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We can use the Quadratic formula to find the value of v. 

a = 5, b = 21, c = v

\(\frac{-21-\sqrt{201}}{10} = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(\frac{-21-\sqrt{201}}{10} = {-21 \pm \sqrt{21^2-4(5)(v)} \over 2(5)}\)

\(\frac{-21-\sqrt{201}}{10} = {-21 - \sqrt{441-20v} \over 10}\)

\(\sqrt{201} = \sqrt{441-20v}\)

\(201 = 441-20v\)

\(20v = 240\)

\(\mathbf{v=12}\)

Therefore, the value of v is 12

Combining all like terms and bringing all terms to one side, we get

\(4x^2 - 8x + 4y^2 + 14y + 12 =0 \). 

Now, completing the square fot both x and y, we get

\(\left(x-1\right)^2+\left(y-\left(-\frac{7}{4}\right)\right)^2=\left(\frac{\sqrt{17}}{4}\right)^2\)

According to the equation of the circle,

we get \(r=\sqrt{\left(\frac{\sqrt{17}}{4}\right)^2}=\sqrt{17}/4\)

Thanks! :)

The only restriction for x is that the value inside the square root must be greater than 0. 

We can write

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

Solving for x, we have

\(-3x^2-x+6 \geq 0\)

Completing the square, we have

\(-3\left(x+\frac{1}{6}\right)^2+\frac{73}{12}\ge \:0\)

\(-3\left(x+\frac{1}{6}\right)^2\ge \:-\frac{73}{12}\)

\(\left(x+\frac{1}{6}\right)^2\le \frac{73}{36}\)

Square rooting both sides,  we get

\(-\sqrt{\frac{73}{36}}\le \:x+\frac{1}{6}\le \sqrt{\frac{73}{36}}\)

\(\frac{-\sqrt{73}-1}{6}\le \:x\le \frac{\sqrt{73}-1}{6}\)

So the domain is \(\frac{-\sqrt{73}-1}{6}\le \:x\le \frac{\sqrt{73}-1}{6}\)

Thanks! :)

We can go through every function to solve this problem. 

First, we have \(x=2x+8\), because we want f(x) to equal x. 

We solve this to find \(x=-8\). Unfortunately, this is not in the range given for x, so the solution is invalid. 

Now, we have \(x=13-5x\).

Solving for this, we have \(x=13/6\). This is in the range given, so it is a valid solution. 

Third, we have \(x=20-14x\). 

We get \(x=4/3\). This is not in the range given, so the solution is invalid. 

Lastly, we have \(x=40+5x\)

Solving for this, we get \(x=-10\). This is not in the range, so it is invalid. 

So, the only solution is \(x=13/6\)

Thanks! :)

Combining all like terms in the equation, 

\(x^2 + (k-9)x + 16\)

In order for this to be the square of a binomial, the coefficient of x has to be 2 times the square root of the third term. 

Take this

\((x+b)^2=x^2+2bx+b^2\)

2b is exactly two times the square root of the third term. 

In this case, we have \(k-9=2\sqrt{16}\)

We have

\(k-9=8\\ k=17\)

So k is 17. 

Thanks! :)

First off, if the 98% of the berries were water, that means that only 2% of the berries weren't water. 

We have \(2/100 \cdot 200 = 4\), so 4 kg weren't water. 

Now, since a lot of it evaportated, that means there is less water. 

HOWEVER, the weight of the non-waters stays the same. 

That means that 4kg is 5% of the total weight. 

We have \(4*20 = 80\), so 80 kg is our answer. 

80 is our final answer!

Thanks! :)

Our first step is to isolate y from the first equation. Moving x to the other side, we get

\(y=25-x\)

Now, we plug this value of y into the second equation. We get

\(6x+3(25-x)\\ 6x+75-3x\\ 3x+75\)

Now, x has to be nonegative, so it cannot be a negative number. The next smallest number is 0. 

Plugging in the value of 0, we get

\(3(0)+75\\ 75\)

So 75 is our final answer!

Thanks! :)

Let's let the two sides of the rectange be a and b. 

From the problem, we have

\(2a + 2b =40 \\ a + b =20 \\ \)

Now, we just sqaure both sides to get

\(a^2 + 2ab + b^2 = 400 \\ a^2 + b^2 + 2ab = 400\)

From here, we have

\(a^2 + 2ab + b^2 = 400 \\ a^2 + b^2 + 2ab = 400 \\ a^2 + b^2 = (10\sqrt{ 2})^2 \\ a^2 + b^2 = 200\\ 200 + 2ab = 400 \\ 100 +ab = 200 \\ ab = 200 - 100\)

So our answer is 100. 

Thanks! :)

We have to make a really important observation to solve this problem. Note that

\(1,005,010,010,005,001 = \\ 1\times 10^{15} + 5 \times 10^{12} + 10\times 10^9 + 10\times 10^6 + 5 \times 10^3 + 1\times 10^0\\ \)

This value is equal to

\((10^3+1)^5\), which is equal to \(1001^5\)

Now, we have to find the largest prime factor of 1001. 

