns.  Can you take it from here?

What is the domain of the real-valued function
q(x) = x/(1 - x^2)?

Hello Guest!

\(x\in \mathbb R\ |\ (-infinite

LaTeX is going crazy, sorry.

\(x\in \mathbb R\ |\) (- infinite < x < - 1) and (-1 < x < 1) and (1 < x < infinite)

  !
 

Hi, I'm pretty sure you got the answer incorrect, but it's still cool that you answered. 

a = k/b 

ab = k

k = -108

a = -108/b

a = -108/12 = 9

Thanks for answering this question, but when I tried this answer, it turned out to be wrong. Thanks anyway!

By rotational symmetry, the number of ways is 6!/6 = 120.

This was answered before.  The answer is 4/7.

Answer is (A), (D).

Since the degree of p(x) is 11, adding any polynomial of lesser degree will result in a polynomial of degree 11.

(The degree of a plynomial is the degree of its hightest degree term.)

(4x + 1) / (x² - 4x - 12)     =     (4x + 1) / [ (x + 2)(x - 6) ]

(4x + 1) / [ (x + 2)(x - 6) ]  =  A(x + 2) / [ (x + 2)(x - 6) ]   +  B(x - 6) / [ (x + 2)(x - 6) ] 

--->    (4x + 1)  =  A(x + 2) + B(x - 6)

--->    4x + 1  =  Ax + 2A + Bx - 6B

--->     Ax + Bx  =  4x     and    2A - 6B  =  1

--->         A + B  =  4

Combining the above two equations:  2A - 6B  =  1     and     A + B  =  4

     A +  B  =  4     --->     x6     --->     6A + 6B  =  24

   2A - 6B  =  1                        --->     2A -  6B  =   1

                                                         --------------------

--->      (adding)                                         8A  =  25     --->     A  =  25/8

Since  A + B  =  4     --->    B  =  4 - B     --->     B  =  4 - 25/8     --->     B  =  7/8

216, 258, 296 are each divisible by 8; therefore, the possitble values of A are 1, 5, and 9.

Their sum is ???

Tank B has  x  liters of water.

Tank A has 4 times as much, so Tank A has  4x  liters of water.

After 30 liters of water were transferred from Tank A to Tank B:

Tank B has  x + 30  liters of water.

Tank A has  4x - 30  liters of water.

Now, Tank B has 24 more liters than Tank A.

--->     Tank B  =  24 + Tank A

--->     x + 30  =  24 + (4x - 30)

--->     x + 30  =  4x - 6

--->     x + 36  =  4x

--->           36  =  3x

--->           12  =  x

Starting amounts:

Tank B  =  12 liters

Tank A  =  4 · 12  =  48  liters

Total amount:  12 liters + 48 liters  =  60 liters.

 How many strawberries were there at first?

Hello Guest!

\(\dfrac{x}{3}+\dfrac{x}{3}+\dfrac{x}{3}=x\\ 3\cdot (\dfrac{x}{3}-4)=\dfrac{x}{3}\\ x-12=\dfrac{x}{3}\\ \dfrac{2x}{3}=12\)

\(\large x=18\)

\({\color{blue}18\ strawberries}\ were\ there\ at\ first.\)

  !

Sharif and his sister have the same amount of money:  Each has  x.

After Sharif spent 75, Sharif now has  x - 75

After his sister spent 37, his sister now has  x - 37

His sister has three times as much money as Sharif:  x - 37  =  3(x - 75)

                                                                                     x - 37  =  3x - 225

                                                                           x - 37 + 225  =  3x - 225 + 225

                                                                                  x + 188  =  3x

                                                                            x + 188 - x  =  3x - x

                                                                                        188  =  2x

                                                                                   188 / 2  =  2x / 2

                                                                                          94  =  x

Sharif started with 94.

Reposted here.

Here is the original

I have just realized that this is a repost.

Next time:

Please give the address of the original thread and ask someone (me if you want) to add to or comment on the original question.

Request them to also leave a note here to say they have done so.  

Thanks.

Since there is a sum of polynomials and the sum is of degree 8, at least one of the polynomials f(x) and g(x) must be of degree 8.

Since the degree of f(x) is 2, the degree of g(x) must be 8.

The coefficients, 3 and 4, have no influence upon this problem.

Guest just made a mistake calculating n.   The rest is right

This is an arithmetic series whose first term is 1, whose last term is 97, and whose common difference is 3.

Find the number of terms and use the formula for the sum of a finite arithmetic series to get your answer.

This is an arithmetic progression.

The first number,  sometimes called 'a' or  T_1 is  1

d is the common difference,    4-1=7-4=....97-94 =  3

L is the last term = 97

You will need to find n (the number of terms) first

Use the formula 

\(T_n=a+(n-1)d\\ 97=1+(n-1)*3\)

you can finish solving that.

Then use the following formula to find the sum of all those n terms.

\(S_n=\frac{n}{2}[a+L]\)

There are a lot of formulas in the AP and GP topic.  You have to memorize them all.

#2:  The positive terms are:  

             1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ... + 1/93 + 1/94 + 1/95

        The negative terms are:

              - 1/6 - 1/7 - 1/8 - ... -1/93 - 1/94 - 1/95 - 1/96 - 1/97 - 1/98 - 1/99 - 1/100

        Cancelling the equivalent positive and negative terms, we are left with:

               1 + 1/2 + 1/3 + 1/4 + 1/5 - 1/96 - 1/97 - 1/98 - 1/99 - 1/100

         Now, it's calculator time!

#3:  If you write out the factors, you have, for one of the factors,  (-12 + 12)  which is 0.

       Therefore, the product is 0.