All Answers 352838 Answers

2 sin A  = 3 cos A     square both sides

4 sin^2 A  = 9 cos^2 A

4 sin^2 A = 9 [ 1 - sin^2 A]

4 sin ^2 A  = 9 - 9 sin^2 A     rearrange as

13 sin^2 A    = 9

sin^2 A  = 9 /13       sinc A is acute and  < 90°   ...take the positive root

sin A  = 3 / √13   =  ( 3 √13 ) / 13

Thank you Heureka, must have just been a typo haha!

Circle  is the incircle of triangle ABC and is also the circumcircle of triangle XYZ. The point X is on line BC, point Y is on overline AB, and the point Z is on line AC. If angle A=40 degrees, angle B=60 degrees, and angle C=80 degrees, what is the measure of angle AYX?

\(\text{Let the center $M$ of the incircle, $\\$called the incenter,$\\$can be found as the intersection of the three internal angle bisectors }\)

\(\begin{array}{|rcll|} \hline \angle XBM &=& \dfrac{60^{\circ}}{2} \\ &=& 30^{\circ} \\\\ \angle XMB &=& 90^{\circ}- \angle XBM \\ &=& 90^{\circ}-30^{\circ} \\ &=& 60^{\circ} \\\\ \angle XMY &=& 2\times \angle XMB \\ &=& 2\times 60^{\circ} \\ &=& 120^{\circ} \\\\ \angle MYX &=& \dfrac{180^{\circ}-\angle XMY}{2} \\ &=& \dfrac{180^{\circ}-120^{\circ}}{2} \\ &=& 30^{\circ} \\\\ \angle AYX &=& 90^{\circ} + \angle MYX \\ &=& 90^{\circ} + 30^{\circ} \\ & \mathbf{=}& \mathbf{120^{\circ}} \\ \hline \end{array}\)

Ummm, ok thanks. Though I think it would've been better to multiply both sides by four. And then, since \(225=3^2\cdot5^2\), there are \(3\cdot3=9\) positive integer divisors of \(225 \). Of these, \(\boxed{5}\) yield ordered pairs of divisors \((2a-15,2b-15)\) for which \(a \geq b\)

2/15  =

1/8  + 1 / 120

1/ 9  +   1 / 45

1/10  +  1  /30

1/ 12  + 1 / 20

I can generate the first through something called "Egyptian Fractions"....I don't know how the other three are obtained

You can read here : http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html

For each of the partitioned sets, calculate the unique permutations using N!/((d₁!)* (d₂!) *...(d_z!)) Where N is the number of elements and d_z is the number of repeated elements (d) of each kind (_Z).  Ex:  for 4,2,2,1,1  it’s 5!/((2!)*(2!)) = 30. This represents 30 uniquely identifiable arrangements.  