We know that \(1001=13*11*7\)

This means the largest prime factor is 13. 

Thanks! :)

There's one important factor we must know to solve this problem. 

We have to know the formula

Dilation Factor = Length of Image/Length of Pre-Image. 

Well, using this equation, we can easily solve the problem!

Plugging in all the values given in the problem, we have

\(5 = 30 / OA\)

\(OA=30/5\)

\(OA=6\)

So our answer is just 6.

Thanks! ;) 

Line EF is parallel to line BC, due to both being lines of the square. Therefore, angle FEA is equal to angle ACB and angle AFE is equal to angle ABC. We can now prove that triangle AFE is similar to triangle ABC due to AA similarity. Let's suppose that the side length of the square is x. Using the ratios of the similar triangles:

\(\frac{3-x}{x} = \frac{3}{1}\)

\(3-x = 3x\)

\(4x=3\)

\(x=\frac{3}{4}\)

So the side length of the square is 3/4. Therefore, the area of the square is (3/4)^2 = 9/16 square units.

OMG THE QUEEN HERSELF RESPONDED!! Queen Mary Sue is absolutely correct (damn gurl, can you read minds??!) Other than that, your answer is ABSOLUTELY COOKING!

Darlin', I think the "Atrocious Slop" Ms. Slayqueen was referring to was this part of your second-to-last line: 

$$2*sin(a)^4*cos(a)^4$$

This can be remedied by placing a backslash like this: \sin(a), \cos(a).

Don't let Ms. Slayqueen's energetic attitude aggravate your sensitivities, hon. You got this!

MarySue

This problem is a bit tricky, but it's solvable using some handy equations. 

First, let's rewrite the inequality. We have

\( \sin^7(a) + \cos^7(a) < 2 \sin^4(a) \cos^4(a) \)

Now, let's let \(x = \sin(a)\) and \( y = \cos(a)\)

Since we have \( 0 < a < (\pi)/(2)\) and  \( 0 < x, y < 1 \), we can write the inequality

\( x^7 + y^7 < 2 x^4 y^4 \)

Now, we can apply the Arithmetic Mean-Geometric Mean Inequality or the AM-GM to x^4 and y^4, we have

\( (x^4 + y^4)/(2) \geq \sqrt{(x^4 y^4) }\\ x^4 + y^4 \geq 2 x^2 y^2 \)

\( x^4 y^4 \leq \left( (x^4 + y^4)/(2) \right)^2\)

Now, also note that

\( x^7 + y^7 < 2 x^4 y^4 \) since x and y have to be between 0 and 1. 

Now, we can prove this same thing with sin(a) and cos(a). 

We have \( 0 < a < (\pi)/(2)\) and the fact that both sin(a) and cos(a) are between 0 and 1. 

Now, using the same logic above, we can clearly see that 

\( \left(\sin(a)\right)^7 + \left(\cos ( a)\right)^7< 2*sin(a)^4*cos(a)^4\)

Thanks! :)

Oh no! My queen NotThatSmart's answer has been blocked by the forces of evil! Maybe she isn't smart enough??!!! HAHA IM JUST KIDDING GURLLL!!!!!

By the way, please fix that ATROCIOUS SLOP on the second to last line so your answer can be an absolute EAT. 

SALAYYYY QUEEN!!!

You go gurlll! Keep on slaving away for those bots! Slaving more like SLAYING!!!! YAS QUEEN!!!!

First, let's note that triangle ABC is a 45-45-90 triangle. 

We also now know that \(AB = BC = 12/\sqrt{ 2} = 6\sqrt{ 2} \)

We have that

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

So by SSS congruence, triangles ABd and CBD are congruent. 

So, this means we have

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

So 18 is our answer!

Thanks! :)

Step one: Find the negative reciprocal of the line where it passes through A

The negative reciprocal of y = x+3 is y = -x+b, and to find b, we can plug in x and y. Doing so gets us "7 = -1 + b", so b is 8. Therefore, the line we were looking for is "y = -x+8"

Step two: Find the point of intersection of the lines.

To find the point of intersection, we simply find what value of x makes both equations true, or, in our case, what value of x makes "x + 3 = -x + 8". Solving for x, we get x = 2.5, so y = 5.5. Therefore, our lines intersect at (2.5, 5.5).

Step three: Find the point's distance from intersection, then flip that.

To go from (2.5, 5.5) to (7, 1), you must move right 4.5 units, then down 4.5 units. We then do the opposite of this, and move up 4.5 units and left 4.5 units from the intersection, bringing us to (2.5 - 4.5, 5.5+4.5) = (-2, 10), which is what we're looking for.

After following all these steps, we can now see that the coordinates of B are (-2, 10).

\(\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 12}\)

\(\frac{1}{4}+\frac{1}{28}+\frac{1}{84}\)

\(\frac{21}{84}+\frac{3}{84}+\frac{1}{84}\)

\(\mathbf{\frac{25}{84}}\)