NotThatSmart

Let's use a handy trick to solve this problem. 

First off, let's set a variable. Let's set \(X=2.\overline{57}\)

For that value of X, then we have that

\(100 X = 257.\overline{57}\)

Now, we subtract x from 100x. Note that the repeated part of the decimal cancels out. Thus, we have

\(\eqalign{100 X &= &\hfill257.57...\cr X &= &\hfill2.57...\cr \hline 99X &= &255\cr}\)

We end up witht he equation

\(99 X = 255\)

Simplifying and isolating x, we get 

\(X = \frac{255}{99}\)

Thus,

The answer is \(2.\overline{57} = 2 \frac{19}{33}\)

Thanks! :)

First, let's turn the decimal into a fraction, unsimplified. We do the process

\(\frac{0.11012}{1}\times \frac{100000}{100000}= \frac{11012}{100000}\)

Now, we must find common factors. 

The GCD of the two numbers are 4, so dividing by 4, we get

\(\frac{11012 \div 4}{100000 \div 4}= \frac{2753}{25000}\)

From here, we can no longer simplify, so

The answer is \( \frac{2753}{25000}\)

Thanks! :)

I didn't follow an algorithm. I legit tried every single possibilitility until I found 1. 

When wee have A, B, C = 1, 0, 1, we get the number \(10011010011101110\)

Converting this into base 10, we have

\((10011010011101110)_2 = (1 × 2^{16}) + (0 × 2^{15}) + (0 × 2^{14}) + (1 × 2^{13}) + \\(1 × 2^{12}) + (0 × 2^{11}) + (1 × 2^{10}) + (0 × 2^9) + (0 × 2^8) + \\(1 × 2^7) + (1 × 2^6) + (1 × 2^5) + (0 × 2^4) + (1 × 2^3) + \\(1 × 2^2) + (1 × 2^1) + (0 × 2^0) = (79086)_{10}\)

Dividing by 7, we get \(79086/7=11298\)

THUS, 

The answer is \((A, B, C) = (1, 0, 1)\)

Thanks! :)

*I'm so tired

First, let's combine all like terms and reaarange so we have the shape of a quadratic formula. 

We get

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

Having a double root means that the descriminant is greater than 0, so we have the equation

\(15^2 - 4 (1)(k) = 0 \\ 225 - 4k = 0 \\ 225 = 4k \\ k = 225 / 4\)

thus, The value of k is \(k = 225 / 4 \)

Thanks! :)

First, let's multiply by the LCM of the denominators. This will get rid of all the denominators and leave us with a nice clean sheet. We get

\(10\left(x+4\right)=5\left(x-3\right)\left(x+5\right)+2\left(x+7\right)\left(x+5\right)\)

Doing the painful process of multiplying out and combining all like terms, we get

\(10x+40=7x^2+34x-5\)

Moving all terms to one side and combining all like terms, we get

\(7x^2+24x-45=0\)

Now, applying the quadratic equation, we find that

\(x_{1,\:2}=\frac{-24\pm \:\sqrt{24^2-4\cdot \:7\left(-45\right)}}{2\cdot \:7}\)

The 2 values of x are \(x=\frac{3\left(\sqrt{51}-4\right)}{7},\:x=-\frac{3\left(4+\sqrt{51}\right)}{7}\)

Thanks! :)

There are actually multiply parabolas that satsify the conditions given.

The simplest one is \(y=2x^2\)

It has a vertex at (0, 0) and faces upwards. 

If this the case, the answer is \(2+0+0 =2\)

Thanks! :)

ABJelly, I believe you were referring to this box in the problem

I understand you don't know how to add an image. 

First off, since both of them equal to y, then we can set the two equations to equal each other. We have

\(\log2 (2x) =\log4 (16 + x)\)

Applying the change of base thereom to the equation, we get

\(\log (2x) / \log2 = \log (16 + x) / \log 4 \\ \log (2x) / \log 2 = \log (16 + x) /\log 2^2 \\ \log (2x) / \log 2 = \log (16 + x) / [ 2 \log 2]\)

Now, we multiply through by log 2. 

\(\log (2x) =\log (16 + x) / 2 \\ 2\log (2x) = \log (16 + x) \\ \log (2x)^2 = \log (16 + x)\)

Now, wait a second. Notice that both logs have the same base now. 

This means there aren't any reprucussions we have to worry about. We simply just take the inside of the parenthesis. 

This implies that

\((2x)^2 = 16 + x \\ 4x^2 - x - 16 = 0\)

Taking the postiive value of x and using the quadratic equation, we get

The answer to the system of equations is \(x = [ 1 + \sqrt{257} ] / 8\)

First, let's make some observations about this question. 

All points 5 units away from point (2, 7) would make a circle with center (2, 7) and radius 5. 

The formula for that circle would look something like \((x-2)^2+(y-7)^2=25\)

Now, combining that with the second equation, and we get a system. 

Subsituting out y with the x values in the line equation, we get

\((x-2)^2+(5x-35)^2=25\)

Simplifying, and we get

\(13x^{2}-177x+602=0\)

Using the quadratic equation, we find two values of x. We get

\(x=7\\ x=\frac{86}{13}\)

Plugging these x values into the second equation where we isolated y, we get thatr \(y=7;y=\frac{66}{13}\)

So our final two answers are (7, 7) and (86/13, 66/13)

Thanks! :)