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Is the expression on the right supposed to be (ab)²  or  a • b²   ?  

\(\frac{Bx-11}{x^2-7x+10}\ =\ \frac{A}{x-2}+\frac{8}{x-5}\\~\\ \frac{Bx-11}{(x-2)(x-5)}\ =\ \frac{A}{x-2}+\frac{8}{x-5}\\~\\ \frac{Bx-11}{(x-2)(x-5)}\cdot(x-2)(x-5)\ =\ \Big(\frac{A}{x-2}+\frac{8}{x-5}\Big)\cdot(x-2)(x-5)\\~\\ Bx-11=\frac{A(x-2)(x-5)}{x-2}+\frac{8(x-2)(x-5)}{x-5}\\~\\ Bx-11\ =\ A(x-5)\ +\ 8(x-2)\)

Assuming the previous equation is true for all values of  x ,  it must be true when  x = 0 , so we can say:

\(B(0)-11\ =\ A(0-5)+8(0-2)\\~\\ -11\ =\ A(-5)-16\\~\\ 5\ =\ A(-5)\\~\\ A\ =\ -1\)

And it must also be true when  x = 5  so we can say:

\(B(5)-11\ =\ A(5-5) + 8(5 - 2)\\~\\ 5B-11\ =\ 24\\~\\ 5B\ =\ 35\\~\\ B\ =\ 7\)

So   A + B   =   -1 + 7   =   6

Find the distance between  \(P_1 \)(3,4) and the line 4x+3y+7=2x-y+5.

Hello Guest!

\(4x+3y+7=2x-y+5\\ 4y=-2x-2\\ \color{blue}y=-0.5x+0.5\)

\(m_1=-0.5\\ m_2=2\\ y=m_2(x-x_1)+y_1\\ y=2(x-3)+4\\ \color{blue}y=2x-2\)

\(-0.5x+0.5=2x-2\\ 2.5x=2.5\)

\(x_2=1\\ y_2=0\)

\(d=\sqrt{4^2+2^2}=\sqrt{4\cdot 5}\)

\(d=2\cdot \sqrt{5}\)

\({\color{blue}The\ distance}\ between\ P_1 (3,4)\ and\ the\ line\ 4x+3y+7=2x-y+5\ \color{blue}is\ 2\cdot \sqrt{5}.\)

  !

WX cannot be larger than XY!!! 

Solve for n:
2/(n + 1) = 1/500

Take the reciprocal of both sides:
(n + 1)/2 = 500

Multiply both sides by 2:
n + 1 = 1000

Subtract 1 from both sides:
 
 n = 999   and  (n+1) ==1,000

Of the numbers:  1, 2, 3, 4, 5, 6,  the two square numbers are  1  and  4

The probability that the first die shows a square number is  2 / 6.

The probability that the second die shows a square number is also  2 / 6.

The probability that both show a square number is (2/6) x (2/ 6)  =  

The general polynomial function of degree 2 can be written as:  ax² + bx + c  =  y

Since it is a monic polynomial:  a  =  1     --->     1·x² + bx + c  =  y  

                                                                                    --->     x² + bx + c  =  y

f(0)  =  4     --->     x = 0  and  y = 4     --->     (0)² + b·(0) + c  =  4     --->     c = 4

f(1)  =  -10     --->     x = 1 and y = -10     --->     (1)² + b(1) + 4  =  -10     --->     b  =  -15

Since  a = 1  and  b = -15  abd  c = 4     --->     x² - 15x + 4  =  y

5^b + 5^b​ + 5^b​ + 5^b​ + 5^b​  =   625^{b - 1} · 25^b

Left-hand side:  5^b + 5^b​ + 5^b​ + 5^b​ + 5^b​  =  5 · 5^b  =  5^{b + 1} 

Right-hand side:  625^{b - 1}  =  (5⁴)^{b - 1}  =  5^{4b - 4}

                                  25^b  =  (5²)^b  =  5^2b

        -->                 625^{b - 1} · 25^b  =  5^{4b - 4} · 5^2b  =  5^{6b - 4} 

Setting the left-had side equal to the right-hand side:  5^{b + 1}  =  5^{6b - 4} 

This means that:  b + 1  =  6b - 4     --->     5  =  5b     --->     b  =  1

[GME] = 3          (Believe it or not!)

