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(x-3)^2 - (x-8)^2      cannot be negative     expand .....to find where the fxn IS zero

x^2 - 6x + 9     -   x^2+16x-64 = 0

10x - 55 =0

10x = 55

x = 5.5    

Taking the largest degree of each side, 6 - 2 = 4. 

=^._.^=

Ohh right! Just tested it with some functions, yes it doesn't always work.

But I noticed it works if and only if the functions are tangents to each other at that point ? 

I don't get this question.  It does not make sense to me. 

If

\(\displaystyle \sum_{k = 1}^n a_k \left( \frac{k}{n} \right)^2 = 1.\\ then\\ \displaystyle \sum_{k = 1}^1 a_1 \left( \frac{1}{1} \right)^2 = 1.\\ so\\a_1=1 \)

Since the sum of any number of terms must equal 1    

\(a_k=0\quad \text{Where k is an integer greater than 1}\)

Obviously, I am not understanding something.

See https://web2.0calc.com/questions/if-f-is-a-polynomial-of-degree-3-such-that-f-0-f_1

No!  They might simply cross over at that point.

Attn: Guest aske

What have you done towards answering?

Have you done any of the expansions?

So why don't you show us how easy it is and aprove one of them.

LCM(10,8) = 40
so 4 dogs, 5 8-buns
LCM(10,12) = 60
so 6 dogs, 5 12-buns

Assuming that Bob have x packets at first.

Morning sale: \( {1 \over 3}{}{}\)x + 2

After morning sale: x - (\( {1 \over 3}{}{}\)x + 2) = \({2 \over 3}{}{}\)x - 2

Afternoon sale: \( {1 \over 4}{}{}\)(\( {2 \over 3}{}{}\)x - 2) + 4 = \( {1 \over 6}{}{}\)x + \( {7 \over 2}{}{}\)
After afternoon sale: \( {2 \over 3}{}{}\)x - 2 - (\({1 \over 6}{}{}\)x + \( {7 \over 2}{}{}\)) = \( {x \over 2}{}{}\)-\( {11 \over 2}{}{}\)
\( {x \over 2}{}{}\)-\( {11 \over 2}{}{}\)= 8                            8 packets left

    x = 27

Use the conditions given to establish the equivalence 

relationship.

So the answer is 27 packets.

BMC + A = 201

BMC + CMA = 180

BMC + C also = 201

3BMC + CMA + C +A = 582, and we know CMA + C + A = 180, so 3BMC = 582-180 = 402, and thus BMC = 134

B = MCB, and finish the last step to get B

16 + 11 - 7 = ___________

The smallest n that works is 2019.

One ordered pair (a,b) satisfies the two equations ab^4 = 384 and ab^5 = 4608. What is the value of a in this ordered pair?     

 ab⁵       4608  

––––  =  ––––    

 ab⁴        384   

   b  =  12  

Plug the value of b into one of the equations and solve for a.  

^.

find the value of k for which the system of equations has a unique solution 2x+ky-9=0 , kx+3y-13=0. ​

Hello Guest!

2x+ky-9=0 , kx+3y-13=0  ​

\(y=(-2x+9)/k\\ y=(-kx+13)/3\)

The system of equations has unique solutions for many values of k.

 \(k\in \{\) \(-1,\)   \(4,\)    \(5,\)  \(etc.\ \}\)

 \(x\in \{\) \(\ 8,\)    \(2.5,\)  \(2\}\)

 \(y\in \{\) \(\ 7,\)     \(1,\)     \(1\}\)

  !

In trapezoid PQRS, Base PQ is parallel to base RS. Let point X  be the intersection of diagonals PR and QS. The area of triangle PQX is 20 and the area of triangle RSX is 48.  Find the area of trapezoid PQRS.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

[PQRS] = (√20 + √48)²

The radius of the larger circle is 4*sqrt(2).

There are 7024 ways the committee could be chosen.

The mathematician solves 96 problems.

There are 24 possible sequences.

Hey there, Guest!

So, \(3x² + 12x - K\).

( constant )/( coefficient of \(x²\))
 

\(P × \frac{1}{P} = \frac{-k}{3}\)

1 = \(\frac{-k}{3}\)

K = -3

Hope this helped! :)
( ﾟдﾟ)つ Bye

Hey there, Guest!

120 divided by 15 = 8

8 times 5 = 48

\(\frac{48}{120}\) equals 40%

40% of the cars were previously damaged in an accident.

(120 ÷ 15) ÷ 6 = 40%.

Hope this helped! :)
( ﾟдﾟ)つ Bye

x = -y^2 - 3y - 5    is a 'sideways'  parabola...for a line vertical line, x= k ,  to intersect at only ONE point it must intersect at the vertex....find the 'x' value of the vertex:

x = -y^2 - 3y - 5

x = -1(y^2 + 3) - 5

x = - (y+1.5)^2 + 2.25 -5   

x = -(y+1.5)^2 - 2.75                        (h,k) = ( -2.75, -1.5)                      k = - 2.75

The doamain is all 'x' except values that cause denominator of any of the terms to = 0

x cannot equal -8     is all

(- inf, -8) U (-8, + inf)

Two integers are relatively prime if they share no common positive factors (divisors) except 1

for 7    1 2 3 4 5 6 8 9 10 11 12 13     15  16 17 18 19 20   22 23 24 25 26 27     29  30 31 32 33 34    36 37 38 39

USing the above list 

for 10: 1    3             9      11      13                17      19             23                27      29      31      33              37       39     count them up !