NotThatSmart

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! ;) 

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! :)