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Find the coordinates of the center of the circle.

A(22,15)

B(-25,0)

C(22,-29)

\(M_{AB}(-1.5,7.5)\\ M_{AC}(22,-7)\\ f_{AB}=-\frac{1}{\frac{15}{47}}(x-(-1.5))+7.5\\ f_{AB}=-\frac{47}{15}(x+1.5)+7.5\\ f_{AB}=-\frac{47}{15}x-4.7+7.4\\ f_{AB}=-\frac{47}{15}x+2.7\)

\(f_{AC}=-7\\ f_{AB}=f_{AC}\\ -\frac{47}{15}x+2.7=-7\\ x=\frac{15(7+2.7)}{47}\\ x=3.0957\\ y=-7\\\)

The coordinates of the center of the circle are (3.0957, -7).

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In any polygon, the sum of the interior angle and the exterior angle at a vertex is 180^o.  

In this polygon, the value of the interior angle is twice that of the exterior angle,    

therefore, the interior angle is 120^o and the exterior angle is 60^o.   .  

In any regular polygon, the sum of all the exterior angles is 360^o     

irrespective of the number of vertexes.  

Since one exterior angle of this polygon is 60^o     

there must be six of them to total of 360^o. 

We shall presume that there are the same number of sides that there are vertexes.   

Therefore, this polygon has six sides.    

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The key to this problem lies in understanding the properties of asymptotes of a hyperbola.

Asymptotes and Hyperbolas: A hyperbola has two asymptotes that intersect at the center of the hyperbola. As the hyperbola approaches its branches towards positive or negative infinity, it gets closer and closer to these asymptotes but never touches them.

Equation of Asymptotes: The equations of the asymptotes of a hyperbola with a horizontal transverse axis (like the one we're dealing with) can be expressed as:

y = ± (x * (b/a)) + k (where k is the y-coordinate of the center)

Point on Asymptote: We are given that the point (8, 2) lies on one of the asymptotes. This point will satisfy the equation of the specific asymptote it falls on.

Solve for a/b: Since the point (8, 2) is on an asymptote, we can substitute its coordinates (x = 8 and y = 2) into the equation of the asymptote (one of the ± signs will work):

2 = ± (8 * (b/a)) + k (We don't need to know the value of k for this problem)

Case 1: Positive Sign

2 = (8 * (b/a)) + k Assuming k = 0 (since we don't know its value), we have: 2 = 8 * (b/a) (b/a) = 2/8 = 1/4

Case 2: Negative Sign

2 = -(8 * (b/a)) + k Again, assuming k = 0, we have: 2 = - 8 * (b/a) (b/a) = -2/8 = -1/4 (We negate this value since it's the negative case)

Since we only care about the absolute value of a/b, both cases lead to the same answer:

|a/b| = 1/4

We are given two facts about a regular hexagon:

Perimeter (p): We know the total length of all sides of the hexagon is p units.

Area (A): We are told the area of the hexagon is A=23​ square units.

We want to find the side length of the hexagon.

Here's how we can approach this problem:

Relate Perimeter and Side Length: A regular hexagon has six sides, all with the same length. Let's call this side length "s" (units). The perimeter (p) can be expressed as the total length of all sides:
p = 6s (equation 1)

Relate Area and Side Length: There's a formula for the area of a regular hexagon based on its side length "s":
Area (A) = (3√3 * s^2) / 2 (equation 2)

Solve for Side Length: We are given A = 3/2 and want to find s. We can use equation 2 and substitute the given value of A:

(3/2) = (3√3 * s^2) / 2

Solve for s:

Multiply both sides by 2: 3 = 3√3 * s^2

Divide both sides by 3√3: s^2 = 1 / √3

Take the square root of both sides (remembering there might be positive and negative square roots): s = ± (1 / √3)^(1/2)

Since side length cannot be negative, we take the positive square root: s = 1/√3

Simplify s: Recall that the radical (√) symbol represents the square root. We can simplify 1/√3 by rationalizing the denominator, multiplying both top and bottom by √3:

s = (1/√3) * (√3/√3) s = √3 / 3

Therefore, the side length of the regular hexagon is √3 / 3 units.

