NotThatSmart

(b) We know there were 52 watermelons and 52 mangoes remaining. 

There are also 130 watermelones and 117 mangoes. 

Thus, we just have

\(130-52=78\\ 117-52=65\)

Adding the two numbers up, we find that \(78+65=143\)

This means they sold a total of 143 watermelons and mangoes. 

Since each cost 3.60, we have

\(3.60 *143 = 514.8\)

Thus, they collected 514.8 dollars. 

Thanks! :)

(a) let's write an equation to solve this problem. 

Let's let w be the total number of watermelons and m be the total number of mangoes. 

Using the problem, we have the system of equations

\(w=m+13\\ 2/5w = 4/9m\)

Subsituting out w with m from the first equation, we get

\(\frac{2}{5}(m+13) = \frac{4}{9}m\)

Solving this equation yields 

\(m=117\). Since there are 13 more watermelons, then \(w=130\)

Next, let's find how many of each there were. Since \(2/5(130) = 52\), we have

\(52*2=104\)

So 104 is our final answer. 

Thanks! :)

Since both numbers we are multiplying are small, we can easily just compute it. 

We find that

\(16*37=592\)

Taking the modulo for 100, we find that

\(592 \equiv 92\pmod{100}\)

So 92 is our final answer/ 

Thanks! :)

Let's make some observations. 

Starting with \(2!\), every other term after that is even, since it contains 2. 

Since we know that \(even + even = even\), then we have \(2! + 3! + \dots + 100! = even\)

Thus, the sequence \(2! + 3! + \dots + 100! \equiv 0 (\mod 2)\)

However, we forgot about 1 term. Since \(1!=1\), and 1 is odd, and we know \(odd + even = odd\), then we have

\(1! + 2! + 3! + \dots + 100! \equiv 1(\mod 2)\)

Thus, the answer is 1. 

Thanks! :)

Let's multiply the polynomials and see what we get. 

We have \((x^3)(x^2 + tx)\)

Distributing in the x^3, we find that we get the expression

\(x^5+tx^4\)

there is actually no x^2 term, meaning t can pretty much be anything. 

As long as t doesn't contain a x^-2 term, then it isn't possible there is an x^2 term. 

Thanks! :)

Let's make some observations about the problem. 

First, let's note that triangle ABC is a 45-45-90 triangle, 

This means that we have the equation \(AB = BC = 12/\sqrt 2 = 6\sqrt 2 \)

Also, we have the relation of

\(AD = AD \\ BD = BD \\ AB = BC\)

Through these equation, we can determine that triangle ABD and triangle CBD are congruent triangles. 

Because of this note, we can write the equation

\( [ ABD ] = (1/2) [ABC] = (1/2) ( 1/2) ( 6\sqrt 2)^2 = (1/4) (72) = 18\)

So 18 is our final answer. 

Thanks! :)

First, let's move all terms to one side and combine all like terms. We get

\(x^2 + (17 - m)x + 4 = 0 \)

If there are two distict real roots, then the descriminant must be greater than 0. 

Thus, we have

\((17 - m)^2 - 4(1)(4) > 0 \\ (17 - m)^2 > 16\)

We must take both intervals for m. 

Taking the first interval, we have

\( 17 - m > 4 \\ 17 - 4 > m \\ m < 13 \)

From the second interval, we have

\( 17 - m < -4 \\ 21 < m \\ m > 21\)

Thus, we have two intervals for m. So our final answer, converting in interval notation, is 

\(m = ( -\infty, 13) U ( 21, \infty) \)

Thanks! :)

(a) For this part, it isn't too hard to multiply the numbers out. 

All 3 numbers are very small, so the product is not large. 

We just have \(9*7*11=693\), so 93 is our answer. 

First, we want to get rid of the nasty fractions. 

Let's do this by multiplying both sides by the LCM of the denominators. 

By doing this, we have

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

Distributing every number in and moving all terms to one side, we have

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

Using the quadratic equation and finding what x is, we have

\(x=\frac{3\sqrt{51}-12}{7}\\ x=\frac{-3\sqrt{51}-12}{7}\)

Thanks! :)