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! :)