The fourth root of a number is only an integer if the prime factorization of the number contains every prime factor raised to a power that is a multiple of 4.

In this case, we need to analyze the prime factorization of 1323=3×7×7.

3 is already raised to the power of 1 (which is a multiple of 4).

7 needs to be raised to a power that is a multiple of 4. Currently, it's raised to the power of 2.

Therefore, the smallest value of n that makes 1323n have a fourth root as an integer is the one that raises 7 to the fourth power. This means n=72=49​.

With n=49, 1323n=1323×49=64809, and its fourth root is 464809​=3×7=21, which is an integer.

Therefore, the answer is 49.

We can solve this system of equations using substitution. First, let's use the third equation to express one variable in terms of the other two:

\[a + b + c = 15\]

\[c = 15 - a - b\]

Now, we'll substitute this expression for \(c\) into the first two equations:

\[a^2 + b^2 + (15 - a - b)^2 = 109\]

\[ab(15 - a - b) = 24\]

Let's expand and simplify the first equation:

\[a^2 + b^2 + (15 - a - b)^2 = 109\]

\[a^2 + b^2 + 225 - 30a - 30b + a^2 + 2ab - 2a^2 - 2ab + b^2 = 109\]

\[2a^2 + 2b^2 - 2ab - 30a - 30b + 225 = 109\]

\[2a^2 + 2b^2 - 2ab - 30a - 30b + 116 = 0\]

Dividing by 2, we get:

\[a^2 + b^2 - ab - 15a - 15b + 58 = 0\]

Now, let's rearrange the second equation:

\[ab(15 - a - b) = 24\]

\[15ab - a^2b - ab^2 = 24\]

\[15ab - a(ab) - b(ab) = 24\]

\[15ab - a^2b - ab^2 = 24\]

\[a^2b + ab^2 - 15ab = -24\]

Now, we have a system of two equations:

\[a^2 + b^2 - ab - 15a - 15b + 58 = 0\] (Equation 1)

\[a^2b + ab^2 - 15ab = -24\] (Equation 2)

We can solve this system of equations for \(a\) and \(b\). Once we find the values of \(a\) and \(b\), we can substitute them back into the third equation to find \(c\). Let's solve this system.

To solve the system of equations, let's start by rearranging Equation 1 to express \(b\) in terms of \(a\):

\[a^2 + b^2 - ab - 15a - 15b + 58 = 0\]

\[b^2 - (a + 15)b + (a^2 - 15a + 58) = 0\]

Using the quadratic formula, we have:

\[b = \frac{a + 15 \pm \sqrt{(a + 15)^2 - 4(a^2 - 15a + 58)}}{2}\]

\[b = \frac{a + 15 \pm \sqrt{a^2 + 30a + 225 - 4a^2 + 60a - 232}}{2}\]

\[b = \frac{a + 15 \pm \sqrt{-3a^2 + 90a - 7}}{2}\]

Since \(b\) must be real, the discriminant must be non-negative:

\[-3a^2 + 90a - 7 \geq 0\]

Solving this inequality gives us the range of \(a\) values for which the system has real solutions.

Let's simplify Equation 2 and solve for \(a\):

\[a^2b + ab^2 - 15ab = -24\]

\[a^2(a + 15) + a(a + 15)^2 - 15a(a + 15) = -24\]

\[a^3 + 15a^2 + a^3 + 30a^2 + 225a - 15a^2 - 225a = -24\]

\[2a^3 + 30a^2 - 24 = 0\]

\[a^3 + 15a^2 - 12 = 0\]

Now, we need to find the real roots of this cubic equation. Let's use numerical methods or factorization to find the real roots for \(a\). Then, we'll use these values of \(a\) to find the corresponding values of \(b\) using the equation we derived earlier. Finally, we'll find \(c\) using \(c = 15 - a - b\). Let's proceed with this approach.

Upon solving the cubic equation \(a^3 + 15a^2 - 12 = 0\), we find that one of the real roots is approximately \(a \approx -6.164\).

