NotThatSmart

First, let's combine all like terms. We get

\(y=-5x^2-6x+10\)

Now, the equation for the x value of the vertex is

\(x=\frac{-b}{2a}\) for quadratics in the form of \(ax^2+bx+c\)

Thus, we have \(x=\frac{-(-6)}{2(-5)} =- \frac{6}{10}\)

Plugging in x into the equation for y, we have 

\(y=\frac{59}{5}\)

Thus the vertex is \(\left(-\frac{3}{5},\:\frac{59}{5}\right)\)

So our answer is \(\left(-\frac{3}{5},\:\frac{59}{5}\right)\)

Thanks! :)

First, let's combine all our like terms. We have

\(x^2+15x+k\)

There are many possibilities for what k may be. 

As long as two factors of k add up to 15, we are good. 

Examples are such as

\(k=14; x^2+15x+14 = (x+14)(x+1)\\ k=56; x^2+15x+56=(x+7)(x+8)\)

k can also be negative, such as

\(k=-100; x^2+15x-100 = (x+20)(x-5)\)

Thanks! :)

First, let's take a look at the first equation we are given. 

\(a+b=1 \)

Squaring both sides, we have

\(a^2 + 2ab + b^2 = 1 \)

Now, we already know a^2 + b^2 is 2, so plugging that in, we have

\(2+2ab = 1\\ 2ab=-1\\ ab=-1/2\)

ab will come into handy later. 

Now, let's notcie something real quick. We have

\(a^3 + b^3 = (a + b) ( a^2 + b^2 - ab)\)

We already know all the terms needed now! We have

\(a^3 + b^3 = (1) ( 2 +1/2)\\ a^3+b^3 = 5/2\)

So our answer is 5/2. 

Thanks! :)

First, let's simplify the equation. We have

\(x^2 - 6x + 13\)

Since the constant is equal to uv, we have the equation

\(uv = 13 \\ 2uv = 26 \)

The coefficient of x is equal to u+v, so we have

\(u + v = 6 \)

Squaring both sides of the equation, we have the equation

\(u^2 + 2uv + v^2 = 36 \\ u^2 + 26 + v^2 = 36 \\ u^2 + v^2 = 10\)

Now, it's time for the fun part of this problem. We can simplify what we want to figure out to

\(u/v^2 + v/u^2 = [ u^3 + v^3 ] / [ uv]^2\)

We already know all these terms! We can easily find them! We have

\(u^3 + v^3 = ( u + v) ( u^2 + v^2 - uv) = (6)(10 - 13) = -18 \\ [uv]^2 = [ 13]^2 = 169\)

Thus, our answer is \(u/v^2 + v/u^2 = [ u^3 + v^3 ] / [ uv]^2 = -18 / 169 \)

So -18/169 is our answer. 

Thanks! :)

Let's note something really quickly. 

If Kenneth ran 2 hours, then he ran 4 kilometers more than Marshall. 

So 4 kilometers is 1/5 of the track's length. 

Thus, we have \(1/5x = 4\) where x is the length of the track. 

We have

\(x=20\)

So the track was 20km. 

So our answer is 20. 

Thanks! :)

There are two seperate cases we need to consider. 

First off, we have orange, purple, orange. 

The probability of this is \(\frac{3}{5}*\frac{2}{4}*\frac{2}{3} = 1/5\)

The other case is when we have purple, orange, purple. 

The probability of this is \(\frac{2}{5}*\frac{3}{4}*\frac{1}{3} = 1/10\)

Adding these two up, we get

\(\frac{1}{5}+\frac{1}{10} = \frac{2}{10}+\frac{1}{10}=\frac{3}{10}\)

So our answer is 3/10 or 30%. 

Thanks! :)

A terrific number must be only divisble by one and itself AND be a perfect square. 

Let's test perfect squares and see what we get. 

First, let's test \(1\). It's only divisble by 1, so it doesn't work. 

Next, we have \(4\). It's divisble by \(1, 2, 4\). So it is a terrific number!

So 4 is our answer. 

Thanks! :)

I'm not sure if  \(y=(\sqrt2 + 1)x\) is correct though. 

Through my understanding, the answer must be \(y=\sqrt{\frac{7}{5}}x\) is the full equation of the line. 

Am I mistaken with this judgement?

I entered the problem with an AI math bot...it confirmed my answer...

Thanks! :)

In order for a number to be terrific, then it must be divisble by 1, itself, and be a perfect square. 

Let's say x is a terrific number. 

It has to be divisble by x and 1, and be a perfect square. 

\(25\) is a perfect example. 

However, let's note that it has to be a perfect square for a prime number, because if the squared number was composite, there would be another divisor. 

Thus, the minimum number of primes is \(1\)

So 1 is our answer. 

Thanks! :)

I'll tackle this problem. 

First, we have two lines given, \(y=3x\) and \(y=2x\)

Let's note the slopes of the two line. The slope for the first line is 3, and the slope for the second line is 2. 

Let's set \(m_1 = 3\) and \(m_2 = 2\)

Using the angle bisector formula, which states that we have

\(m=\sqrt{\frac{m_1m_2+1}{m_1+m_2}}\) where m1 and m2 are the slopes of the two lines, we have

\(m = \pm \sqrt{\frac{3*2+1}{3+2}}\\ m = \pm \sqrt{\frac{7}{5}} \)

So our final answer is \(m = \pm \sqrt{\frac{7}{5}} \)

Thanks! :)