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This one is also answered now :)

Thanks very much guest.  

ok I'll do the proof in full

Consider minor arc LB

This subtends the angle LRB at the circumference.  And subtends angle LNB at the centre

So     < LNB= 2*< LRB

 $Let \quad \angle LRB=\theta \qquad so \qquad \angle LNB=2\theta$

Consider minor arc RD

This subtends the < RBD at the circumference.  And subtends

So    

$Let \quad \angle RBD=\alpha \qquad so \qquad \angle RND=2\alpha$

$Consider\quad \triangle RBE\\ \angle LRB=\angle RBE+\angle REB \qquad \text{(Exterior angle of a triangle = sum of opposite interior angles)}\\ \theta=\alpha+\angle REB\\ \angle REB=\theta-\alpha\\ \angle REB=\frac{1}{2} (2\theta-2\alpha)\\ \angle REB=\frac{1}{2} (\angle LNB-\angle RND)\\ $

Consider the relation y = −5|x − 4| + 7.

What are the coordinates of the vertex, V, and which direction does it open?

The line represented by y = 2x + 1 and a line perpendicular to it intersect at R(2, 5).

What is the equation of the perpendicular line?

$\begin{array}{|rcll|} \hline y &=& 2x + 1 \quad m = 2 \quad & \quad m_{\perp} = -\frac{1}{m} = - \frac12 \\\\ y_{\perp} &=& -\frac12\cdot x + b \quad & \quad R=(2,5) \\ 5 &=& -\frac12\cdot 2 + b \\ 5 &=& -1 + b \\ 6 &=& b \\ y_{\perp} &=& -\frac12\cdot x + 6 \\ \hline \end{array}$

The equation of the perpendicular line is $y = -\frac12\cdot x + 6$

A stock market analyst observes the following for the price of two stocks that he owns, one of which is growing in a linear fashion (arithmetic) and the other of which is growing at an exponential rate (geometric).

Stock A:

$a_n = 15n+50 \\ a_{40} = 15\cdot 40 + 50 \\ a_{40} = $650$

Stock B:

$a_n = 13\cdot 1-11^{n-1} \\ a_{40} = 13\cdot 1-11^{39} \\ a_{40} = $761.27$

Assuming these stock values continue to grow in the same manner until retirement,

which stock option will be worth more in 40 years,

and how much more (rounded to the nearest cent) will this stock be worth per share?

Stock B;  $761.27- $650 = $ 111.27

What is the general-term equation, an, for the arithmetic sequence below,

and what is the 23rd term of this sequence?
$\begin{array}{|l|r|r|r|} \hline n & 1 & 2 & 3 \\ \hline a_n & 2 & -3& -8 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline a_n &=& a_1 + (n-1)\cdot d \quad & | \quad a_1 = 1 \\ && \quad & | \quad d = a_2-a_1\\ && \quad & | \quad = -3-2 \\ && \quad & | \quad = -5 \\ a_n &=& 2 + (n-1)(-5) \\ a_n &=& 2 -5n + 5 \\ a_n &=& 7 -5n \\\\ a_{23} &=& 7 - 5\cdot 23 \\ a_{23} &=& -108 \\ \hline \end{array}$

-15 +(30 + 18+ ....... )

-15+( sum of a GP with a=30, r=0.6 n=10)

$-15+\frac{a(1-r^n)}{1-r}\\ =-15+\frac{30*(1-0.6^{10})}{1-0.6}\\ =-15+\frac{30*(1-0.6^{10})}{0.4}\\$

-15+(30*(1-0.6^10))/0.4 = 59.5465 m

This is to the 10th bounce but it does not include the 10th bounce

Feel free to ask questions :)

One endpoint of a line segment is at (−3, 1).

The line is bisected by placing the midpoint of the line segment at (5, −3).

What are the coordinates of the other endpoint?

$\begin{array}{|lrcll|} \hline \frac{-3+x}{2} &=& 5 \\ -3+x &=& 10 \\ x &=& 13 \\\\ \frac{1+y}{2} &=& -3 \\ 1+y &=& -6 \\ y &=& -7 \\ \hline \end{array}$

The coordinates of the other endpoint is (13, -7)

x=352600

How many terms are in the arithmetic sequence 5, 1, −3, …, −99?
Hint: an = a1 + d(n − 1), where a1 is the first term and d is the common difference

$\begin{array}{|rcll|} \hline 5, 1, -3, \ldots, -99 \\ a_1 &=& 5 \\ a_2 &=& a_1 + d \\ d &=& a_2 - a_1 \\ d &=& 1-5 \\ d &=& -4 \\\\ a_n &=& a_1 + (n-1)\cdot d \quad & | \quad a_1 = 5 \quad a_n=-99 \quad d = -4 \\ -99 &=& 5 + (n-1)\cdot (-4) \quad & | \quad -5 \\ -104 &=& (n-1)\cdot (-4) \quad & | \quad : (-4) \\ \frac{-104}{-4} &=& n-1 \\ \frac{104}{4} &=& n-1 \\ 26 &=& n-1 \quad & | \quad +1 \\ 27 &=& n \\ \hline \end{array}$

In the arithmetic sequence 5, 1, -3, …, -99 are 27 terms.

The right size

Given the way the problem is written Guest#1 is correct!

However, if the problem should have said "Max gave 2/5 of his beans away to his brother, and consumed 1/3 of his remaining beans. He takes the remaining amount of beans and places it evenly into 3 containers. If each container had 2 beans how many beans did he have to start (in the beginning)?", then:

Let B be the original number:

After giving away 2/5B max is left with 3/5B.

After eating 1/3 of this Max is left with (2/3)*3/5B → 2/5B

This equals 2*3 → 6

so 2/5B = 6

B = 6*5/2  → 15

So Max originally had 15 beans.

2 out of 10 digits are greater than 7, so the probability of choosing these is 2/10.  Since there are 25 students the number likely to choose a number greater than 7 is (2/10)*25 or 5.

.

Quote from Wikipedia:

"A t-test is any statistical hypothesis test in which the test statistic follows a Student's t-distribution under the null hypothesis. It can be used to determine if two sets of data are significantly different from each other."

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No problem. Try to figrue out what you in putted wrong.

thanks for your anwser...

You may be inputting the problem wrong.

6/2(2+1) = 3(3) = 9

(1/2) * (1/3) * (1/2) = 1 / 12

Chance of Dairy being ice cream: (1/2)

Chance of Fruit being pinneapple: (1/3)

Chance of Topping being granola: (1/2)

Remember that θ is the angle that a line from the origin to the point makes with the positive x axis. That is, the line to the right of the y-axis.

To find r, we can use the Pythagorean theorem.

(-52)² + 75² = r²

$\sqrt{2704+5625}$ = r

$\sqrt{8329}$ = r

r ≈ 91.26

To find θ, we can take the arctan(75/-52) ...

arctan(75/-52) ≈ -55.27º

But..the calculator gives us an angle that lies in the fourth quadrant.

The tangent values repeat every 180º ...so just add 180º to get the angle that has the same tangent in the second quadrant.

-55.27º + 180º = 124.73º ≈ θ

So...the polar coordinates are (91.26, 124.73º)  

No problem @MagikalRedNiteTiger. Now that AP testin is over, my mentality has improved... Yet l'm scarred.

1/12

dude really that's a stretch and really firefox also a major stretch