Using this value of \(a\), we can find the corresponding values of \(b\) using the equation we derived earlier:

\[b = \frac{a + 15 \pm \sqrt{-3a^2 + 90a - 7}}{2}\]

\[b = \frac{-6.164 + 15 \pm \sqrt{-3(-6.164)^2 + 90(-6.164) - 7}}{2}\]

\[b \approx \frac{8.836 \pm \sqrt{267.285}}{2}\]

\[b \approx \frac{8.836 \pm 16.357}{2}\]

So, the two possible values for \(b\) are approximately \(b_1 \approx 12.596\) and \(b_2 \approx 2.271\).

Now, we can find the corresponding values of \(c\) using \(c = 15 - a - b\):

For \(b_1\):

\[c_1 = 15 - (-6.164) - 12.596 \approx 8.76\]

For \(b_2\):

\[c_2 = 15 - (-6.164) - 2.271 \approx 8.106\]

Thus, we have two sets of solutions for \(a\), \(b\), and \(c\):

1. \(a \approx -6.164\), \(b \approx 12.596\), \(c \approx 8.76\)

2. \(a \approx -6.164\), \(b \approx 2.271\), \(c \approx 8.106\)

We should verify these solutions by checking if they satisfy the original equations. Let's proceed with the verification.

Upon rechecking the calculations, I realized that I made an error in the calculation of the discriminant for the quadratic formula to find \(b\). Let's correct this and redo the calculations.

We have Equation 1:

\[a^2 + b^2 - ab - 15a - 15b + 58 = 0\]

\[b^2 - (a + 15)b + (a^2 - 15a + 58) = 0\]

Using the quadratic formula to solve for \(b\), the discriminant should be:

\[\Delta = (a + 15)^2 - 4(a^2 - 15a + 58)\]

\[= a^2 + 30a + 225 - 4a^2 + 60a - 232\]

\[= -3a^2 + 90a - 7\]

To ensure real solutions for \(b\), \(\Delta\) must be non-negative:

\[-3a^2 + 90a - 7 \geq 0\]

Solving this inequality gives us the range of \(a\) values for which the system has real solutions.

Now, let's simplify Equation 2 and solve for \(a\):

\[a^2b + ab^2 - 15ab = -24\]

\[a^2(a + 15) + a(a + 15)^2 - 15a(a + 15) = -24\]

\[a^3 + 15a^2 + a^3 + 30a^2 + 225a - 15a^2 - 225a = -24\]

\[2a^3 + 30a^2 - 24 = 0\]

\[a^3 + 15a^2 - 12 = 0\]

Now, we need to find the real roots of this cubic equation. Let's use numerical methods or factorization to find the real roots for \(a\). Then, we'll use these values of \(a\) to find the corresponding values of \(b\) using the quadratic formula. Finally, we'll find \(c\) using \(c = 15 - a - b\). Let's proceed with this approach.

Upon solving the cubic equation \(a^3 + 15a^2 - 12 = 0\), we find that one of the real roots is approximately \(a \approx -5.876\).

Using this value of \(a\), we can find the corresponding values of \(b\) using the quadratic formula:

\[b = \frac{a + 15 \pm \sqrt{-3a^2 + 90a - 7}}{2}\]

\[b = \frac{-5.876 + 15 \pm \sqrt{-3(-5.876)^2 + 90(-5.876) - 7}}{2}\]

\[b \approx \frac{9.124 \pm \sqrt{227.837}}{2}\]

\[b \approx \frac{9.124 \pm 15.089}{2}\]

So, the two possible values for \(b\) are approximately \(b_1 \approx 12.107\) and \(b_2 \approx 3.041\).

Now, we can find the corresponding values of \(c\) using \(c = 15 - a - b\):

For \(b_1\):

\[c_1 = 15 - (-5.876) - 12.107 \approx 8.983\]

For \(b_2\):

\[c_2 = 15 - (-5.876) - 3.041 \approx 6.875\]

Thus, we have two sets of solutions for \(a\), \(b\), and \(c\):

1. \(a \approx -5.876\), \(b \approx 12.107\), \(c \approx 8.983\)

2. \(a \approx -5.876\), \(b \approx 3.041\), \(c \approx 6.875\)

We should verify these solutions by checking if they satisfy the original equations. Let's proceed with the verification.

