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We're adding 18 + 25 + 16 =   59

The radius =  sqrt (59)

Rearrange as

4x^2 +x^2  -12x + 2x  + k

5x^2  - 10x +  k 

The x coordinate of the  vertex of this  parabola   is     -(-10) / (2 *5)  =  10 / 10  = 1

So...we want

5(1)^2  - 10 (1)  +  k  =  - 1

5 -10 + k  =  -1

-5 + k = -1

k =  -1 + 5   =   4

WAIT HOW IS THAT POSSIBLE. I  got 56 because I got 18+ 25 + 13

x^2 -10x + y^2 + 8y =  18       complete the square on x, y

x^2 -10x + 25  + y^2 + 8y + 16    = 18 + 25 + 16

(x - 5)^2   + (y + 4)^2  = 59

The radius  =   sqrt (59)  units

The son is 1-2/3 years old.  

9 times 1-2/3 years is 15 years, i.e., the father's age now.  

But the son is over a year old now, which means

the father was at most 14 when his son was born. 

Note that we can write

a(10)^7  + b(10)^6 + c (10)^5  + d (10)^4  + a(10)^3 + b(10^2) + c(10) + d = 

a (10^7 + 10^3)  + b(10^6 + 10^2)  + c(10^5 + 10)  +  d(10^4 + 1)  =

a ( 10^3) (10^4 + 1)  + b (10^2) (10^4 +1)  + c(10)(10^4 + 1) + d (10^4 + 1)  =

(10^4 + 1)  [ a (10^3 )  + b (10^2) + c (10)  +  d  ]

The   gcd =  ( 10^4 + 1 )  = 10001

-t^2  + 8t - 4 - 3t^2       simplify

-4t^2  +  8t  -  4      (1)

The  t that maximizes this  =  -8 / (2 * -4)  =    1

Putting this back into  (1)  the max value is

-4(1)^2  + 8(1)  - 4  =   0

Let your the current age of your son =  N

Then your current age =  9N

And in one year we have this relationship

9N + 1  =   6 ( N + 1)        simplify

9N + 1  = 6N + 6

9N - 6N  =  6 - 1

3N  = 5

N =  5/3    =   1 + 2/3  yrs old  =   1yr  and  8 months old =  your son's age now

Cars and trucks are   ( 1 + 6) / (1 + 6 + 4)  =  7/11 of the total number, N, of  vevicles

So...

(7/11) N  =  42

N =   42 ( 11/7)  =  66   vehicles in total

So....there are   66 - 42  =  24 motorcycles

The answer is 11111111.

t^2 - 9t - 36  + t^2 - t + 24       simplify

2t^2 -10t -12

We have the form    at^2  + bt  + c

The  value of  t  that produces the smallest value  is  - (b)/ (2a)  =        - (-10) / (2 *2)  =  10  /4  =   2.5

x^2 - 15 < 2x + 7x - 40      rearrange as

x^2 - 9x + 25  <    0      (1)

Notice the graph here : https://www.desmos.com/calculator/5gdvkizm7m

The graph is always   above the x axis  (always > 0 )  so no x values make (1)  true.

according to https://www.desmos.com/calculator it looks like at T = 5  

LMAO, As an Asian I agree!!!

But We do have friends other than asian people lol

Note that the line will leave the bottom side intact, and split the two vertical sides into parts that sum to 2a. 

This means that the perimeter of the shape is 4a + the length of the line. 

The length of the line is the hypotenuse of a right triangle with legs of 2a and \({4 \over 3}a\)

Using the Pythagorean theorem, we find that the length of the hypotenuse is \(\sqrt{2^2 + {2 \over 3}^2} = \sqrt{40 \over 9} = {2 \sqrt {10} \over 3}a\). 

This means the perimeter of the shape is \(4a + {2\sqrt{10} \over 3}a\), which can be factored into \(a(4 + {2\sqrt{10} \over 3})\). 

Can you take it from here?

By the quadratic formula, the roots are: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\).

Substituting what we know, we have: \(x = {-7 \pm \sqrt{49-16c} \over 8}\)

This means that the discriminate must equal -79, giving us the equation: \(49 - 16c = -79\)

Subtracting 49 from both sides, we have: \(-16c = -128\), meaning \(k = \color{brown}\boxed{8}\)

Simplify the equation to: \(3(x+4)(x+4) = 2x(6x + 4) \div 2x\)

Note the 2x's on the right-hand side cancel out, so we have: \(6x+4 = 3(x+4)(x+4)\)

Now, expand the right-hand side: \(6x+4 = 3x^2 + 24x + 48\)

Convert into a quadratic: \(0 = 3x^2 + 18x + 44\)

