NotThatSmart

First, let's set up variables to solve this question. 

Let's let \(d = a_2 - a_1\)

The conditions in the problem give us the system of equations

\(\begin{cases} a_1 + d = -1\\ a_1 + 3d = \dfrac13 \end{cases}\)

Now, solving this equation, we get that

\(d = \dfrac23\) and \(a_1 = -\dfrac53\)

So the final answer is -5/3. 

Thanks! :)

First, let's calculate the distance Will travels before grace even leaves. 

Since he traveled for 45 minutes, we have

\((45 min)(50 m/min) = 2250 m\)

So when Grace starts, she and Will have

\((2800 m) – (2250 m) = 550 m\) to cover. 

Let's let t be the amount of time it takes for the two to catch up to each other after Grace starts rowing. 

We have the equation

\((50)(t) + (30)(t) = 550 \\ 80t = 550 \\ t = 550/80 = 6.875\)

This is approximately 6 minutes and 52 seconds. 

So the clock time they meet is 6 min 52 sec after Grace starts.

Grace started at 2:45, so add 6 min 52 sec and the clock will read 2:51:52 pm when they meet  

Thus, our final answer is \(2:51:52\)

Thanks! :)

We can complete this problem in two different ways. 

The first tactic is to essentially compare this root to the quadratic equation of \(5x^2+21x+v = 0\)

From this quadratic, we can identify that a = 5, b = 21, c = v. 

We have the equation

\(\frac{-21-\sqrt{201}}{10} = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \frac{-21-\sqrt{201}}{10} = {-21 \pm \sqrt{21^2-4(5)(v)} \over 2(5)}\\ \frac{-21-\sqrt{201}}{10} = {-21 - \sqrt{441-20v} \over 10}\\ \sqrt{201} = \sqrt{441-20v}\\ 201 = 441-20v\\ v=12\)

This is a bit complicated and takes a lot of computations, but it does give us the correct answer. 

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The second tactic is to use conjugations of square roots. 

This is because the conjugate root theorem states that if a root of a polynomial is a square root \(a+\sqrt b\), then its conjugate, \(a-\sqrt b\) is also a root

We can apply that to this problem. If \(\frac{-21-\sqrt{201}}{10}\)  is a root, then \(\frac{-21 + \sqrt {201}}{ 10 }\) is also a root. 

The product of the roots is \([ (-21)^2 - 201 ] / 100 = 240/100 = 2.4\)

However, in the quadratic, we also have that \(v/5\) is also equal

Thus, we have 

\(v/5 = 2.4\\ v = 12\)

SO 12 is the final answer. 

Thanks! :)

First of all, let's focus on the second equation. 

We can factor the right side of the equation to get \((x+y)(x^2+xy+y^2)=162+x^2+y^2\)

Since x+y is equal to 10 from the first equation, the equation becomes

\(10(x^2+xy+y^2)=162+x^2+y^2\)

\(10x^2+10xy+10y^2= 162+x^2+y^2\)

\(9x^2+9y^2+10xy=162\)

Now, isolating x in the first equation, we get

\(x=10-y\)

Now plugging this value of x into the second equation, we get

\(9(10-y)^2+9y^2+10(10-y)y=162\\ 9(100-20y+y^2)+9y^2+100y-10y^2=162\\900-180y+9y^2+9y^2+100y-10y^2=162\\ 8y^2-80y+900=162\\8y^2-80y+738=0\)

However, plugging this into the quadratic formula, notice that unofruntately, the descriminant is less than 0, menaing the eventual solutions are NONREAL numbers. 

Thanks! :)

Alright. Let's first put y in terms of x. 

We get \(y=25-x\). 

Subbing this value back into the second equation, we get 

\(3x+75\)

Since x can't be negative, the smallest possible value of x is 0. 

When x is 0, we have \(3(0)+75=75\)

So 75 is our final answer. 

Thanks! :)

I believe the rhombus is a square for main reasons. 

For one thing, the opposite angles of a rhombus are 90 degrees, so we have two 90 degree angles.  

The adjacent angle also should be 90 degrees, so all 4 angles are 90 degrees. 

Therefore, we just have a square. 

We have \(12^2 = 144\)

So the area is 144. 

Thanks! :)

From the problem, we can write the system

 where t is toast, b is bagel and m is muffin. 

\(2T + 1B = 3.15 \\ 1M + 1B = 3.50 \\ 1T + 2B + 3M = 8.15\)

From the first two equations, we have

\([3.15 - B ] / 2 = T \\ M = 3.50 - B\)

Subbing this in to the third equation, we have 

\([ 3.15 - B ] / 2 + 2B + 3 [ 3.50 - B ] = 8.15 \\ [3.15 - B ] + 4B + 6 [ 3.50 -B ] = 16.30 \\ -3B + 3.15 + 21 = 16.30 \\ -3B = 16.30 - 3.15 - 21 \\ -3B = -7.85 \\ B = 7.85 / 3 ≈ $2.62\)

This is about 262 cents or 261 cents and 2/3. 

Thanks! :)