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We just subtract  2  from both sides.....

GAC  forms a right triangle with the right  angle  at G

The other two sides are equal sides of the cube such that  GC = GA

So this forms an isosceles right riangle....so angle GAC = 45°....and the  sin 45°  = sqrt (2) /  2

HAC is a little more difficult to  see.....let's call the side of the  cube = 1  = HA

AC =  the diagonal  across the face of the  cube  = sqrt (2)

And  HC is the major diagonal of the  cube =  sqrt (3)

Then HAC must form a right triangle  because

HA^2 + AC^2  = HC^2

 1^2    +  (sqrt 2)^2  = (sqrt 3)^2

  1    +   2  =  3

And angle HAC is opposite the  hypotenuse in this  triangle  so its  measure = 90°....and sin HAC = sin 90° = 1

Rearrange the inequality as

2x^2 + 8x - 6 < 0              divide through  by  2

x^2 + 4x - 3  <  0

This  does not factor

Let's turn it  into  an equality

x^3 + 4x - 3  = 0       

The equation on the  left is a parabola which  turns upward....using the Quadratic Formula we  can  find the roots of this eqaution......the answer to the  problem  will lie  between these roots  (because the parbola will be  below the  x axis betwwen  these values, e.g, it will be < 0 )

1st  Root  =    ( -4 - sqrt [ 4^2 - (4 * 1 * -3) ] ) /  2 =      [ -4 - sqrt ( 28) ] / 2 =  [ -4 - 2sqrt (7) ] / 2 = -2 -sqrt (7)

The 2nd Root will be the conjugate of the  first root = -2 + sqrt (7)

Solution    x =  ( -2 - sqrt (7)   ,   -2 + sqrt (7) )

Hey, i appreciate you so much. Thank you NotydoleM :)

just to understand more clearly, how did you get from 6x²-x<2 to 6x²-x-2<0 

First make it 

\(6x^2-x-2<0 \)

Then factor

\((3x-2)(2x+1)<0\)

Then solve if the inequality was an equality

Then add the inequality back in

x (the wired e sign) (-1/2,2/3)

hope this helps :)

You're right, there's a clever trick you can use to solve this problem efficiently.

Start by rearranging the given equation: 
\[ab - 6a + 5b = 37 \Rightarrow ab - 6a + 9b - 4b = 37 \Rightarrow (a+9)(b-4) = 37.\]

Since \(37\) is a prime number, it only has two factors: \(1\) and \(37\). Thus, the possible pairs \((a+9, b-4)\) are \((1,37)\) and \((37,1)\).

Case 1: \(a+9 = 1\) and \(b-4 = 37\)
This implies \(a = -8\) and \(b = 41\). However, since both \(a\) and \(b\) need to be positive integers, this case is not possible.

Case 2: \(a+9 = 37\) and \(b-4 = 1\)
This implies \(a = 28\) and \(b = 5\).

Now, calculate \(|a-b|\):
\[|a-b| = |28 - 5| = 23.\]

So, the smallest possible value of \(|a-b|\) is \(23\).

thanks!

To solve this question, you need to use the concept of mixing solutions with different concentrations to achieve a desired concentration. Here's how to start approaching each part:

(a) The goal here is to find the amount of red potion that must be added to a certain amount of blue potion to achieve a specific concentration. Let's denote the amount of red potion to be added as \(x\) mL. The concentration of the red potion is 1 magical syrup per volume, and the concentration of the blue potion is \(\frac{1}{2}\) magical syrup per volume. The final concentration we want is 1 magical syrup per volume.

Set up the equation based on the concentration formula:

\[\text{Concentration} \times \text{Volume} = \text{Concentration} \times \text{Volume}.\]

For the red potion:

\[1 \cdot x = \text{magical syrup in red potion}.\]

For the blue potion:

\[\frac{1}{2} \cdot 100 = \text{magical syrup in blue potion}.\]

Since the total volume after mixing is \(100 + x\) mL, and we want the concentration to be 1 magical syrup per volume, the equation becomes:

\[1 \cdot (100 + x) = (\text{magical syrup in red potion} + \text{magical syrup in blue potion}).\]

Solve for \(x\) using this equation.

(b) Similar to part (a), you need to find the amounts of red potion and blue potion to be combined to achieve a certain concentration. Let \(x\) be the amount of red potion (in mL) and \(y\) be the amount of blue potion (in mL). The concentration of the red potion is 1 magical syrup per volume, the concentration of the blue potion is \(\frac{1}{2}\) magical syrup per volume, and the final concentration is 1 magical syrup per volume.

Set up an equation for the amount of magical syrup:

\[1 \cdot x + \frac{1}{2} \cdot y = 1 \cdot (x + y).\]

Solve for \(y\) in terms of \(x\) using this equation, and then substitute this into the total volume equation to solve for \(x\).

(c) Think about the concentrations involved. If you have a red potion with a concentration of 1 magical syrup per volume and a blue potion with a concentration of \(\frac{1}{2}\) magical syrup per volume, is there any way to combine them to achieve a concentration of 1 magical syrup per volume?

Remember that concentrations represent the amount of magical syrup in a certain volume, and you're trying to mix these concentrations to achieve a desired result.