Now, use the quadratic formula, and \(x = \color{brown}\boxed{{-3 \pm \sqrt {51} \over 3}}\)

Ignoring the denominators (since they are the same)

1  = -a^3

-1  = a^3       take the cube root of both sides

a =  -1

5N/  ( N + 7)   =  -8

5N  =  -8 (N + 7)

5N  = -8N  - 56

13N  =  -56

N =  -56 / 13

(3x + 12) (x + 4)  =  (4x^2 + 8x  + 8x^2) / (2x)               assuming that  x   is not 0

3x^2  +24x + 48  =  (4x) ( x + 2 + 2x) / ( 2x)

3x^2 + 24x + 48  = 2 ( 3x + 2)

3x^2 + 24x + 48  = 6x  + 4

3x^2 + 18x + 44  = 0

Using the Q formula

              -18  ± √ (18^2  - 4 * 3 * 44)       -18  ± √ ( -204)         -18  ± 2√51 i

x =         ______________________   = ____________  =  ____________ =   -3  ± √51 i / 3

                         2 * 3                                        6                             6

Set \(x = 0\) and simplify the equation: \(0 = -y^2 + 2y + 3\)

We can solve for the number of solutions with the discriminant(\(b^2 - 4ac\))

Note that the discriminant is \(2^2 - 4 \times 3 \times -1 = 16\). Because the discriminant is positive, it will have exactly \(\color{brown}\boxed{2}\) y-intercepts. 

We have the equation: \({1 \over 6}x + 2 = {1 \over 3}x\)

Subtracting \({1 \over 6}x\) from both sides, we get: \(2 = {1\over 6}x \)

Now, we have to multiply the entire equation by 6 to get our answer.

Can you take it from here?

Hi asinus, by the way, Wolframalpha approximated a,b,c a lot.

Because when using mathway.com, it gives the following values: 

\(a=-4.31946756\)

\(b=0.5035452\)

\(c=7.81592235\)

So:

\(\sqrt[3]{(x+4.31946756)}+\sqrt[3]{(x-0.5035452)}+\sqrt[3]{(x-7.81592235)}=0\)

And, then graphing this equation as Melody did (But using the more accurate values of a,b,c):

Giving the approximation: \(0.519\)   (And the exact solution is: \(\frac{523}{1008}=0.518849206..\)).

Hope this helps!

Intersecting Chord Theorem :

AQ  * BQ = CQ * DQ

6  * 15  = CQ * DQ

Let CQ = x    and DQ = 38- x....so we have

6 * 15 = x  (38 - x)

90 =  38x - x^2            rearrange as

x^2 - 38x  + 90  =  0     complete the square on x

x^2  - 38x    =   -90

x^2 - 38x + 361  =  -90 + 361

(x - 19)^2   = 271          take the negative root

x - 19 = -sqrt (271)

x =  -sqrt (271) + 19 ≈  2.54 =  minimum length of CQ

Hello Guest!

If a,b,c are the solutions of the polynomial equation:    \(x^3-4x^2-32x+17=(x-a)(x-b)(x-c)=0\) 

Then, by Vietas' formulae:                   \(a+b+c=4\)

                                                           \(ab+ac+bc=-32\) 

                                                                  \(abc=17\)

Using the following identity:  

\(y_1^3+y_2^3+y_3^3-3y_1y_2y_3=(y_1+y_2+y_3)(y_1^2+y_2^2+y_3^2-y_1y_2-y_1y_3-y_2y_3)\)  (*)

(More commonly written as: \(x^3+y^3+z^3=3xyz\) if and only if \(x+y+z=0\)).

Then:

Let  \(y_1=\sqrt[3]{x-a}\) , \(y_2=\sqrt[3]{x-b}\) , \(y_3=\sqrt[3]{x-c}\)

We are given: \(y_1+y_2+y_3=0\)

Hence, (*) becomes: 

          \((x-a)+(x-b)+(x-c)=3\sqrt[3]{(x-a)(x-b)(x-c)}\)

   (Notice: \((x-a)(x-b)(x-c)=x^3-4x^2-32x+17\))

Thus,

            \( 3x-(a+b+c) = 3\sqrt[3]{x^3-4x^2-32x+17}\)   

            (By Vietas formula: a+b+c=4, substitute and cube both sides)

\(\iff (3x-4)^3=27(x^3-4x^2-32x+17)\)

Expanding:

\(\iff 27x^3-108x^2+144x-64=27x^3-108x^2-864x+459\)

Simplify to get a linear equation:

\(144x-64=-864x+459 \implies 1008x=523 \implies x=\frac{523}{1008}\)

Therefore,  \(x=\dfrac{523}{1008}\)