\(\begin{array}{|r|r|r|r|r|r|r|r|r|} \hline \overset {..}\smile &G & I & N & G & E & R&A &L &E & \overset {..}\smile \\ \hline 10 & & & & & & & & & & 1\\ \hline 9 & 1 & & & & & & & & & 2 \\ \hline 8 & 2 & & & & & & & & & 2 \\ \hline 8 & 1 & 1 & & & & & & & & 3 \\ \hline 7 & 3 & & & & & & & & & 2 \\ \hline 7 & 2 & 1 & & & & & & & & 6 \\ \hline 7 & 1 & 1 & 1 & & & & & & & 4 \\ \hline 6 & 4 & & & & & & & & & 2 \\ \hline 6 & 3 & 1 & & & & & & & & 6 \\ \hline 6 & 2 & 2 & & & & & & & & 3 \\ \hline 6 & 2 & 1 & 1 & & & & & & & 12 \\ \hline 6 & 1 & 1 & 1 & 1 & & & & & & 5 \\ \hline 5 & 5 & & & & & & & & & 1 \\ \hline 5 & 4 & 1 & & & & & & & & 6 \\ \hline 5 & 3 & 2 & & & & & & & & 6 \\ \hline 5 & 3 & 1 & 1 & & & & & & & 12 \\ \hline 5 & 2 & 2 & 1 & & & & & & & 12 \\ \hline 5 & 2 & 1 & 1 & 1 & & & & & & 20 \\ \hline 5 & 1 & 1 & 1 & 1 & 1 & & & & & 6 \\ \hline 4 & 4 & 2 & & & & & & & & 3 \\ \hline 4 & 4 & 1 & 1 & & & & & & & 6 \\ \hline 4 & 3 & 3 & & & & & & & & 3 \\ \hline 4 & 3 & 2 & 1 & & & & & & & 24 \\ \hline 4 & 3 & 1 & 1 & 1 & & & & & & 20 \\ \hline 4 & 2 & 2 & 2 & & & & & & & 4 \\ \hline 4 & 2 & 2 & 1 & 1 & & & & & & 30 \\ \hline 4 & 2 & 1 & 1 & 1 & 1 & & & & & 30 \\ \hline 4 & 1 & 1 & 1 & 1 & 1 & 1 & & & & 7 \\ \hline 3 & 3 & 3 & 1 & & & & & & & 4 \\ \hline 3 & 3 & 2 & 2 & & & & & & & 6 \\ \hline 3 & 3 & 2 & 1 & 1 & & & & & & 30 \\ \hline 3 & 3 & 1 & 1 & 1 & 1 & & & & &15 \\ \hline 3 & 2 & 2 & 2 & 1 & & & & & & 20 \\ \hline 3 & 2 & 2 & 1 & 1 & 1 & & & & & 60 \\ \hline 3 & 2 & 1 & 1 & 1 & 1 & 1 & & & & 42 \\ \hline 3 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & & 8 \\ \hline 2 & 2 & 2 & 2 & 2 & & & & & & 1 \\ \hline 2 & 2 & 2 & 2 & 1 & 1 & & & & & 15 \\ \hline 2 & 2 & 2 & 1 & 1 & 1 & 1 & & & & 35 \\ \hline 2 & 2 & 1 & 1 & 1 & 1 & 1 & 1 & & & 28 \\ \hline 2 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & 9 \\ \hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \overset {..}\smile &G & I & N & G & E & R&A &L &E & \downarrow \\ \hline \end{array}\\ \small \text {Sum of all unique permutations } \hspace{2.85cm} \fbox {512}\\\)

EDIT: Corrected Errors

GA

Great thank you so much!

In right triangle DEF, we have angle E = 90 degrees and tan D = 3*sin D. What is sin F?

tan D  = 3 sin D

sin D  / cos D  =   3 sin D        

Since  0° < D < 90°, then sin D  >  0

So...divide both sides by sin D and we have that

1 / cos D  = 3  .......rearrange as

cos D   = 1/3

Since D and F  are  complementary, then

cos D  = 1/3  =  sin  F

"at a rate of ???? cubic meters per hour."

Hmmmmmmmmmm....something seems to be missing!

Thank you so much CPhill!

4.

In one period, the sine will be negative  from   180°  to 270°   and from 270° to 360°

So.....  360 + 360  + 270  = 990°

So....the sin x  = -0.31  =  [ 2 + 2  + 1 ]  =   5 times  from 0°  to 990°

We need to find the radius of the Earth  as follows :

25100  = 2 pi * radius

25100 / [ 2 pi]  = radius

12550 / pi   =  radius  [in miles]

At  45°,  the radius of the smaller circle parallel to the equator running through Lena will be =

12550  /  [ √2 pi]  miles

So....the circumference  of this smaller circle will be   =

2 pi   [ 12550 / (√2 pi ) ] =

√2 * 12550  ≈  17,748 miles  = 17,700 miles

My guess is 24

You have two places to start

House - car

car - House       then the next 3 spaces have 6 possibilities

2 x 6 = 12 ways.        My guess.

3:  ABCD is a regular tetrahedron (right triangular pyramid). If M is the midpoint of line CD, then what is cos (angle AMB)?

Let AB  =  side of the tetrahedron  = s

Then BM =  AM  =   s * sqrt(3)/2

Using the Law of Cosines, we have that

AB^2  = BM^2 + AM^2  -  2 (BM)(AM) cos AMB

s^2  =   (3/4)s^2 + (3/4)s^2  - 2 ( 3/4)s^2 cos AMB

s^2  = (3/2)s^2  - (3/2)s^2  cos AMB

- (1/2)s^2   =  -(3/2)s^2  cos AMB

[ 1/2]s^2  / [ (3/2) s^2 ]  = cos AMB

[ 1/2]  / [3/2] = cos AMB

(1/2)( 2/3) = cos AMB

1/3   = cos AMB 

2:  In triangle GHI, we have GH = HI = 25 and GI = 30. What is sin (angle GHI)?

