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First, let's focus on the expression we need A and B to be equal to. 

In order to find what we need, we need to be able to compare coefficients. let's convert it to standard form. 

We have

\((x - 2)(x + 2)=x^2-4\)

However, the constans don't match. Thus, we need to divide what we have by -4 to get a constant of +1. 

We have

\(-\frac{1}{4}x^2+1\)

Now. we cam compare coefficients. Since A is the coefficient of x^2, we have

\(A=\frac{-1}{4}\)

Since B is the coefficient of x, we have \(B=0\)

Thus, we finally have 

\(A+B=-\frac{1}{4}+0 = -\frac{1}{4}\)

So -1/4 is our final answer. 

Thanks! :)

Let's create a system of equations to solve this question. 

First, we know that the two points must be on the equation \(x^2+y^2=25\)

We also know that x must be 4. 

Thus, plugging in x=4 into the first equation, we can find y values. We have

\(4^2+y^2=25\\ y^2=25-16\\ y=\pm\sqrt9\\ y=3, y=-3\)

Thus, the two points A and B are \((4, 3), (4, -3)\)

Since they share an x value, the difference between the y values is the distance. We have

\(3-(-3)=6\)

Thus, 6 is our final answer. 

Thanks! :)

We just have

\(64 \cdot 3,816 = 244,224 \)

\(244224 \pmod{ 64} \equiv 0\)

244224 works. 

Thanks! :)

One ordered pair $(a,b)$ satisfies the two equations $ab^4 = 48$ and $ab^3 = 24$. What is the value of b in this ordered pair?   

                                                     ab⁴  =  48       

                                                     ab³  =  24    

divide both sides of the first by     

respective equal values of the     

second to obtain                              b = 2     

the problem doesn't ask for     

the value of a, but it's                       a = 3      

_.    

A school orders $99$ textbooks, all for the same price.   

When the bill for the total order comes, the first and last   

digits are obscured. What are the missing digits?   
$_5801.3_   

Consider that the price per book is multiplied by 99.   

And, multiplying by 99 incorporates multiplying by 9.    

The digits of the product of a number multiplied by 9    

will always sum to a total that is evenly divisible by 9.    

5 + 8 + 0 + 1 + 3 = 17  so the sum of the two missing     

numbers will have to add 1 or 10 or 19, etc. to that 17.     

Using brute force, I reckoned that 15801.39 is evenly

divided by 99.  So the missing numbers are 1 and 9.    

And the cost of each textbook was 159.61.    

_.    

First, let's move all terms to one side and combine all like terms. We get the equation

\(x^{2}+10x+9=0\)

Now, notice that we can easily factor the quadratic equation. We get

\((x+9)(x+1)=0\)

This gives us the values for x of 

\(x=-1\\ x=-9\)

Thus, our answer is -1, -9.

Thanks! :)

First , simplify this equation to:

 x^2 + 15x + k = 0      Where   a = 1    b = 15    c = k 

   if the discriminant  (b^2 - 4ac)   is equal to zero, it will be a 'double root' 

             b^2 - 4ac = 0 

                15^2 - 4 (1)(k) = 0 

                  225 = 4k 

                    k = 225/4 

For the first problem:

To address the problem, we will use the properties of similar triangles since lines \( DE \) and \( EF \) are parallel to sides \( BC \) and \( CD \) of triangle \( ABC \).

Since \( DE \) is parallel to \( BC \) and \( EF \) is parallel to \( CD \), triangle \( ADF \) is similar to triangle \( ABC \).
The sides of similar triangles are proportional, so we can set up the following proportions:

\[
\frac{AD}{AB} = \frac{AF}{AC} = \frac{DF}{BC}
\]

We know that \( AF = 7 \) and \( DF = 2 \). Let \( BD = x \). The entire length \( AB \) can be expressed as:

\[
AB = AD + DF = AD + 2
\]

Let’s designate \( AD = x \). Hence, we have \( AB = x + 2 \).

Considering the structure of triangle \( ADF \) in relation to triangle \( ABC \), we also need to calculate \( AC \) in terms of \( AE \):

Since \( EF \) is parallel to \( CD \), triangles \( AEF \) and \( ACD \) are also similar, resulting in the relationship:

\[
\frac{AE}{AC} = \frac{AF}{AB} = \frac{DF}{DC}
\]

But for our immediate task of determining the length \( BD \) just using \( AF \) and \( DF \), we will not utilize this second proportion at the moment.

As both triangles \( ADF \) and \( ABC \) are similar due to the parallel lines, we can set the ratios of the line segments as follows:

\[
\frac{AD}{AB} = \frac{AF}{AC}
\]

But we know \( AB = AD + DF \), substituting this gives us:

\[
\frac{x}{x + 2} = \frac{7}{7 + 2} = \frac{7}{9}
\]

Cross-multiplying yields:

\[
9x = 7(x + 2)
\]

Expanding the right-hand side:

\[
9x = 7x + 14
\]

Subtract \( 7x \) from both sides:

\[
2x = 14
\]

Solving for \( x \):

\[
x = \frac{14}{2} = 7
\]

Thus, we find that \( AD = 7 \).

Since we established \( BD = DF = 2 \), thus:

\[
BD = 7 - 2 = 5
\]

So the length of \( BD \) is:

\[
\boxed{5}
\]

Find a six-digit multiple of $64$ that consists only of the digits $2$ and $4$.   

64 x 3,816 = 244,224     

_.   

Isn't it 100~

this is close to 1 *10^100, BUT it's actually 9.9999... * 10^99

If given     a - 1/a = 0 

                   then   a = 1/a    

Sub this into the first equation 

    a^3 - 1/a^3   =    (1/a)^3 - 1/(a^3)   =  1/a^3 - 1/a^3 = 0       

                                                               Did I miss something here??

\(f(x) = \sqrt{x - \sqrt{x}}.\)

This is impossible. No such 3 digit numbers exist for x.

The answer is \(36\)

Let's make some observations of the problem. 

First, note that triangle ABC is a 45-45-90 right triangleThis means that\( AB = BC  =   12/\sqrt 2 =  6\sqrt 2 \)

Also, we have that

\(AD = AD\\ BD = BD \\ AB = BC \)

This means that triangles  ABD  and CBD  are congruent

Thus, we have \([ ABD ]   =  (1/2) [ABC]   =  (1/2) ( 1/2) ( 6\sqrt 2)^2  =  (1/4) (72)   = 18\)

So our aswer is 18. 

Thanks! :)

Going form R to S : 

  X   goes from -6  to 15    a distance of 21      2/3 of this is 14       add 14 to -6 to get    x = 8

  Y   goes from 19 to -5     a distance of -24     2/3 of this is -16      add -16 to 19 to get  y = 3

(8,3) 

The coordinates of the point are (2/3*(-6) + 1/3*(15), 2/3*19 + 1/3*(-5)) = (1,11).