BuilderBoi

\({x^2 \over x^2} = \color{brown}\boxed{1}\)

I am assuming the score goes from 0 - 10.

Because 8 is the mode, it must occur the most often. We also know that there are 15 10's in the data set, there must be at least 16 8's in the data set. 

To minimize the mean, we need to include as many 0's as possible and limit the amount of "high" numbers. This means there will be exactly 16 8's

Also note that because we know the median is 9, the 20th and the 21st numbers must both be 9. 

If we order them in a list (lowest to highest), we have 15 perfect scores (26th to the 40th). We will also have exactly 7 9's (20th to the 25th), so the median will be 9. We also will have 16 8's (4th to 19th), to make the mode 8. This means that the remaining 3 spots will be 0's (1st to 3rd), representing the days Qinna was too lazy to do her homework. 

So, the mean is \(((10 \times 15) + (7 \times 9) + (16 \times 8)) \div 40 \approx \color{brown}\boxed{8.5}\)

The formula for the area of a circle is \(\pi r^2\), where the radius is 6. 

Can you take it from here?

maybe you could check your question before posting it 

Note that \(3 - i\sqrt3 = 3 - \sqrt{-3}\).

Now, recall the quadratic formula: \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\). Substituting \(a = 1\), we have \(3 - \sqrt {-3} = {-b - \sqrt{b^2-4c} \over 2}\) (take the minus root, not the plus)

Multiplying both sides by 2 gives us: \(6 - 2 \sqrt {-3} = -b - \sqrt{ b^2 - 4ac}\). Simplify the left-hand side: \(6 - \sqrt{-12} = -b - \sqrt {b^2 - 4ac}\).

Note that \(6 = -b\), meaning \(b = -6\). Plugging this into the discriminant, we have \({36 - 4c} = -12\) (disregard the square root because both have them)

Solving, we find that \(c = 12\).

Thanks for the feedback!!

Note that because all the bases in the term \(7^4 \times 2^4\) are raised to the 4th power, the 4th root of this is an integer. 

So, we basically have: \({7^4 \times 2^4 \over 7^x \times 2^y }= 7^2 \times 2^1\)

Start with \(7^4 \over 7^x\) . We want this to equal \(7^2\). 

Recall that when you are dividing exponents with the same base, you subtract the exponents (\(3^4 \div 3^2 = 3^{4 - 2} = 3^2\)).

Applying that here, we have \(7^4 \div 7^x = 7^2\), thus \(x = 2\). (\(7^4 \div 7^2 = 7 ^ {4 - 2} = 7^2\))

Now, doing the same thing to \(2^4 \over 2^y\). We want this to equal \(2^1\), so \(y = 3\). (\(2^4 \div 2^3 = 2^{4 - 3} = 2^1\)).

So, the answer we are looking for is \(7^2 \times 2^3 = \color{brown}\boxed{392}\)

Hope this helps!

Wait a minute... 

In \(\triangle BCD\), the hypotenuse is 17, while one of the legs is 18. 

That isn't a triangle...

The prime factorization of 98 is \(7^2 \times 2\). 

Because we are taking the fourth root, everything must be raised to a power of a multiple of 4, (4, 8, 12, etc.)

This means that the smallest value of n that works is \(7^2 \times 2^3 = \color{brown}\boxed{392}\) 

The minimum occurs at \(-{b \over 2a} = -{-8 \over 4} = 2\). 

Now, substitute this into the equation: \(\text{___} = 2\times 2^2 - 8 \times 2 + 19\)