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Let's first focus on the slope of the line introduced. 

First, note that the slope of a line is in the form \(\frac{y_2-y_1}{x_2-x_1}\)

Thus, plugging in the two points we have, we get

\( [ b^2 - a^2 ] / [ b - a ]=2\)

Now, using the difference of squares thereom, we can simpplify the top to

\( [ (b -a) (b + a) ] / (b -a) = 2\)

The b-a cancels out, leaving us with

\( b + a = 2\)

So 2 is our answer. 

Thanks! :)

Let's use a handy trick to solve this problem. We first have

\(ab -3a + 4b = 131 \)

Now, let's take the coefficients of a and b, which are -3 and 4. We multiply them and take the product and add it to both sides. We get

\(ab - 3a + 4b + (4 * - 3) = 131 + (4 * -3) \)

Now, we simplify and factor the left side of the equation. We get that

\(ab - 3a + 4b - 12 = 119 \\ (a + 4) ( b - 3) = 119 \)

Now, let's focus on the factors of 119. We have

\(1 , 7 , 17, 119\)

Clearly, 7 and 17 will get us the minimal value, so plugging that in, we have

\(( 13 + 4) ( 10 - 3) → |a - b | = |13 - 10 | = 3 \)

so 3 is our final answer. 

Thanks! :)

First, let's zoom in and focus on the equation we are given. Let's square it. We get

\((a^2 - 2 + 1/a^2) = 0\)

This will achieve the equation

\(a^2 - 1/a^2 = -2\)

Next, let's focus on what we are trying to find. We know that

\(a^3 - 1/a^3 = ( a - 1/a) ( a^2 + 1 + 1/a^2)\)

WAIT! The first term in what we must find is 0!

That means our answer is just 0. 

So 0 is our answer!

Thanks! :)

We can write a handy equation to solve this problem. 

First, let's note that every number satsifying the conditions given can be written in the form

\(171x+5\) where x is an integer. 

We set this to equal 1000 to get the largest possible x can be, so we have

\(171x+5 = 1000\)

We are trying to solve for x right now. We have

\(171x=995\)

Dividing both sides, we find that 

\(x=5.81871345029\)

We must round down, so 5 must be our answer. 

So our final answer is 5. 

Thanks! :)

The answer is 8. 

Each of the conjugate of the roots are also roots, so we have 8. 

First, let's make some observations about the problem. 

First, let's note that if \(\angle BXA = 90\), then we have \(\angle BXC = 90\)

Triangle BXA is a right triangle, so we can use Pythoagorean Thereom, we simplfy have

\(BC^2 = CX^2 + BX^2 \\ 15^2 = 5^2 + BX^2\\ 225 = 25 + BX^2 \\ 200 = BX^2\)

Square rooting both sides, we finally reach a conclusion. We finally have

\(BX = \sqrt{200} = 10\sqrt 2 \)

So our final answer is 10\sqrt2. 

Thanks! :)

First, let's make some observations about the problem. 

Let's note that triangle ABC is 45-45-90 triangle. 

This means that we can easily find some other lengfths. We have

\(AB = BC = 12/\sqrt 2 = 6\sqrt 2\)

First, let's also note that we als have

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

this finally means that we know that triangle ABD and triangle triangle CBD are congruent triangles. 

Thus, we can finally we have

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

So 18 is our final answer. 

Thanks! :)

2 times.   

Let the first machine cycle two times and then shut it off, for a total of 10 added.   

Let the second machine cycle three times and then shut it off, for a total of 9 removed.   

10 – 9 = 1 candy left in the bottom    

_.

Class size is at most 60.  The most students that could not   

be arranged in more than one row would be 59 students.  

_.

In how many ways can the numbers 1, 2, 3, 4, 5, 6 be arranged in a row,    

so that the product of any two adjacent numbers is at least 5?    

You must have the 1 between the 5 and the 6.   

The rest of the numbers can be anywhere.   

So, consider two cases:   

(1) Permutation of   2  3  4  (5 1 6) ..... ₄P₄  =  24   

(2) Permutation of   2  3  4  (6 1 5) ..... ₄P₄  =  24   

Add them together                 total  .....  48 ways   

_.

To find the number of 7-digit numbers where the sum of the digits is divisible by 10, we can use generating functions.

