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3x - 4 > -2x + 1    |+2x

5x      > 1             |+4

5x      > 5             | : 5

  x      > 1

In rectangle EFGH, EH = 3 and EF = 4. Let M be the midpoint of \(\overline{EF}\),
and let X be a point such that MH = MX and \angle \(MHX = 72^\circ\), as shown below.
Find \(\angle XGH\), in degrees.

\(\begin{array}{|rcll|} \hline 2x &=& 180^\circ-2*72^\circ \quad | \quad : 2\\ x &=& 90^\circ-72^\circ \\ \mathbf{x} &=& \mathbf{18^\circ} \\ \hline \end{array}\)

This is very simple. All you need is a basic trig and the pythagorean theorem.  

You will want to find the values of "c" and "a" so you must use the given values to find this.

Using the given values and some minor rearranging we can get following the 2 equations:

\(-3 = a+c\)

\(60 = 64a + c\)

We can now put these 2 equations into matrix form and rearrange to give the unknowns. See below.

\(\begin{pmatrix} 1 & 1\\ 64 & 1 \end{pmatrix} * \begin{pmatrix} a\\ c \end{pmatrix} = \begin{pmatrix} -3\\ 60 \end{pmatrix} \)

\(\therefore\begin{pmatrix} a\\ c \end{pmatrix} = {\begin{pmatrix} 1 & 1\\ 64 & 1 \end{pmatrix}}^{-1} * \begin{pmatrix} -3\\ 60 \end{pmatrix}\)

I used a calculator to perform this operation, I got:

\(a = 1, c = -4\)

to check that this works we can sub these values into one of the original equations, i'll use one for an example:

\(-5 = 1*({1}^{3})-2(1)-4\)

\(-5 = -5\)

This means that the values hold and the equation is:

\({x}^{3}-2x-4\)

Now to find the possible values of x you must allow the equation to equal zero and factor and use quadratic formula:

\(0 = {x}^{3} - 2x-4\)

factor out (x-2):

\((x-2)(x^2 + 2x + 2) = 0\)

from this we can gather taht a possible value for x is 2.

I wont proceed to do the calculations here, however using the quadratic formula, you can gather the remaining 2 possible answers.

the 3 possible values of x is:

\(x=2, x=-1+i, x=-1-i\)

hope this was helpful :)

would it simply be:

\(\theta = {tan}^{-1}(-\frac{6}{7})\)

\(\theta = -40.6 + 180\)

or am i missing something here!?

There are four unequal, positive integersa, b, candNsuch thatN= 5a+ 3b+ 5c.
It is also true that N= 4a+ 5b+ 4c and N is between 131 and 150.
What is the value of a+b+c?

\(\begin{array}{|rcll|} \hline N &=& 5a+ 3b+ 5c \\ -N &=& 4a+ 5b+ 4c \\ \hline 0 &=& a-2b+c \qquad \text{or} \qquad \mathbf{a+c= 2b} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline N &=& 5a+ 3b+ 5c \\ N &=& 5(a+c) + 3b \quad | \quad a+c= 2b \\ N &=& 5*2b + 3b \\ \mathbf{N} &=& \mathbf{13b} \qquad \text{, where $b$ is an integer} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{Multiples of $13$ between $131$ and $150$:} \\ \hline 13\times10 &=& 130 \\ 13\times11 &=& 143\ \leftarrow \ \text{ the only one} \\ 13\times12 &=& 156 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{N} &=& \mathbf{13b} \quad | \quad \mathbf{N=143} \\ 143 &=& 13b \quad | \quad : 13 \\ 11 &=& b \\ \mathbf{b} &=& \mathbf{11} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a+c} &=& \mathbf{2b} \quad | \quad \mathbf{b=11} \\ a+c &=& 2*11 \\ \mathbf{a+c} &=& \mathbf{22} \\ \hline a+b+c &=& (a+c) +b \\ a+b+c &=& 22 +11 \\ \mathbf{a+b+c} &=& \mathbf{33} \\ \hline \end{array}\)

Arc WY works out to (170 - 110)/2 + 30 = 60 degrees.

I'm not sure about the approach you took, but have you tried multiplying both sides by the denominator(\(9\sin{\theta}-\cos{\theta}\)) from the beginning? Maybe that would get you somewhere, since you know that \(r * \cos{\theta} = x \) and \(r * \sin{\theta} = y\)

Not sure if you're asking for the solution, but if you are, then you can keep reading I guess.

Using the method mentioned above:

\(9r\sin{\theta} - r\cos{\theta} = 7\)

Using our substitutions, we rewrite this as:

\(9y - x = 7\)

From here on out, it gets pretty intuitive. Dividing and rearranging to get our desired form, we get:

\(9y = x+7\)

\(y = x/9 + 7/9\)

or 

\(y = \frac19 x + \frac79\)

Hope this helped!

Are you serious, Guest? I'm pretty sure you can work this one out.

If not, the answer is -1, 0, 1, 2, 3.

Because -1 +0 +1 +2 +3 = 5.

Yes, the answer is 0.32.

This is because the probability for 1 shoe in the closet is 0.12, and the probability for 2 shoes in the closet is 0.2.

Add those chances up and you get 0.32 as the probability that a teenager has fewer than 3 shoes in their closet.

Hey look in case the teacher was playing a prank I wanted to make it clear that I am in no way liable LOL.

lmao you arent wrong

In total there are 35 students.

5 students made an A, so 5/35 is the fraction form.

5/35 as a percentage is 14.29%.

Since There are 5 people getting an A in 35, 5/35 = 1/7

You can calculate decimal yourself.

Note that this may be incorrect. From a mathmatical point of view, no, but in real life, yes.

Grades are too hard to calculate. THere are WAY too much variables.

A einstein kid could be depressed and get a D.

A idiotic kid could have bribed a teacher and gotten an A.

Thus you must note that this is theoretical probability based on information given, and strictly this. This probability can change year to year, trimester to trimester, or even day to day! 

A straight A kid can flunk a test and get a F since his friend betrayed him.

You never know.

x= 1/2, 0

Just trust me.

nice response, I would have left this guy to think about this easy question lmao.

Hi. I have offered a hint on another thread containing the same question. It should be very helpful.

tysm!!! 

great answer, guest! :)

d=0;a=1; b=1;c=3*a + b; if(c==100, goto5, goto7);d=d+1;printd,"-", a, b; a++;if(a<200, goto3, 0);a=1;b++;if(b<200, goto3, discard=0;print d
n   x  y
1 - 33 1
2 - 32 4
3 - 31 7
4 - 30 10
5 - 29 13
6 - 28 16
7 - 27 19
8 - 26 22
9 - 25 25
10 - 24 28
11 - 23 31
12 - 22 34
13 - 21 37
14 - 20 40
15 - 19 43
16 - 18 46
17 - 17 49
18 - 16 52
19 - 15 55
20 - 14 58
21 - 13 61
22 - 12 64
23 - 11 67
24 - 10 70
25 - 9 73
26 - 8 76
27 - 7 79
28 - 6 82
29 - 5 85
30 - 4 88
31 - 3 91
32 - 2 94
33 - 1 97

100x mod 997 =1, solve for x
x =997c + 668, where c=0, 1 ,2 ,3......etc.
The smallest x = 668

The length of a rectangle is 4 meters more than three times its width.  The perimeter of the rectangle is 56 meters.  What is the width?

w + 3w + 4 = 56/2

4w + 4 = 28

4w = 24

w = 6