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I;m using x  instead of theta  (it doesn't matter )

Note  that      sin (-x)  = - sin (x)     and  sec (-x )  =  sec (x)

sin (-x) tan (x)        - sin (x) sin x / cos x           - sin^2 x

___________   =  __________________  =   ____________   =  - sin^2 x

  sec (-x)                       sec (x)                          (1/cos x) cos x

tan A  + sec A =  2     

sin A  / cos A  +  1/cos A  =  2   multiply both sides by cos A

sin A + 1   =  2cosA        square both sides

sin^2A + 2sin A + 1 = 4cos^2A

sin^2A + 2sinA + 1 =  4 ( 1 - sin^2A)

sin^2A + 2sin A + 1  = 4 - 4sin^2 A

5sin^2 A + 2sin A - 3   =  0    factor as

(5sin A - 3 ) (sin A + 1 )  =  0

So either   sin A + 1  = 0     ⇒  sin A = - 1      ⇒ A  = 3pi/2  not acute

Or

5sin A - 3  =  0

sin A  = 3/5

So we have a 3 - 4  - 5  right triangle

So

cos A   =  4/5

Proof

tanA  + sec A   =     3/4 + 5/4  = 8/4  =  2

As follows:

Hi guest!
 

So first we can start by adding 8 to both sides of the equation. 

\(x^2 + y^2 - 10x + 8y = 8\)

Let's start by completing the square for \(x^2-10x\)

\((x-5)^2\). But, we have an extra \(+25\) term, so we have to subtract 25.

If we complete the square for \(y^2+8y\), we get \((y+4)^2\). But, we have an extra \(+16\) term, so we have to subtract 16.

So, our equation now looks like 

\((x-5)^2+(y+4)^2-16-25=8\)

\(\boxed{(x-5)^2+(y+4)^2=49}\)

I'm assuming you know the form of the circle, so you should be able to do the rest from here easily. You got this! :)

I hope this helped you, guest!

:)

One question per post please.

1. This is mostly geometry believe it or not.

The value of K in the circle equation would be the blue line squared. Or in other words, the blue line is \(\sqrt{k}\)

The value of BC in the tangent equation would just be \(k\)

We know ABC is a 45 - 45 - 90 triangle (If the coefficients of x and y are 1 when graphed in standard form for the red line). The blue line is the perpendicular bisector. 

Based on pythagorean theorem: \(2 *\sqrt{k}^2 = k^2\) (two legs are \(\sqrt{k}\) and hypotenuse is BC)

\(2k=k^2\)

\(k(k-2)=0\)

\(\boxed{k=2}\)

Hi guest!

This is actually really simple if you just plug everything in!

\(8 \odot 6 = 8 + \frac{3(8)}{2(6)}\)

I think you can solve it from here! You got this! 

:)

I'm getting (a,b,c) = (2,4,2).

great job Guest! 

thanks for helping!!! I really appreciate it :)))

Hi guest!
 

The formula for a series is \(\frac{a_1+a_n}{2}\cdot n\) 

n = how many terms there are

So, we know \(a_1=10\) since it's the first term (in this case let's save the 7 for the end, since it doesn't follow the pattern)

\(a_n=479\) since it's the last term

There are 68 terms because there are 479-10=469, 469/7 = 67 terms, but you have to add 1 since we didn't count 10, so there are 68 terms.

Now, all we have to do is plug the equation in!

 \(\frac{10+479}{2} \cdot 68=16626\). We have to add that first 7 too, so \(16626+7=\boxed{16633}\).

I hope this helped you, guest!

:)

Sorry, I did have both of them but didn't read the question correctly:

(189, 300, 300, 300, 300, 300, 300, 300, 300, 300) >>>Total Number = 2889
(0, 300, 600, 900, 1200, 1500, 1800, 2100, 2400, 2700)>>Total Sum = 13500

2) The center of the circle is (0, 7/4).

The relfected point is (12/5, 51/5).

1 / 2
1 / 3
1 / 4
1 / 5
1 / 6

2 / 3
2 / 4 - Reducible
2 / 5
2 / 6 - Reducible

3 / 4
3 / 5
3 / 6 - Reducible

4 / 5
4 / 6 - Reducible

5 / 6
Total =15 - 4 Reducibles =11 / 15 - The probability that the fraction cannot be reduced.

The probability is 23/80.

thanks Guest! even though this isn't the answer, your explanation was the part I was stuck on! (I couldn't figure out how many 1s, 2s, 3s, etc. there were) 

So the answer is 1+2+3...9 =45 *300 = 13500.\(\)

(a) Let P be the intersection of AD and MN, and let Q be the intersection of AL and MN.  Then by Menelaus's Theorem, Q is the midpoint of PK.  Since AD and LK are parallel, triangles ADL and LKQ are similar.  And since Q is the midpoint of PK, QK = PK/2 = DL/2.  Therefore, from the similar triangles, AD/LK = 2.

(b) Since QK = DL/2 and QG = LG/2, triangles GDL and GKQ are similar.  Then KG = DG/2, so by part (a), triangles ADG and LKG are similar.

(c) In part (b), we found that triangle GDL and GKQ were similar.  Then angle QKG = angle LDG, so D, G, and K are collinear.  And by the similarity in part (b), DG/GK = 2.  Easy!

Each digit is written 99 times, so the sum is (0 + 1 + 2 + ... + 9)*99 = 4455.

189 , 300 , 300 , 300 , 300 , 300 , 300 , 300 , 300 , 300 >>Total = 2889

sorry that's not correct... but thanks for helping :)

7 + sum_(n=1)^68(7 n + 3) = 16,633

The probability is 1/7.

6.54 r   +   7.28 (6- r)       =   7.1  (6)         solve for r= the # raisins then  nuts = 6-r 

sorry, that's wrong too... thanks for taking the time to help though :)