Let's represent the generating function for a single digit as:

$(x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9)$

The generating function for a 7-digit number can then be represented as:

$(x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9)^7$

We need to find the coefficient of $x^{10}$ in this expression in order to find the number of 7-digit numbers where the sum of the digits is divisible by 10.

Using mathematical software or the binomial theorem, we can calculate this coefficient to be 8,343.

Therefore, there are 8,343 7-digit numbers where the sum of the digits is divisible by 10.

It is impossible for thius to happen. 

We have

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

No such integers exist. 

Thanks! :)

first, let's focus in the equation we are given. 

Squaring both sides of the equation, we get that

\((a^2 + 2 + 1/a^2) = 0\)

This means that we have \(a^2 + 1/a^2 = -2\)

Now, let's focus on what we are trying to find. 

Let's notice that 

\(a^3 - 1/a^3 = ( a - 1/a) ( a^2 + 1 + 1/a^2) \)

Wait a minute! we already have all the terms needed. We know both values to solve the problem. 

Since the first term is 0, we just have \((0)(-1) = 0\)

So 0 is our answer. 

Thanks! :)

First, let's find how many pans of cookies they need to make. 

We just have \(216/15=14.4\). since they can't make partial pans, we just round up to 15 pans. 

Now, since they made 15 pans, they're going to need \(15 \cdot 3 = 45\) tablespoons of butter to complete the task. 

Divide that by 8, we get \(45/8 =5.625\)

This rounds up to 6, so they need 6 sticks of butter. 

So 6 is our answer. 

Thanks! :)

First, look at the first equation. Squaring both sides, we have

\(a^2 + 2ab + b^2 = 16\)

Now, we have two equations to work with. We have

\(a^2 + b^2 = 16 - 2ab \\ a^2 + b^2 = 6 + 2ab \)

Subtracting the second equation from the first equation, we get

\(0 = -10 + 4ab \\ 10 = 4ab \\ ab = 10 / 4 = 5/2\)

Now, let's acknowledge something about a^3+b^3. Note that

\(a^3 + b^3 = (a + b) ( a^2 + b^2 - ab) \)

Wait! we already have all the terms needed to solve the problem!

We have

\( \\ (4) ( 6 + 2ab - ab) = \\ (4) ( 6 - ab) = \\ (4) ( 6 - 5/2) = \\ (4) ( 7/2) = \\ 14\)

So 14 is our answer. 

Thanks! :)

There are actually multiple parabolas that actually satisfy these conditions. 

However, let's set the origin or \((0, 0)\) as the vertex. 

This would mean the parabola is facing upwards. 

One parabola that wokrs is \(y = 2x^2 \). 

If that's the case, then we have \(2+0+0 = 2\) as our answer. 

So 2 is our answer, i guess. 

Thanks! :)

Let's write a system of equations to solve this problem.. 

Let's set b as bagel. 

Let's let m be muffin.

Let's let t be toast.

We have the system

\(2T + 1B = 3.15 \\ 1M + 1B = 3.50 \\ 1T + 2B + 3M = 8.15\)

We want to isolate b. From the first 2 equations, we can easily isolate the other two variables to only contain b, We have

\([3.15 - B ] / 2 = T \\ M = 3.50 - B\)

Now, we subsitute these two into the third equation. We get

\([ 3.15 - B ] / 2 + 2B + 3 [ 3.50 - B ] = 8.15 \\ [3.15 - B ] + 4B + 6 [ 3.50 -B ] = 16.30 \\ -3B + 3.15 + 21 = 16.30 \\ -3B = 16.30 - 3.15 - 21 \\ -3B = -7.85 \\ B = 7.85 / 3 ≈ $2.6.67\)

This is about 262 cents or 261 and 2/3 cents. 

Thanks! :)

First, let's combine all like terms and simplify. We have

\(x^2 + 15x + k \) as the simplified version. 

To have a double root means that the descriminant is 0. 

The descriminant is in the form of \(b^2-4ac\), so we have the equation

\(15^2 - 4 (1)(k) = 0 \\ 225 - 4k = 0\\ 225 = 4k \\ k = 225 / 4\)

Thus, our answer is 225/4. 

Thanks! :)