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f(x)  = g(x)    when  x  =  1, 7, 11

interesting ??

EP: I think you "overthought" this one !! : 

4n = n - 24

3n = - 24

n = - 8

Check:

-8 * 4 = - 8 - 24

- 32 = - 32

If I  understand your question:

x * 4x = x-24

4x^2=x-24

4x^2-x+24=0

x=0.125+2.4463i         x=0.125−2.4463i

Thank you

The 'break even' is when the two functions are equal...so set them equal

20 + 30h = 10 h^2 - 80     re-arrange and solve for h

10h^2-30h -100 = 0        use Quadratic formula to find h = 5   or -2 (throw out)

Remember that the formula for perimeter is 2L+2W. In this case, that would be 2(2x-4)+2(2x+3). If you add those numbers, you get the perimeter as 10x-2. So, the answer is C.

Hint: 

We know that the perimeter of a rectangle is \(2*(length+width)\) -why? (The perimeter is the sum of the outer lines of a figure, I.e. not the inside unlike area, so we add all lines)  Since rectangle has 2 lengths and 2 widths thus adding a length and a width and multiply it by 2 is  a correct formula.

Solution

given length=3x-4

given width=2x+3

Finding perimeter is just applying the rule,

2(3x-4+2x+3)

simplify the bracket

2(5x-1)=10x-2 

So C. is the correct choice.

Thanks, heureka.......that is a more general way to derive this....!!!!

I would just say that their closing speed is 100 + 80 = 180 km/hr

    30 minutes (which is 1/2 hour)  before they meet would be 1/2 of this  = 90 km

a=1;b=11;c=b%10;d=int(b/10);f=d*100+c;e=if(f%b==0, goto6, goto7);printf,b;a++;if(a<1000,goto2,0);a=1;b=b+1;if(b<1000,goto2,0)

OUTPUT:
15          105
18          108
 

please get rid of your meaningless dollar signs basically you copied, in the title:)

\(P(x)\) is a polynomial with real coefficients such that \(P(2+i)=4-3i\). Find \(P(2-i)\).

\(\begin{array}{|rcll|} \hline \mathbf{P(x)} &=& \mathbf{ax+b} \\\\ P(2+i) &=& a(2+i)+b \\ P(2+i) &=& 2a+ai+b \\ P(2+i) &=& 2a+b+ai \quad | \quad P(2+i) = 4-3i \\ 4-3i &=& \underbrace{2a+b}_{=4}+\underbrace{a}_{=-3}i \qquad \text{compare} \\\\ \mathbf{a} &=& -3 \\\\ 2a+b &=& 4 \\ 2(-3)+b &=& 4 \\ -6+b &=& 4 \\ \mathbf{b} &=& 10 \\\\ \mathbf{P(x)} &=& \mathbf{-3x+10} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{P(x)} &=& \mathbf{-3x+10} \\ P(2-i) &=& -3(2-i)+10 \\ P(2-i) &=& -6+3i +10 \\ \mathbf{P(2-i)} &=& \mathbf{4+3i} \\ \hline \end{array}\)

Two trains started moving one from point A and the other from point B against each other at the same time
for a trip length of 240 km.
The first train speed is 80 km/h and the second one is 100 km/h.
What is the distance between the two trains 30 minutes before they meet?

\(\begin{array}{|rcll|} \hline y_B &=& -100t+240 \\ y_A &=& 80t \\ \hline \end{array}\)

meeting time

\(\begin{array}{|rcll|} \hline \mathbf{y_A} &=& \mathbf{y_B} \\\\ 80t_{meet} &=& -100t_{meet}+240 \\ 180t_{meet} &=& 240 \\ t_{meet} &=& \dfrac{240}{180} \\ \mathbf{t_{meet}} &=& \mathbf{ \dfrac{4}{3}\ h } \qquad ( t_{meet}= 80\ \text{minutes} )\\ \hline \end{array}\)

\(t_{meet}- 30\ \text{minutes} = 50\ \text{minutes}\)

\(\begin{array}{|rcll|} \hline y_B &=& -100 \left(\dfrac{50}{60}\right)+240 \\ y_A &=& 80\left(\dfrac{50}{60}\right) \\ \hline y_B - y_A &=& -100 \left(\dfrac{5}{6}\right)+240 - 80\left(\dfrac{5}{6}\right) \\ &=& -180 \left(\dfrac{5}{6}\right)+240 \\ &=& -150+240 \\ &=& \mathbf{90\ \text{km}} \\ \hline \end{array}\)

The distance between the two trains 30 minutes before they meet is \(\mathbf{90\ \text{km}}\)

80km/h*0.5h+100km/h*0.5h=40km+50km=90km     (30min=0.5h)

Don't be so lazy and work it out yourself!

