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f(x)  = 3x^2 - 2x + 4

g(x) = x^2 - kx - 6

f(10)  =  3(10)^2 - 2(10) + 4  =  284

g(10)  =  (10)*2 - 10k - 6  =  100 - 6 - 10k  =  94 - 10k

f(10) - g(10)    = 10    ....so....

284  - [94 - 10k]  =  10

190 + 10k  =  10

10k  = -180

k = -18

x + y + z  = 6       (1)     

1/x + 1/y + 1/z = 2     ⇒  [ yz +  xz + xy] / xyz  = 2  ⇒  [ xy + xz + yz]  = 2xyz      (2)

                x + y          y + z            x + z

Find        ____  +   ______   +  ______    =

                 z               x                   y

xy ( x + y)   + yz( y + z)  +  xz ( x + z)

________________________________  =

                     xyz

xy  ( 6 - z)   + yz ( 6 - x)   +  xz ( 6 -y)

______________________________      =

                   xyz

6xy - xyz  + 6yz - xyz  +  6xz - xyz

_______________________________  =

                    xyz

6 [ xy  + xz +  yz ] -  3 [ xxz ]

_______________________   =

                   xyz

6  [ 2xyz ]   - 3 [xyz]

________________   =

          xyz

xyz [ 6*2  - 3 ]

____________  =

     xyz

6*2  - 3  =

12 - 3  =

9

srry the answer is not 84/5 thanks anyways :)

Good job, Rom!

\(\dfrac 1 x + \dfrac 1 y + \dfrac 1 z = 2\\ \dfrac{x+y+z}{x+y+z}\left( \dfrac 1 x + \dfrac 1 y + \dfrac 1 z\right) = 2\\ \dfrac 1 6 \left(1+\dfrac{y+z}{x} + 1 + \dfrac{x+z}{y}+1+\dfrac{x+y}{z}\right) = 2\\ 3 + \left(\dfrac{y+z}{x} + \dfrac{x+z}{y}+\dfrac{x+y}{z}\right) = 12\\ \left(\dfrac{y+z}{x} + \dfrac{x+z}{y}+\dfrac{x+y}{z}\right) = 9\)

Im not sure why you can't respond to any problems, you probably have to create an account to fix the problem

Here is my hint:

Plug in 10 for x, for the equation f(x).

Then plug in 10 for x, for th equation g(x)

You equation should be something like this:

(3(10^2) - 2(10) + 4) - 10^2 - 10k -6 =10

Solve for k

Using dimensional analysis, we have

34 in  x  [ 1 ft] / [12 in]  x [ 30.5 cm] / [1 ft ]      cancel units and we are left with

34 x  [30.5 /12] cm  =

34 * 30.5  /12   cm   ≈  86.42 cm

You're on the right track  !!!!

10b+5v+5z=55   ⇒  5v + 5z  =  55 - 10b    (1)
3b+5v+5z=34      ⇒  5v + 5z  = 34  - 3b    (2)
5b+6v+3z=42         (3)

Note that   (1)  must = (2)....therefore

55 - 10b  =  34 - 3b   rearrange as

55-34  = 7b

21 = 7b 

b = 3

Knowing this  (1)  becomes 

10(3) + 5v + 5z  = 55

30 + 5v + 5z = 55

5v + 5z  = 25    [divide through by 5 ]

v + z =  5     →    z = 5 - v     (4)

Sub (4)  into (3)  and sub 3 for b   and we have that

5(3) + 6v + 3(5 - v )  = 42

15 + 6v + 15 - 3v  = 42

30 + 3v  = 42

3v = 12

v = 4

So  z = 5 - v  =  5 - 4   = 1

So    { b, v , z}   =  ( 3, 4 , 1}

OK, man  !!!

Thank you for your quick answer!!