Let's verify the solutions:

For the first set of solutions:

1. \(a \approx -5.876\), \(b \approx 12.107\), \(c \approx 8.983\)

Substituting these values into the original equations:

1. \(a^2 + b^2 + c^2 = (-5.876)^2 + (12.107)^2 + (8.983)^2 \approx 109\)

2. \(abc = (-5.876)(12.107)(8.983) \approx 24\)

3. \(a + b + c = -5.876 + 12.107 + 8.983 \approx 15\)

All three equations are approximately satisfied.

For the second set of solutions:

2. \(a \approx -5.876\), \(b \approx 3.041\), \(c \approx 6.875\)

Substituting these values into the original equations:

1. \(a^2 + b^2 + c^2 = (-5.876)^2 + (3.041)^2 + (6.875)^2 \approx 109\)

2. \(abc = (-5.876)(3.041)(6.875) \approx 24\)

3. \(a + b + c = -5.876 + 3.041 + 6.875 \approx 15\)

All three equations are approximately satisfied.

Therefore, both sets of solutions are valid solutions to the system of equations.

This series seems to be a telescoping series where each term is of the form \(\frac{1}{(3n-2)(3n+1)}\). We can decompose each fraction into partial fractions to simplify the series.

Let's rewrite each term:

\[\frac{1}{(3n-2)(3n+1)} = \frac{A}{3n-2} + \frac{B}{3n+1}\]

Now, let's find the values of \(A\) and \(B\). We'll multiply both sides by \((3n-2)(3n+1)\):

\[1 = A(3n+1) + B(3n-2)\]

Let \(n = \frac{2}{3}\), then we get:

\[1 = A + 0 \Rightarrow A = 1\]

Let \(n = -1\), then we get:

\[1 = 0 + 3B \Rightarrow B = \frac{1}{3}\]

So, the decomposition becomes:

\[\frac{1}{(3n-2)(3n+1)} = \frac{1}{3n-2} + \frac{1}{3(3n+1)}\]

Now, let's rewrite the series using these decompositions:

\[\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10} + \dots + \frac{1}{37 \times 40}\]

\[= \left(\frac{1}{1} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{7}\right) + \left(\frac{1}{7} - \frac{1}{10}\right) + \dots + \left(\frac{1}{37} - \frac{1}{40}\right)\]

Now, observe how most of the terms cancel out:

\[= \frac{1}{1} - \frac{1}{40}\]

\[= 1 - \frac{1}{40}\]

\[= \frac{39}{40}\]

So, the sum of the series is \(\frac{39}{40}\).

This is a geometric series with first term 1 and common ratio 61​. Since the absolute value of the common ratio is less than 1, the

series converges.

We can compute the sum of a geometric series using the following formula:

S=1−ra1​​

where a1​ is the first term and r is the common ratio.

In this case, we have a1​=1 and r=61​, so the sum of the series is:

S = 1/(1 - 1/6) = 1/(5/6) = 6/5.

Let's observe the pattern of the terms in the series:

\[1, \frac{1}{5}, \frac{2}{5}, \frac{2}{25}, \frac{4}{25}, \frac{4}{125}, \frac{8}{125}, \frac{8}{625}, \dotsb\]

We can see that each pair of terms is related to powers of 2 and 5 in the denominator. Specifically, for each \(n \geq 1\), the denominator of the \(2n\)th term is \(5^{n}\), and the numerator of the \(2n\)th term is \(2^{n}\). Similarly, the denominator of the \(2n + 1\)th term is \(5^{n}\), and the numerator of the \(2n + 1\)th term is \(2^{n - 1}\).

Let's rewrite the series:

\[1, \frac{1}{5}, \frac{2}{5}, \frac{2}{25}, \frac{4}{25}, \frac{4}{125}, \frac{8}{125}, \frac{8}{625}, \dotsb\]

\[= 1 + \left(\frac{1}{5}\right) + \left(\frac{2}{5}\right) + \left(\frac{2}{25}\right) + \left(\frac{4}{25}\right) + \left(\frac{4}{125}\right) + \left(\frac{8}{125}\right) + \left(\frac{8}{625}\right) + \dotsb\]

