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Wait, who's that?

oh no! Alan, it is blocked by moderator! :(

Oh ok thanks!

g(x) = x + 1;

g(y) = y + 1

g(x + y) = x + y + 1

g(x)*g(y) = x*y +x + y + 1

g(x+y) + x*y = x + y + 1 + x*y

wages = hourly rate x hours worked per day x number of days 

I think it means like the following Melody:

Consider n = 9.  The factors of 9 are 1, 3 and 9. That is, there are three factors, so P(9) = 3, which means P(9)^2 = 9.

On the other hand the factors of 6 are 1, 2, 3 and 6 so P(6) = 4, hence P(6)^2 > 6

The question is asking for all the values of n that are like my first example (I think!).

I am confused, what is squared?

The radius CANNOT  be 7 units !!!

r = [sqrt( 13² + 1² )] / 2 = 6.519202405 units  

The area of the shaded region:>               A = ( r²pi ) - (7² + 6²) / 2 = 91.0176878 u² 

The usual method for this sort of problem, where the number of unknowns exceeds the number of equations by one, is to assign some arbitrary value to one of the unknowns and to then find the other unknowns in terms of that value.

With this problem then, start by letting x = t, (t an arbitrary integer). 

Then,

\(\displaystyle w(t+2)+t=348,\text{ so, }w=\frac{348-t}{t+2}.\)

\(\displaystyle w = \frac{350 -(t+2)}{t+2}=\frac{350}{t+2}-1.\)

w is to be an integer, so it's necessary that t + 2 should divide into 350 without remander.

350 has factors 2, 5, 5, 7 so t + 2 can take one of the values, 2, 5, 7, 10, 14, 25, 35, 50, 70, 175 and 350,

so t can take one of the values, 0, 3, 5, 8, 12, 23, 33, 48, 68, 173, 348.

Similarly, 

\(\displaystyle y(t+3)+4t=373, \text{ so, }y=\frac{373-4t}{t+3}=\frac{385-4(t+3)}{t+3}=\frac{385}{t+3}-4.\)

385 has prime factors 5, 7, 11, so t + 3 has to take one of the values 5, 7, 11, 35, 55, 77, 385,

in which case t has to take one of the values 2, 4, 8, 32, 52, 74, 382.

Comparing the two sets of possibles for t, the only one that appears in both is t = 8, leading to the unique solution

w = 34, x = 8, y = 31, z = 8.

Using the same routine with the original problem finds two possible solutions to the three equations in four unknowns problem,

w = 24, x = 20, y = 6, z = 14,

and 

w = 74, x = 6, y = 20, z = 4.

The fourth equation then narrows this down to a unique solution.

Ok.  Here is one way to do it:

49(pi) - (1/2)(7)(7) -(1/2)(6)(6)

A and B are angles in the interval \(0 < A, B < 45^\circ\).
If \(\cos(A + B) = \dfrac{4}{5}\) and \(\sin(A - B) = \dfrac{5}{13}\), find \(\tan(2A)\).

\(\begin{array}{|rcll|} \hline \cos(A + B) &=& \dfrac{4}{5} \\ A + B &=& \arccos\left( \dfrac{4}{5} \right) \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \sin(A - B) &=& \dfrac{5}{13} \\ A - B &=& \arcsin\left( \dfrac{5}{13} \right) \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (A + B) +(A - B) &=& \arccos\left( \dfrac{4}{5} \right)+\arcsin\left( \dfrac{5}{13} \right) \\ 2A &=& \arccos\left( \dfrac{4}{5} \right)+\arcsin\left( \dfrac{5}{13} \right) \\ \tan(2A) &=& \tan\Bigg( \arccos\left( \dfrac{4}{5} \right)+\arcsin\left( \dfrac{5}{13} \right) \Bigg) \\ \tan(2A) &=& \tan\Bigg( 36.8698976458^\circ+22.6198649480^\circ \Bigg) \\ \tan(2A) &=& \tan\Bigg( 59.4897625939^\circ \Bigg) \\ \tan(2A) &=& 1.69696969697 \\\\ \mathbf{\tan(2A)} &=& \mathbf{\dfrac{56}{33}} \\ \hline \tan(2A) &=& \tan\Bigg( 59.4897625939^\circ \Bigg) \\ 2A &=& 59.4897625939^\circ \\ \mathbf{A} &=& \mathbf{29.7448812969^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (A + B) -(A - B) &=& \arccos\left( \dfrac{4}{5} \right)-\arcsin\left( \dfrac{5}{13} \right) \\ 2B &=& \arccos\left( \dfrac{4}{5} \right)-\arcsin\left( \dfrac{5}{13} \right) \\ \tan(2B) &=& \tan\Bigg( \arccos\left( \dfrac{4}{5} \right)-\arcsin\left( \dfrac{5}{13} \right) \Bigg) \\ \tan(2B) &=& \tan\Bigg( 36.8698976458^\circ-22.6198649480^\circ \Bigg) \\ \tan(2B) &=& \tan\Bigg( 14.2500326978^\circ \Bigg) \\ \tan(2B) &=& 0.25396825397 \\\\ \mathbf{\tan(2B)} &=& \mathbf{\dfrac{16}{63}} \\ \hline \tan(2B) &=& \tan\Bigg( 14.2500326978^\circ \Bigg) \\ 2B &=& 14.2500326978^\circ \\ \mathbf{B} &=& \mathbf{7.12501634890^\circ} \\ \hline \end{array}\)

Question 1.

A mathematician works for \(t\) hours per day and solves \(p\) problems per hour, where \(t\) and \(p\) are positive integers and \(1< p < 20\).
One day, the mathematician drinks some coffee and discovers that he can now solve \(3p+7\) problems per hour.
In fact, he only works for \(t-4\) hours that day, but he still solves twice as many problems as he would in a normal day.
How many problems does he solve the day he drinks coffee?

∠IJL = 90 - ∠KJL

∠IKJ = 22 degrees

This is your question:

WXYZ is a parallelogram where angle A is half of a right angle QX = 10, and XY = 14, What is the area of the parallelogram.

Can you solve it?

Oh Q is supposed to say W I typed it wrong

thank you!

the hint helps a lot, i understand now :D

my brain works like a dead fish...so i'm really slow at understanding haha, it takes a while. 

6x +8 + 4x + 2 = 180        solve for x

The answer is (A).