We can use the Law of Cosines to solve this   [although there may be other ways, too]

We have that

8^2  =  3^2 + 6^2  - 2 (3)(6) cos  A        rearrange  as

[8^2  - 3^2 -6^2] / [ -2 (3) (6) ]  = cos A     simplify

[ 19] / [ -36]  = cos A

-19/36  = cos A        and we can find the angle A  with the cosine inverse  (arccos)

arccos ( -19/36)  = A  ≈  121.86°

P.S.  - this angle is obtuse  ( > 90°)....and a triangle may only have one angle > 90°...so...we know it's the largest

Thank you so much :)

https://web2.0calc.com/questions/find-constants-nbsp-nbsp-and-nbsp-nbsp-such-thatfor-all-nbsp-nbsp-such-that-nbsp-nbsp-and-nbsp-give-your-answer-as-the-ordered-pair-nb

If  "a"  is the  x coordinate of P, then "a^2"   will be the y coordinate of P

The angle of inclination =     arctan (a^2 / a) =  θ

And  r  = √ [ x^2 + y^2 ] =  √ [a^2 + a^4 ]

The x coordinate will then become  :   r cos  θ =     √ [a^2 + a^4] cos [ arctan (a^2/a) ] 

And the y coordinate  will become :  r sin  θ  =      √ [a^2 + a^4] sin [ arctan (a^2/a) ] 

This image might help  :

Angle POC in the right triangle POC is the angle of elevation = θ......and OP  = the hypotenuse   = r

The x  coordinate is r cos θ     and the y coordinate is  r sin  θ

b)     rx + s

       ______

        2x + t

Using the same info

If the vertical asymptote  = 3....we need to solve this for t :   

 2(3) + t  = 0 →  6 + t  = 0 → t = -6

If the horizontal asymptote = -4, the we need to solve this for r :

r/2  = -4      multiply both sides by 2

r  = -8

And since (1,0)  is on the graph,  then we can solve this for s :

-8 (1)  + s  = 0

-8 + s  = 0

s = 8

So

f(x)   =      -8x + 8

               _______

                 2x - 6

Thanks

(a)     ax + b

       ______

         x + c

If the vetical asymptote  =  3    then we want to solve this :  3 + c  = 0   →   c   = -3

If the horizontal asymptote  = - 4    then   a/1  = -4  →  a   =-4

Lastly  ......if the x intercept is (1, 0)  then  we need to solve this : 

-4(1) + b  = 0  →  -4 + b = 0 →  b  = 4

So  f(x)   =          -4x + 4

                       _________

                            x  - 3

HAHAHA!!!!.......something large....something VERY large!!!!!

Triangle CPQ  is a right triangle with legs CP (= R)  and PQ (= d)  .  CQ is the hypotenuse.

And we can use the tangent to find PQ.....we have that

tan ( θ/2)   =   PQ / R

tan (39/2)  = d / 964      multiply both sides by 964

964 tan (19.5)  = d  ≈  341.4  ft

For number four, I recommend proceeding with casework with n=2, n=3, n=4, etc.

2^(1000!):
1 - Calculate 1000! to its full accuracy or to 2,568 digits. There are "arbitrary precision" calculators on the Internet that can calculate such big numbers. Or, if you know a programming language, it can easily be calculated as well. 

2 - The above calculation comes to :
4.02387260077093773543702433923....... x  10^2567.

3 - Calculate log(2) base 10, to  at least the same accuracy as in (1) above, or even to 3000 decimal places, which would give: 0.30102999566398119521373889472449....etc.

4 - Multiply (2) above x (3) above also to a precision of about 3000 digits. It should start with:
1.2113063515624881214916220007003..... x  10^2567

5 - The fractional part of (4) above should start with:
0.8024233892872257177919623......etc.

6 - Raise the frational part in (5) above to the power of 10, or 10^ 0.8024233892872257177919623.....etc.

7 - The above should give you the first digits of 2^(1000!), which are:

8 - 63448796572785746651.......etc.

9 - Count the first 10 digits from the left:
6,344,879,657.........etc.

10 - And that is the END!!.

What is the largest perfect square less than 12345654320?

Just take the square root of   12345654320  ≈  111110.9

So....the largest perfect square  =  (111110)^2 =   12345432100

Thank you so much, c-phil!

Ironically, just as you answered one of the questions, I figured out one of them too.

Suppose a and b are positive integers and \(\dfrac{a}{4}+\dfrac{b}{3}=\dfrac{13}{12}\). What is the value of a^2+b^2?

a/4   + b/3  = 13/12

[ 3a + 4b]           13

_______   =      ___

    12                  12

This implies that  3a + 4b  =  13

If we want positive integer values.....then a = 3  and b  = 1

So

a^2 + b^2   =

3^2 + 1^2   =

9 +  1  =

10