NotThatSmart

This problem actually has a very simple answer. 

Let's note the two sidelengths given, We have \(WX = 2\)  and \(XY = 3\)

In order to find the area of WXYZ, we must find the height and base. 

The base of the rectange is \(WX\) and the height of rectangle is \(XY\)

The problem gave all the dimensions we need to already solve this problem. We have

\(WX \cdot XY = 2 \cdot 3 = 6\)

So our final answer is 6. 

Thanks! :)

There are a few ways you can tackle this problem. 

The first way is reccomended. We find that after distributing, how many pairs create a coefficient of x. 

We first have x and 14. \(14 \cdot x = 14x\)

Next, we have 1 and -23x. \(1 \cdot -23 = -23x\)

These are the only two ways to achieve x. Now, we simply add then up. 

\(14x - 23x = -9x\)

Orrrrrrrrrrrrrrrrrrrrrrrrrrr, you can distribibute EVERYTHIG and take 100 years, but here u go :)

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

Thanks! :)

According to the Triangle Angle Sum Theorem, the sum of all interior angles of a triangle is always 180 degrees. 

Using this information, we can write the equation

\(2x + 3y + 8x + 15y + 4x - 2y = 180\\ 14x + 16y = 180\\ 7x+8y = 90\)

Now, if we were to say that x and y can be any number, then there an infinite amount of choices. 

However, if we were to only limit x and y to integers, then \(x=6; y= 6\) is the only solution. 

Hope this helps. 

~NTS

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! :)

So for this problem, let's use the arithmetic series formula.

We find the sum is \(\frac{(140+1)(70)}{2}\), which is \(141*35.\) Now that we have a factorization, we can easily find the greatest prime divisor.

The prime factorization is \(3*5*7*47\), so the greatest prime divisor is 47.

Thanks! :)

Let's know that XZW will form a triangle

\(\angle ZXW = 90^\circ \\ \angle XWZ = 60^\circ\)

Thus, \(\angle XZW = 180 - 90 - 60 = 30\)

30 degrees is our answer. 

Thanks! :)

An obtuse angle is an angle that measures in between \(90^\circ\) and \(180^\circ\)

Here's an image that lists all types of angles.

Notice that B is the only one that fits for the obtuse angle. 

A would be an acute angle since it is less than 90 degrees. 

C and D would be reflexive angles since they are more than 180 degrees, less than 360 degrees. 

The hour hand will move at a rate of 

\(360°  \space \text{in} \space 12 \space \text{hours} \\ 30° \space \text{in}  \space 1 \space \text{hour} \\ 30°/ 60  =  0.5° \space \text{in}  \text{ one minute}\)

So  at   4:15, the hour  hand has  moved   \(4 (30°) +  15(.5°)  =  [120+ 7.5]° = 127.5°\)from 12 o'clock

The minute hand  will move  at a rate of \(360°/ 60  = 6°\) per minute

So at 4:15 it will have moved  \([15* 6]°  = 90°\)  from the top of the hour

So...the smaller angle formed by the hands\(  = 127.5 - 90 = 37.5^\circ\)

May have made an error....

Thanks! :)

Let's note that 1/17 is a repeating decimal. 

\(1/17 = 0.\overline{ 0588235294117647}\)

There are 16 repeating digits in the decimal. 

Dividing 4000 by 16, we get

\(4000/16 = 250\)

Since it is divisble, then we can conclude that the last digit of the repeating decimal is the correct answer. 

Thus, 7 is the final answer. 

Thanks! :)

Let's first focus on the first equation given.

Since we know that \((x+y)^2 = x^2+2xy+y^2\), squaring both sides on the first equation, we get that

\(x^2+2xy+y^2=4\)

Reaaranging the second equation a bit, we find that

\(x^2+xy+y^2=5\)

Wait! These two equations are quite similar to each other. Subtracting the second equation from the first equation, we get that

\(x^2+2xy+y^2=4\\ x^2+1xy+y^2=5 \space \space \space \space -\)

______________________

\(xy=-1\)

Now, let's look at the second equation again. We know that

\(x^2+xy+y^2=5\\ x^2+y^2=5-xy\\ x^2+y^2=5-(-1) = 6\)

So the answer is 6...i think

Thanks! :)

\(Slope = [ b^2 - a^2 ] / [ b - a ] = [ (b -a) (b + a) ] / (b -a) = b + a = 2\)

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! :)

For this problem, we have to go throug every function and find when f(x) = x. 

First off, we have \(f(x)=2x+8\). 

We have to solve \(2x+8=x\), in which we get \(x=-8\). However, -8 isn't in the range \(1 \le x \le 2\), so it isn't a valid solution. 

Next, we have \(f(x) = 13 - 5x\)

Solving for \(13-5x=x\), we get \(x=13/6\). Indeed, 13/6 is in the range \(2 < x \le 3\), so it is a valid solution. 

After, we have \(f(x) = 20 - 14x\)

Solving for \(20-14x=x\), we get \(x=4/3\), which is not a valid solution since it's not in the range \(3 < x \le 4\). 

Lastly, we have \(f(x)= 40 + 5x\). 

We have \(40+5x=x\), in which we get \(x=-10\), which is not a valid solution. 

So, we only have x = 13/6

Thanks! :)

For there to be no real solutions, the discriminant needs to be NEGATIVE

The descriminant is \(b^2 - 4ac <0 \)  with  \(b = k ;   \space a = 5   ; \space c = 8\)

Plugging it all in, we have

\(k^2 - 4(5)(8) <0 \\ k^2 < 160\)

\(k <   12.65\)      

We are trying to find the largest INTEGER, which would be 12. 

Thanks! :)

\(XZW \) will form a triangle \(\triangle XZW \)

\(\angle ZXW  = 90^\circ \\ \angle XWZ = 60^\circ\)

So \(\angle XZW =  180^\circ - 90^\circ - 60^\circ  =   30^\circ\)

Thanks! :)

Let's first take a look at the fraction 1/17. 

It has 16 repeating digits in the form of \(0. \overline{0588235294117647}\)

We simplfy have to find which digit the 4000th lies on. 

First, let's do 4000/16. 

We get

\(4000/16=250\)

So the 4000th digit is the last digit of the repeating decimal, which is 7. 

So 7 is our answer. 

Thanks! :)