NotThatSmart

We can first note somethings really important to solve this problem. 

We can find the percentage of milk chocolates there is from the ratio. We have

\(25\% \cdot 7/5 = 5\% \cdot 7 = 35\%\)

Thus, 35% of the box is milk chocolate. Finally we know the amount of orange chocoaltes there are. We have

\(100\% - 25\% - 35\% = 40\%\)

Thus, 40% of the box is orange chocolates. Now, we can use the third condiditon stated in the problem. 

Since dark and orange chocolates formed 78 pieces, let's find the percentage it is in total. We add the two up and we get

\(35\% + 45\% = 75\%\)

Thus, 75% of the box is equal to 78 pieces. We can now right the equation

\(3/4x = 78\) where x is the total amount of pieces. Thus, we have

\(x = 78 * 4/3\\ x = 26*4\\ x = 104\)

Thus our answer is 104 chocolates. 

So choice 1 is our answer .

Thanks! :)

To solve this problem, let's note something really important first. 

Since in the problem, we have \( AD = AB \), then we can conclude that \(\angle ABD = \angle ADB\)

Thus, we can easily solve the problem from here. We can write the equation

\( ABD = (180 - 17) / 2 = 81.5° \)

Thus, our answer is 81.5

Thanks! :)

First, let's multiply the terms out. We get that

\(x^3(x^2+tx) = x^5+tx^4\)

There is no x^2 in this polynomial. 

However, it is possible if t has a x^-2 to it, then we get x^2. 

Other than that, no other terms work. 

Thanks! :)

To solve this problem, let's focus on the ways we can get x in the problem. 

First, we have x and 14. 

We get \(14x\) from this. 

Next, we have 1 and \(-23x\)

Multiplying these two together, we get

\(-23x\)

Adding these two together, we get \(14x-23x=-9x\)

We could also expand everything out. Using the Distributive Property, we get

\(x^{7}-7x^{6}+10x^{5}-13x^{4}+8x^{2}-9x+14\)

Thus, the coefficient is -9. 

So our answer is -9. 

Thanks! :)

In this case, note that the only way x is possibily not in the domain is if the denominator of a fraction is 0. 

First off, the smallest fraction 3/x. x can't be 0 since 3/0 is undefined. 

Next, we have

\(1+3/x \neq 0\\ 3/x\neq-1\\ -x \neq 3\\ x \neq -3\)

Thus, x cannot be negative 3. Lastly, let's focus on the last fraction. We have

\(1+\frac{2}{1+3/x} \neq 0\\ \frac{2}{1+3/x}\neq-1\\ 2\neq-1-3/x\\ 3\neq -3/x\\ x \neq -1\)

So x also can't possibly be -1. 

Thus, we have 3 following solutions. We have x cannot be 0, -1 and -3. 

So our answer is 0, -1, -3. 

Thanks! :)

The only way for the system to have infinitely solutions is where we have two exactly congruent equations. 

We first notice that we have

\(6a + 2b = k + 3a + mb\)

Subtracting 3a from both sides will achieve us the same exact left sides of the equations. We have

\(3a+2b=2\\ 3a+2b = k+mb\)

Setting the right sides of the equation to equal each other, we have

\(k+mb=2\)

Now, let's note that if m was any number other than 0, the two equations would fail to match each other. Therefore, m must be 0. 

For this value of m, we have k is 2. So we have

\(m=0\\ k=2\)

So our final answer is m = 0 and k = 2. 

Thanks! :)

First, let's put y in terms of x to make our lives easier. 

We have

\(y=25-x\)

Now, we subsitute it into the second expression and see what we get, 

We have

\(6x+3(25-x)\\ 6x+75-3x\\ 3x+75\)

Now, since x must be nonegative, the smallest value it can be is 0. Thus, we have

\(3(0) + 75\\ 75\)

So 75 is our answer. 

Thanks! :)

There are many ways to solve this problem, but I prefer casework. 

There are two cases. 

The first case is that we get two even numbers from the lists. We have

\(\frac{\binom22}{\binom52} = 1/10\)

The second case we have is that we one odd number and we have one even number. Using permutations, we get

\(\frac{\binom31}{\binom52} = 6/10\)

Adding these two together, we get

\(1/10+6/10=7/10\)

So our final answer is 0.7

Thanks! :)