[GME] = [ABC] / 24

12/sqrt(5)

when x > 0

the solution is always 1

when x < 0

the solution is always -1

when x = 0

the solution is indeterminate

so therefore it's 2

Thanks for answering guest but you need to put your latex in a latex box for it to render properly.

Hi Straight,

Everyone with a username is a member.

So you are already a member.   

Alan, asinus and I are members like you but we are also moderators.

hi Juriemagic,

If they are the same parabola then they are the same equation.

You need to check the answer,     

\(f(x)= a(x^2+4x+4)+c\\ f(x)= ax^2+4ax+(4a+c)\\~\\ g(x)=2x^2+(2b+5)x+5b\\~\\ \text{equate coefficients}\\ a=2\\~\\ 2b+5=4a\\ 2b+5=8\\ 2b=3\\ b=1.5\\ \\~\\ 4a+c=5b\\ 8+c=7.5\\ c=-0.5\\ \)

Ax^2 +4ax+ 4a+c     And

2x^2  +x(5+2b)  +5b       Are supposed to be the same 

a=2      5+2b=4a   So b = 3/2       4a+c=5b     So c =    You can finish from here i am sure 

From x + 3y = -5, y = -x/3 - 5, so the slope of the line is -1/3. The slope of the new line is also -1/3, so y = -x/3 + B.  Pugging in x = 2 and y = -7, we get -7 = -2/3 + B, so B = -19/3.

Then the line is y = -x/3 - 19/3.  Then 3y = -x - 19, so 3y + x = -19.  We want the right-hand side to be 3, so we mutiply both sides by -3/19: -9/19*y - 3/19*x = 3.  Therefore, A + B = -9/19 - 3/19 = -12/19.

This is NOT the correct solution for the question asked.

We are told two rolled-dice sum to a prime.  How the dice roll to produce this prime sum is irrelevant.

Equivalent restatement of question:

Given that the sum of two numbers on fairly-rolled dice is a prime number, what is the probability that the sum is three (3)?

Solution:

Calculate the Relative Probabilities of its occurrence by weighting the occurrence of each possible prime by its Absolute Probability.

\(\begin{array}{|rccc|} \hline & Sum &\text {Absolute Probability} &\text{Relative Probability }\\ &2 &2.78\% &6.67\% \\ &3 &5.56\% &13.34\% \\ &5 &11.11\% &26.67\% \\ &7 &16.67\% &40.00\% \\ &11 &5.56\% &13.34\% \\ \hline \end{array}\\ \text{ }\\ \; \;\text {The probability is } \mathrm {13.34\%}\\ \)

GA

--. .-

Leon bought 3X/4+8.   remaining in the box X/4-8

Hariz gets 1/2(X/4 - 8) + 4 = X/8. Remaining X/8 -8

Jackson Gets 1/3(X/8 -8) +2   Remainig 2/3(X/8 -8) - 2

2/3(X/8 - 8) -2 = 10 

X= 208

Thanks,

Krishna Korukonda

Thank You

hey straight
thats not correct
if you substitute, it ain't 600 

answer is 

600-0.03 =  599.97

Hello Guest,

How many integer solutions for (a, b) satisfied the following equation

a^2+b^2=ab^2

\(a^2+b^2=ab^2\)

\((a, \mbox { } b) = (2, \mbox {} \pm \mbox { } 2) \)

\(​​​​(a, \mbox { } b) = ( \pm \mbox { } 2, \mbox { } 2)\)

        \((a, \mbox { } b) = (0, \mbox { } 0)\)

Hope this was helpful . . . ^^

 !

The area of the circle is 48*pi.

$[GME] = 8$