To find the domain of the function \(f(x) = \sqrt{\sqrt{x^2-16}-\sqrt{x^2 - 9}}\), we need to consider the values of \(x\) that ensure that the expression inside the square roots is valid. Since we cannot take the square root of a negative number in the real number system, both expressions inside the square roots must be non-negative.

First, let's consider the expression inside the first square root: \(\sqrt{x^2-16}\). For this expression to be non-negative, we require \(x^2 - 16 \geq 0\), which means \(x^2 \geq 16\). This gives us the condition \(x \leq -4\) or \(x \geq 4\).

Next, let's consider the expression inside the second square root: \(\sqrt{x^2-9}\). For this expression to be non-negative, we require \(x^2 - 9 \geq 0\), which means \(x^2 \geq 9\). This gives us the condition \(x \leq -3\) or \(x \geq 3\).

Now, let's combine the conditions from both square roots:

- For \(\sqrt{x^2-16}\) to be non-negative, we need \(x \leq -4\) or \(x \geq 4\).
- For \(\sqrt{x^2-9}\) to be non-negative, we need \(x \leq -3\) or \(x \geq 3\).

To satisfy both conditions simultaneously, we need \(x \leq -4\) or \(x \geq 4\), and \(x \leq -3\) or \(x \geq 3\). This can be simplified to \(x \leq -4\) or \(x \geq 4\).

Therefore, the domain of the function \(f(x) = \sqrt{\sqrt{x^2-16}-\sqrt{x^2 - 9}}\) is the set of all real numbers except those in the interval \([-4, -3]\) and \([3, 4]\). In interval notation, the domain is \((- \infty, -4) \cup (-3, 3) \cup (4, \infty)\).

sqrt ( x^2 -16) - sqrt ( x^2 -9)   must be ≥ 0

So

sqrt ( x^2 - 16)  - sqrt (x^2 -9)  ≥ 0

sqrt ( x^2 - 16) ≥ (x^2 - 9)                square both sides

x^2 - 16 ≥ x^2 - 9

-16 ≥ -9

Not true for any  reals......so  no  domain on  reals  exists

tan z  =   opp / adj  =  3 /  1

The hypotenuse =  sqrt [ 3^1 + 1^1 ] =  sqrt (10)

cos z  = adj / hyp  =    1 /sqrt (10)  =  sqrt (10) / 10

In a right triangle XYZ where angle X is 90 degrees, we can use the information provided to find the cosine of angle Z.

Given that tan Z = 3, we know that:

\[ \tan Z = \frac{\text{opposite}}{\text{adjacent}} = \frac{YZ}{XZ} = 3.\]

Since triangle XYZ is a right triangle, we can use the Pythagorean theorem:

\[ XY^2 + YZ^2 = XZ^2.\]

Since angle X is 90 degrees, we have:

\[ XZ^2 = XY^2 + YZ^2 = YZ^2 + YZ^2 = 2YZ^2.\]

Solving for YZ:

\[ YZ^2 = \frac{XZ^2}{2}.\]

Since we know the value of YZ/XZ (which is the tangent of angle Z):

\[ \tan Z = \frac{YZ}{XZ} = 3.\]

Squaring both sides:

\[ YZ^2 = 9XZ^2.\]

Now, substituting the value of YZ^2 from the Pythagorean theorem:

\[ 9XZ^2 = \frac{XZ^2}{2}.\]

Solving for XZ:

\[ \frac{XZ^2}{2} = 9XY^2.\]

\[ \cos Z = \boxed{\frac12}.\]

This answer is ChatGPT-generated and incorrect. I believe you have already been told to stop using AI's to answer questions. If you can't figure it out, that's fine but don't spread misinformation using ChatGPT. SpectraSynth is currently going to town spamming AI answers. Don't contribute to the problem.

We are given that $n = \overline{ABC}_8$ and $n = \overline{CBA}_6$, where the subscripts represent the bases. Let's convert both representations to base $10$ and set them equal to find the value of $n$.

In base $8$, the number $n$ can be represented as:
\[n = A \cdot 8^0 + B \cdot 8^1 + C \cdot 8^2.\]

In base $6$, the number $n$ can be represented as:
\[n = C \cdot 6^0 + B \cdot 6^1 + A \cdot 6^2.\]

Equating the two expressions for $n$:
\[A + 8B + 64C = C + 6B + 36A.\]

Simplify the equation:
\[35A + 2B = 56C.\]

Since $A$, $B$, and $C$ are all non-negative integers, we can observe that the maximum value for $C$ is $1$ (since the left side should not exceed the right side by too much). Substituting $C = 1$ gives:
\[35A + 2B = 56.\]

Now, we need to find non-negative integer values of $A$ and $B$ that satisfy this equation. The largest possible value for $A$ is $1$, and for $B$ is $17$. When $A = 1$ and $B = 17$, the equation is satisfied.

So, the largest possible value of $n$ in base $10$ is obtained by substituting $A = 1$, $B = 17$, and $C = 1$ into the base $8$ representation:
\[n = 1 \cdot 8^0 + 17 \cdot 8^1 + 1 \cdot 8^2 = 1 + 136 + 64 = \boxed{201}.\]