NotThatSmart

Here, let me give this problem a shot. 

First, let's try to simplify this problem a bit. We can use Simon's Favorite Factoring Trick. 

Let's factor the right side a bit for n. We get

\(n = ab + 2a + 3b \\ n = a(b + 2) + 3b \)

Now, since we want b to be factored as well, let's add 6 to both sides. This will come in handy later. 

We get

\( n + 6 = a(b + 2) + 3(b + 2) \\ n +6 = (a + 3)(b + 2)\)

Now, this problem has transformed for numbers that match the conditions

\(76\leq (a+3)(b+2) \leq 96\)

The only numbers that don;t satsify these conditions are numbers with only two factors and one is less than 3. Thus, we can calculate each number. Here, I did it for you! :)

Thus, since there are a total of 21 numbers and 7 are not valid, we have 21 - 7 = 14 as our final answer. 

So 14 should be the answer. I'm not confident on the answer, I might have errored a bit. However, I think it's CLOSE to the right answer. Maybe 12 or 15, but 14 should be right. 

Thanks! :)

First, let's note that

\(3 + 4 + 5 = 3 *4\)

\(13 + 14 + 15 = 3 * 14\)

This pattern will keep going. Thus, we just have

\(3 [ 4 + 14 + 24 + ....... + 994 ]\)

No. of terms =  100

Thus, our final answer is

\(3 [ 994 + 4 ] [ 100 / 2 ] = 149,700\)

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! :)

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! :)

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! :)

First, we can write a handy equation to solve for the constant and find the 3 terms. 

Let's let the constant added to every number be x. 

Since the three terms for a geometric series, we have the equation

\(\frac{100+x}{60+x} = \frac{140+x}{100+x}\)

Now, when we corssmultiply and then expand everything out, we get

\((x+100)(100+x)=(x+60)(140+x)\\ 200x+x^{2}+10000=140x+x^{2}+{8400+60x}\\ 200x+x^{2}+10000=200x+x^{2}+8400\)

Now, we bring all terms to one side of the equation. We have

\(200x+x^{2}+10000-x^{2}-200x-8400=0\\200x+1600-200x=0 \\ 1600=0\)

However, this statement is obviously not true, meaning that x is invalid. 

This also means that there are NO solutions to this given problem. 

*Note, I may have made a mistake. Not sure. 

Thanks! :)

If angle XWZ = 60°, angle WXY = 180° - 60° = 120° because XY and WZ are parallels.

However, this means that Angle ZXW must be less than 120°.

In the problem though, it clearly defines ZXW as 120, meaning this problem is invalid. 

Thanks! :)

Hi! Although I don't have much expertience witht he calculator, I would reccomend websites like 

Symbolab or Mathway

for the job. Although their explenations are simple, their solutions are almost always correct. 

I wouldn't reccomend it, but sometimes AI bots such as chatgpt also come up with the correct solution, although MUCH more inaccurate. 

Also, it wouldn't hurt to solve the problem yourself to verify which website is most accurate! :)

I hope this helped!

Thanks! :)