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! :)