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Let us denote the center as point o.

So angle NOL is 180-90-15=75 and so 5x=180-75 so 5x=105 ad x=21

21 is our solution.

5x and the right angle+15 are vertical angles so they are equal. 

5x=90+15

5x=105

x=21

(p,q) = (8,-5).

(a) f(x) = (x + 3)/(x - 3).

(b) f(x) = (x + 3)/(2x - 6).

to solve this you need to calculate 525*14 and substract it from $31770, you have $24420

 now the amount of tickets for rear and near the front seats is the same

you need to obtain the average value from all the tickets, (14+19)/2.

16.5 is the average. you divide 24420 in 16.5 and you have 1480 tickets sold.

740 tickets from  seats near the front and (740+525)= 1265 tickets for the rear seats.

740 tickets in front 

1265 tickets in back 

The midpoint divides the line into two equal lengths so 8x+11 must be equal to 12x-1.

8x+11=12x-1

4x=12

x=3.

First we convert ax+by=3 into slope intercept form: -ax/b+3/b=y

We know that it is parralell to x+3y=-5 (-x/3-5/3=y in slope intercept form) so the slopes are the same. There for the slope is -1/3. 

Putting this into the 1st equations we get -1/3x+3/b=y and also putting in the point (-7,2) we get 7/3+3/b=2. 

From this we can solve that b=-9

Plugging everything (including b and point (-7,2)) back in the original equation ax+by=3, we get -7a-18=3 so a=-3

And thus b-a=-9-(-3)=-9+3=-6

So b-a is equal to -6 that is our answer

Lets first denote x and y as the 2 numbers, so this means:

1.x^2+y^2+225

2.(x+y)^2=441

Expanding the second equation we get that x^2+2xy+y^2+441

Subtracting the 2 equation from the 1st one we get 2xy=441-225 2xy=216 and xy=108

The problem asks for the reciprocal of thier product so 1/xy=1/108

And thus 1/108 is our final answer

2. From x + 3y = -5, y = -x/3 - 5, so the slope of the line is -1/3. The slope of the new line is also -1/3, so y = -x/3 + B.  Pugging in x = 2 and y = -7, we get -7 = -2/3 + B, so B = -19/3.

Then the line is y = -x/3 - 19/3.  Then 3y = -x - 19, so 3y + x = -19.  We want the right-hand side to be 3, so we mutiply both sides by -3/19: -9/19*y - 3/19*x = 3.  Therefore, B - A = -9/19 - 3/19 = -12/19.

3. The solution is (-5/3,inf).

Thanks for helping out Mathulator. Unfortunately, \($4^{50}$\) is incorrect. 

Hi Alan, I was just wondering. The questions say that a and b are complex numbers but your solutions are not complex numbers. 

Thank you so much Alan!

Thanks guest!

Deleted as it was nonsense! See Guest#5 answer below.

Hi melody!

I just fixed them. It was because there was an extra space at the end, which caused a lot of problems...

Thank you for telling me!

Like so:

Hi, Nacirema!

Numbers 7 and 13 represent the areas of 2 semicircles.

The triangle shown is a right triangle. The semicircles have the sides of the triangle as diameters. The areas of two of the semicircles are shown. What is the area of the third semicircle?

1/    {sqrt[(7 * 2) / pi]} * 2 = 4.222008246           ( short leg )

2/    {sqrt[(13 * 2) / pi]} * 2 = 5.753627392         ( long leg )

3/     {[sqrt(4.222008246² + 5.753627392²)] / 2}² *pi / 2 =  20 u²     ( area of largest semicircle )        

Here is the answer:

Find i^1+i^2+i^3+...+ i^97 + i^98+i^99.

\(\begin{array}{|rcll|} \hline s &=& i^1+i^2+i^3+\ldots+ i^{97} + i^{98}+i^{99} \\ i*s &=& i^2+i^3+\ldots+ i^{97} + i^{98}+i^{99}+i^{100} \\ \hline s-i*s &=& i^1 - i^{100} \\\\ s(1-i) &=& i - i^{100} \\\\ s &=& \dfrac{i - i^{100}}{1-i} \\\\ s &=& \dfrac{i - i^{2*50}}{1-i} \\\\ s &=& \dfrac{ i - \left(i^2\right)^{50} }{1-i} \quad | \quad i^2 = -1 \\\\ s &=& \dfrac{ i - \left(-1\right)^{50} }{1-i} \\\\ s &=& \dfrac{ i - 1 }{1-i} \\\\ s &=&- \dfrac{ (1 - i) }{(1-i)} \\\\ \mathbf{s} &=& \mathbf{-1} \\\\ \mathbf{i^1+i^2+i^3+\ldots+ i^{97} + i^{98}+i^{99} } &=& \mathbf{-1} \\ \hline \end{array}\)

We see that adding 8 units to the x coordinate and adding -8 to the y coordinate to point a gives us point b. Therefore we must just multiply 3/4 to both the 8 and -8. This gives us adding 8*3/4=6 units to the x coordinate of a and -8*3/4=-6 to the y coordinate of a to get our final point.
 And thus (-2,1) is our final answer

That is ok Jenny. I am not terribly comfortable with limits myself.

If you do not understand it is quite reasonable for you to ask again.

So long as you do the best you can,to get as much as possible, from already provided answers, more questions are fine.

First We look at the traingle and it has legs of 6 and 8. We know that the area of the triangle is b*h/2, substitting the values of 6 and 8 as b and h we get that the area of the traingle is 6*8/2 is 24ft^2

Next we look at the semi circle it has a diameter of 10 so it has a radius of 5. We also know that the area of a semi circle= half a circle is pi*r^2/2 so substuting the raduis of 5 in we get pi*25/2=25pi/2.

Seeing the problems details we use 3.14 as pi and we get 25*3.14/2=39.25ft^2 as the semi circle area.

Finally we add the traingle area and the semi circle area together 24+39.25=63.25ft^2

63.25 ft^2 is our final answer