What happened to 'Thanks very much'   That would have been an appropriate response!

How many even natural-number factors does \(n = (2^2)(3^1)(7^2)\) have?

\(\begin{array}{|llcll|} \hline 2^{\color{red}2} * & 3^{\color{blue}1} * 7^{\color{green}2} \\\\ ({\color{red}2}+1 ) *& ({\color{blue}1}+1) * ({\color{green}2}+1) = 3*2*3 &=& 18~ \text{divider} \\ & ({\color{blue}1}+1) * ({\color{green}2}+1) = 2*3 &=& 6~ \text{odd divider} \qquad (\text{odd*odd=odd})\\ \hline &&& 18-6 = \mathbf{12}~ \text{even divider} \\ \hline \end{array} \)

All you need is a bottom row:

Circle diameter   >  D = 1

Radius                >  R = 0.5

Triangle side       >  a = 4D + 2[R / tan(30°)]     a = 5.732    a/2 = 2.866

Triangle height    >  H = sqrt[ a² - (a/2)²]      H = 4.964

Triangle area       >  A = H * a/2              A = 14.227 u²   

Thanks Heureka.

I will have to think about this one before I will be able to claim full understand.

Find all solutions of the equation \(|x^2-20x-1|=20\).

a)
\(x^2-20x-1 = 20\)

\(\begin{array}{|rcll|} \hline x^2-20x-1 &=& 20 \\ x^2-20x-21 &=& 0 \\\\ x &=& \dfrac{20\pm\sqrt{400-4(-21)} }{2} \\ x &=& \dfrac{20\pm\sqrt{484} }{2} \\ x &=& \dfrac{20\pm 22 }{2} \\ \\ x_1 &=& \dfrac{42}{2} \\ \mathbf{x_1} &=& \mathbf{21} \\\\ x_2 &=& \dfrac{-2}{2} \\ \mathbf{x_2} &=& \mathbf{-1} \\ \hline \end{array}\)

b)
\(x^2-20x-1 = -20\)

\(\begin{array}{|rcll|} \hline x^2-20x-1 &=& -20 \\ x^2-20x+19 &=& 0 \\\\ x &=& \dfrac{20\pm\sqrt{400-4(19)} }{2} \\ x &=& \dfrac{20\pm\sqrt{324} }{2} \\ x &=& \dfrac{20\pm 18 }{2} \\ \\ x_3 &=& \dfrac{38}{2} \\ \mathbf{x_3} &=& \mathbf{19} \\\\ x_4 &=& \dfrac{2}{2} \\ \mathbf{x_4} &=& \mathbf{1} \\ \hline \end{array}\)

How do you know that guest?

 The question does not specify either way.

Then again maybe the candies are cut into bits and shared. The question does not specify that condition either.

hint for 1: Keep in mind that you have two variables, each of which is chosen randomly from [0, 1].

any hints for part c?

Let f(x) be a polynomial with integer coefficients such that \(f(5) = f(7) = 20\).
What is the smallest possible value of \(|f(0)|\)?

a) 

\(f(x) = 20 + (5-x)(7-x) \qquad \Big(~ f(x)=x^2-12x+55 ~\Big)\)

\(\begin{array}{|rcll|} \hline f(0) &=& 20 + (5-0)(7-0) \\ f(0) &=& 20 + 35 \\ f(0) &=& 55 \\ \mathbf{|f(0)|} &=& \mathbf{55} \\ \hline \end{array}\)

b)
\(f(x) = 20 - (5-x)(7-x) \qquad \Big(~ f(x)=-x^2+12x-15 ~\Big)\)

\(\begin{array}{|rcll|} \hline f(0) &=& 20 - (5-0)(7-0) \\ f(0) &=& 20 - 35 \\ f(0) &=& -15 \\ \mathbf{|f(0)|} &=& \mathbf{15} \\ \hline \end{array}\)

The smallest possible value of \(|f(0)|\) is 15