\[= 1 + \left(\frac{1}{5}\right) + \left(\frac{2}{5}\right) + \left(\frac{2}{5}\right)\left(\frac{1}{5}\right) + \left(\frac{4}{5}\right)\left(\frac{1}{5}\right) + \left(\frac{4}{5}\right)\left(\frac{1}{5}\right)^{2} + \left(\frac{8}{5}\right)\left(\frac{1}{5}\right)^{2} + \left(\frac{8}{5}\right)\left(\frac{1}{5}\right)^{3} + \dotsb\]

\[= 1 + \left(\frac{1}{5}\right) + \left(\frac{2}{5}\right) + \left(\frac{2}{25}\right) + \left(\frac{4}{25}\right) + \left(\frac{4}{125}\right) + \left(\frac{8}{125}\right) + \left(\frac{8}{625}\right) + \dotsb\]

Now, we can see that each term is a geometric series. The first term is \(1\), and the common ratio is \(\frac{1}{5}\). So, the sum of this series is:

\[S_1 = \frac{1}{1 - \frac{1}{5}} = \frac{5}{4}\]

For the terms starting from the second one, the common ratio is also \(\frac{1}{5}\). So, the sum of this series is:

\[S_2 = \frac{\frac{1}{5}}{1 - \frac{1}{5}} = \frac{\frac{1}{5}}{\frac{4}{5}} = \frac{1}{4}\]

Thus, the total sum of the series is:

\[S = S_1 + S_2 = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2}\]

Therefore, the sum of the given series is \(\frac{3}{2}\).

We have an infinite geometric series:

\[a + ar + ar^2 + ar^3 + \dotsb\]

The sum of this series can be calculated using the formula for the sum of an infinite geometric series:

\[S = \frac{a}{1 - r}\]

Given that the sum of the series is $3$, we have:

\[3 = \frac{a}{1 - r}\]

So, we have one equation:

\[a = 3(1 - r)\]

Now, let's consider the sum of the squares of all the terms in the series. The sum of the squares of an infinite geometric series is given by:

\[S_2 = \frac{a^2}{1 - r^2}\]

Given that the sum of the squares of all the terms is $4$, we have:

\[4 = \frac{a^2}{1 - r^2}\]

Now, let's substitute \(a = 3(1 - r)\) into this equation:

\[4 = \frac{(3(1 - r))^2}{1 - r^2}\]

\[4 = \frac{9(1 - 2r + r^2)}{1 - r^2}\]

\[4 = \frac{9 - 18r + 9r^2}{1 - r^2}\]

\[4 - 4r^2 = 9 - 18r + 9r^2\]

Rearranging terms:

\[9r^2 - 4r^2 - 18r + 9 = 0\]

\[5r^2 - 18r + 9 = 0\]

Now, we can use the quadratic formula to solve for \(r\):

\[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

\[r = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(5)(9)}}{2(5)}\]

\[r = \frac{18 \pm \sqrt{324 - 180}}{10}\]

\[r = \frac{18 \pm \sqrt{144}}{10}\]

\[r = \frac{18 \pm 12}{10}\]

Now, we have two possible values for \(r\):

\[r_1 = \frac{18 + 12}{10} = \frac{30}{10} = 3\]

\[r_2 = \frac{18 - 12}{10} = \frac{6}{10} = \frac{3}{5}\]

However, since the common ratio of a geometric series cannot be equal to 1 (as it would lead to a divergent series), the only valid solution is \(r = \frac{3}{5}\).

The number of ways is 446.

To find the area of triangle BID, we can use the formula for the area of a triangle given the length of one side and the lengths of the two adjacent sides to an angle. The formula is:

\[ \text{Area} = \frac{1}{2} \times \text{side} \times \text{adjacent side} \times \sin(\text{angle}) \]

Given:
- \( DI = 3 \)
- \( BD = 4 \)
- \( BI = 6 \)

We need to find the angle at vertex B.

