NotThatSmart

First, we have the fact that the axis of symmetry as 2. 

Since the distance (4, 7) from x = 2 is the same as (?, 7), we can easily find another point. 

We find that (4, 7) is two units away from the line x = 2, so we want to find another point that has y value of 7 and is also 2 units away. 

We find that (0, 7) matches all these characterisitics, thus being our answer!

Thanks! :)

First, let's write this in vertex form. 

We have \(y = a(x-h)^2 + k\) where \((h,k) \) denotes the vertex. From the graph, we can write, \(y=a(x-3)^2+1\) since the vertex is at (3, 1)

Now, we have to find a. We can use the y intercept to help. We have 

\(7 = a (0 - 3)^2 + 1 \\ 6 = 9a \\ a = 6/9 = 2/3\)

We find that when \(y=2/3(x-3)^2+1\), we have a y intercept at (0, 7), meaning we found our equation. 

Now, we simply have to expand it. Expaning, we get \(y = \frac{2}{3} x^ 2 − 4 x + 7\). 

This means \(abc = 2/3(-4)7 = -56/3\)

-56/3 is our answer!

Thanks! :)

I'm not sure if it's just me, but I don't see a graph. 

The equation of a circle is \((x – h)^2 + (y – k)^2 = r^2\) where \((h, k)\) is the coordinates of the center of the circle and r is the radius. 

Thanks! :)

\(f(x) \text{ being invertible means that if }f(x)=f(y) \text{ then }x=y\\ \text{so we are looking for points where }x^2 = x^4\\ \)

There are 3 such points. 

We have \(x=0, 1, -1\)

This means they intersect at 3 points. 

I'm not sure if this is correct, though. 

Thanks! :)

Hello!

So, first, let's review. We know that 

\(f(a) = \frac{\frac13 (3a+7)}{3a+7} + \frac{32/3}{3a+7}\) . I believe you understand this part!

Now, for the 1/3 in the second part where it starts getting confusing, let's first note that \(\frac{(ax+b)}{(x+b)} = \frac{a}{1} = a\). Since the polynomial is equivalent, we can cancel them out. That's what we are doing here. Canceling out the 3a + 7, we get the remaining 1/3. 

Now, for the \(\frac{32}{9a+21}\). Let's note that \(\frac{a}{b} = \frac{a}{b} \cdot \frac{c}{c}\). We want to get rid of the 3 in the numerator, so we multiplythe numerator and denominator by 3. 

We get \(\frac{32/3}{3a+7} = \frac{32/3}{3a+7} \cdot \frac{3}{3} = \frac{32}{3(3a+7)} = \frac{32}{9a+21}\)

So there, that's how we got from the second step to the third step. 

Good luck from here!

I hope I answered your question!

Thanks! :)

We get two seperate equations from the problem. 

We have

\( a^3 + 3ab^2 = 679 \\ 3a^3 - ab^2 = 615\)

Let's multiply the second equation by 3 so that we can get 3a^3 in both equations

\(3a^3+9a^2=2037\)

\(3a^3-ab^2=615\)

Now, subtract the second equation from the first equation. We get

\(10ab^2 = 1422 \\ ab^2=142.2\)

Now, sub this value back into the first equation to get that

\(a^3+3(142.2)=679 \\ a^3+426.6=679 \\ 252.4=a^3 \\ a=\sqrt[3]{252.4} \\ a \approx 6.19\)

Now we have a, we can find b. 

\(b^2 = (142.2)/(6.19) \\ b^2 \approx 22.97 \\ b \approx \sqrt{22.97} \\ b \approx 4.79\)

So, we have

 \(a - b \approx 6.19 - 4.79 \\ a - b \approx 1.40\)

So 1.4 is our answer

Thanks! :)

First, we have \(16*5*1/10 = 8\)

\( \log_{1/2} (8 ) =-3\)

so -3 is our answer

Thanks! :)

First, let's note that

\(2^3 + 3^1 = 11, 2^3 + 3^6 = 737 = 11 \cdot {67} \\ 2^3 + 3^{11} = 177155 = 16105 \cdot 11\)

Now, from this, we can tell that

\(m = 10a + 3 \\ n = 5b + 1\)

where a and b are different numbers. 

This means m has \( 3, 13, 23, ..., 593 = 60 \) values

And n has \(1, 6, 11, ..., 596 = 120 \) values. 

This means that there are \(60 * 120 = 7,200 \) cases that work.

In total, there are \(600 * 600 = 360,000\) different cases. 

Now, we just have \(7200/360000 = 72/3600 = 2/100 = 1/50\)

This is a 2% chance. 

Thanks! :)

First, let's apply some log rules to simplify our equation. We get

\(\log _3\left(x\right)=x\) from this. 

We know \(\mathrm{If}\:f\left(x\right)=g\left(x\right),\:\mathrm{then}\:a^{f\left(x\right)}=a^{g\left(x\right)}\)

From this, we have \(3^{\log _3\left(x\right)}=3^x\)

Which is the same as \(x=3^x\)

There are NO SOLUTIONS to this equation. 

Thanks! :)