NotThatSmart

I found the original question from where this was posted. 

The missing number is 3/2. 

If the number is 3/2, then we can do the problem the following. 

Since there are 3/2 times more sophomore, using f for number of freshmen, we can write

\(f+3/2f = 1\\ 5/2f=1\\ f = 2/5\)

So 2/5 of the team is freshmen. 

The other \(1-2/5=3/5\) of the team are sophomores.

So 3/5 is our answer!

Thanks! :)

We want to solve the equation for

\(a\mathbin{\spadesuit} (a + 1) = 9\)

Now, we plug in the function, and we get the equation \(9=\frac{a^2}{a+1}\)

Now, we simplfy solve the equation. We get

\(9a+9=a^2\)

\(a^2-9a-9=0 \)

Using the quadratic equation, we get

\(a=\frac{3\sqrt{13}+9}{2}\\ a=\frac{-3\sqrt{13}+9}{2}\)

So our answer is

\(a=\frac{3\sqrt{13}+9}{2}\\ a=\frac{-3\sqrt{13}+9}{2}\)

Thanks! :)

We could probably use casework to solve this problem.

The first case to look for is only two digit numbers. 

We have 23 and 32, so there are 2 possibilities. 

Next, consider 3 digit numbers. 

We have 311, 131, 113 for 3 more. 

We also have 221, 212, and 122 to add 3 more on. 

Next, consider 4 digit numbers. 

The only numbers that work are numbers with three 1s and one 2. There are 4 ways we could do this. 

Lastly, we have 5 digit numbers. The only one we have to consider is \(11111\), one possibility. 

Now, we add up the numbers. We get \(2+3+3+4+1 = 13. \)

So 13 is our answer!

Thanks! :)

First, let's note that \((a \cdot b) (3) = a ( b(3) ) \)

Now, we just work from inside to outside. We have

\(b(3) = 4 -(3)^2 + 4(3)^4 = -5 + 324 = 319 \\ a (319) = 2(319) + 5 - 3(319)^2 + (319)^3 = 32,157,119\)

We do the same exact thing with the second part of the equation. We get

\((b \cdot a)(3) = b (a (3)) \\ a(3) = 2(3) + 5 - 3(3)^2 + 3^3 = 11 \\ b (11) = 4 - (11)^2 + 4 (11)^4 = 58447\)

We have the two values needed now. All we have to do is to subtract them. We get

\((a ° b)(3) - (b ° a)(3) = 32,157,119 - 58447 = 32,098,672 \)

So 32,098,672 is our final answer!

Thanks! :)

We have

\((6032)_{12} = (6 × 12^3) + (0 × 12^2) + (3 × 12^1) + (2 × 12^0) = (10406)_{10} \)

How it works is that for every digit of 6032, we mulitply it by a power of 12 starting from 0. 

So our final answer is 10406. 

Thanks! :)

We can write

\((987)_{16} = (9 × 16^2) + (8 × 16^1) + (7 × 16^0) = (2439)_{10} \)

We know this because we multiply each digit of 987 by 16 raised to the power of a digit starting with 0. 

We multiply 9 with 16^2 and 8 with 16^1 and 7 with 16^0 as a result. 

This gets us our answer of 2439. 

Thanks! :)

First, we start off by doing \(35/8\). This gets us a quotient of 4 an remainder 3. 

Now, we divide the quotient by 8.

We get \(4/8\). This gets us a quotient of 0 and a remainder of 4. 

So, we tank the second digit with the first digit to get \(43\)

So our final answer is 48. 

Thanks! :)

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! :)