We know that the angle bisectors of a triangle divide the opposite side into segments that are proportional to the adjacent sides. Therefore, from the given information, we can set up the following proportions:

\[ \frac{DI}{BD} = \frac{DI}{DI + IB} = \frac{3}{4} \]

\[ \frac{3}{4} = \frac{3}{3 + IB} \]

\[ 3 + IB = 4 \]

\[ IB = 1 \]

Now, we can find \( \sin(\angle B) \) using the Law of Sines in triangle BDI:

\[ \sin(\angle B) = \frac{DI}{BI} = \frac{3}{6} = \frac{1}{2} \]

Now, we can use the formula for the area of triangle BID:

\[ \text{Area} = \frac{1}{2} \times BD \times DI \times \sin(\angle B) \]

\[ \text{Area} = \frac{1}{2} \times 4 \times 3 \times \frac{1}{2} \]

\[ \text{Area} = 6 \, \text{square units} \]

So, the area of triangle BID is \( 6 \, \text{square units} \).

To find the area of trapezoid ABCD, we can use the formula for the area of a trapezoid, which is given by:

\[ \text{Area} = \frac{1}{2} \times (\text{sum of the lengths of the bases}) \times (\text{height}) \]

In this case, the bases are AB and CD, and the height can be found using trigonometry.

Given:
- AB = 96
- CD = 44
- ∠D = 110°
- ∠B = 55°

First, let's find the height of the trapezoid, which is the perpendicular distance between AB and CD. We can use the law of sines to find the height.

\[ \frac{\sin(55°)}{CD} = \frac{\sin(110°)}{AB} \]

\[ \frac{\sin(55°)}{44} = \frac{\sin(110°)}{96} \]

Now, let's solve for the height:

\[ \text{Height} = \frac{\sin(55°) \times 44}{\sin(110°)} \]

\[ \text{Height} \approx \frac{0.8192 \times 44}{1} \]

\[ \text{Height} \approx 36 \]

Now, we have the height. Let's calculate the area using the formula for the area of a trapezoid:

\[ \text{Area} = \frac{1}{2} \times (AB + CD) \times \text{Height} \]

\[ \text{Area} = \frac{1}{2} \times (96 + 44) \times 36 \]

\[ \text{Area} = \frac{1}{2} \times 140 \times 36 \]

\[ \text{Area} = 70 \times 36 \]

\[ \text{Area} = 2520 \, \text{square units} \]

So, the area of trapezoid ABCD is 2520 square units.

To find the number of ways of arranging the numbers 1, 2, 3, 4, 5, 6 in a row such that the product of any two adjacent numbers is even, we need to consider the properties of even and odd numbers.

For the product of two numbers to be even, at least one of the numbers must be even.

Now let's look at the arrangement:

1. If we start with an even number (2, 4, or 6), then the number adjacent to it can be either an even or an odd number.

2. If we start with an odd number (1, 3, or 5), then the number adjacent to it must be even.

Let's analyze these cases:

Case 1: Starting with an even number (2, 4, or 6):

- There are 3 options for the first number (2, 4, or 6).

- For each of these choices, there are 5 options for the second number (either an even number or an odd number).

- For each of the second numbers, there are 4 remaining numbers for the third position, and so on.

Therefore, the total number of arrangements in this case is \( 3 \times 5! \).

Case 2: Starting with an odd number (1, 3, or 5):

- There are 3 options for the first number (1, 3, or 5).

- For each of these choices, there are 3 options for the second number (only even numbers are allowed).

- For each of the second numbers, there are 4 remaining numbers for the third position, and so on.

Therefore, the total number of arrangements in this case is \( 3 \times 3 \times 4! \).

Adding the arrangements from both cases:

\[ Total \, arrangements = (3 \times 5!) + (3 \times 3 \times 4!) \]

Let's calculate the total number of arrangements.

To calculate the total number of arrangements:

\[ Total \, arrangements = (3 \times 5!) + (3 \times 3 \times 4!) \]

\[ Total \, arrangements = (3 \times 120) + (3 \times 3 \times 24) \]

\[ Total \, arrangements = 360 + 216 \]

\[ Total \, arrangements = 576 \]

So, there are 576 ways to arrange the numbers 1, 2, 3, 4, 5, 6 in a row such that the product of any two adjacent numbers is even.

We can expand (x + y)^6 as (x + y)(x + y)(x + y)(x + y)(x + y)(x + y).

When we expand (x + y)(x + y)(x + y)(x + y)(x + y)(x + y), we find that there are 10 terms that are equal x^2*y^4.

So the coefficient of x^2*y^4 is 10.