                                       H

                              25             25

                             G       30          I

This ttrangle is isosceles

Using the Law od Cosines

30^2   =  2 (25)^2 - 2(25)^2  cos A

-350 / [ -1250 ]  = cos A

cos A  =  7/25

sin A  = sqrt  [ 1 - cos^2 A ]  = sqrt [ 1  - (7/25)^2 ]  = sqrt [ 625 - 49] / 25  = 

sqrt [ 576] / 25   =    24 /25

1:  If A is an acute angle such that tan A + sec A = 2, then find cos A.

tan A  + sec A   = 2

sin A / cos A  +   1 / cos A  = 2

(sin A + 1 ) /cos A   =  2   ......  since A  is acute, then A < 90...so cos A is not 0

So ...multiply both sides by cos A

sin A + 1  = 2 cos A          square both sides

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

sin^2A + 2sinA + 1  = 4(1- sin^2 A) ..... simplify

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

(5 sin A  - 3 ) ( sin A + 1)  =  0

So....we have two possible solutions

5  sinA - 3 =  0               sin A + 1  = 0

5 sin A  = 3                     sin A  = -1

sin A  = 3/5                     A  =  270°    (reject)

Since the sin A  = 3/5....then cos A  =   sqrt [  1 - ( 3/5)^2 ]  = sqrt [ 25 - 9] / 5  =

sqrt (16)/5   =   4/5

Since 98 =2 x 7^2, it, therefore, follows that if we multiply 98 x [2^3 x 7^2] =98 x 392 =[2^4 x 7^4]^1/4 = 2 x 7 = 14. So, the number you are looking for is 392, which gives 14 as an integer.

By the Triangle Inequality, the sum of any two sides is greater than the remaining side

So we have these inequalities :

11 + n  > `17      subtract  11 from both sides

n  >  6

And

11 + 17  > n    

28 > n  ⇒    n < 28

So....combining these, we have

6 < n < 28

So......the number of possible integer lengths for the remaining side  =

[28  - 6]  - 1  = 

22 - 1  =  

21

Do you want the graph of this  ???

OK, The partitions of(10) =42 distinct partitions as follows:

This is called the "Partition Function = P(10)"

10 = 10
9 + 1 = 10
8 + 2 = 10
8 + 1 + 1 = 10
7 + 3 = 10
7 + 2 + 1 = 10
7 + 1 + 1 + 1 = 10
6 + 4 = 10
6 + 3 + 1 = 10
6 + 2 + 2 = 10
6 + 2 + 1 + 1 = 10
6 + 1 + 1 + 1 + 1 = 10
5 + 5 = 10
5 + 4 + 1 = 10
5 + 3 + 2 = 10
5 + 3 + 1 + 1 = 10
5 + 2 + 2 + 1 = 10
5 + 2 + 1 + 1 + 1 = 10
5 + 1 + 1 + 1 + 1 + 1 = 10
4 + 4 + 2 = 10
4 + 4 + 1 + 1 = 10
4 + 3 + 3 = 10
4 + 3 + 2 + 1 = 10
4 + 3 + 1 + 1 + 1 = 10
4 + 2 + 2 + 2 = 10
4 + 2 + 2 + 1 + 1 = 10
4 + 2 + 1 + 1 + 1 + 1 = 10
4 + 1 + 1 + 1 + 1 + 1 + 1 = 10
3 + 3 + 3 + 1 = 10
3 + 3 + 2 + 2 = 10
3 + 3 + 2 + 1 + 1 = 10
3 + 3 + 1 + 1 + 1 + 1 = 10
3 + 2 + 2 + 2 + 1 = 10
3 + 2 + 2 + 1 + 1 + 1 = 10
3 + 2 + 1 + 1 + 1 + 1 + 1 = 10
3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10
2 + 2 + 2 + 2 + 2 = 10
2 + 2 + 2 + 2 + 1 + 1 = 10
2 + 2 + 2 + 1 + 1 + 1 + 1 = 10
2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 10
2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 10

Now you can re-arrange these partitions into any order you want such as:3+3+2+2, 3+2+2+3, 2+2+3+3, 2+3+2+3.......etc. I would estimate that you would have more than 100 such re-arrangements.

2.75/3.5  x 42 mi =33 miles

So, the distance in inches on a map, between Tinsel Town and Emerald City is 3.5 inches, and the actual distance is 42 miles.

And, the problem is asking us to find the actual distance between Emerald City and Diamond Bluff, if it's 2.75 inches apart.

 We can first solve for 1 inch, then multiply the result by 2.75 inches, to find how far apart 2.75 inches is in miles(on a map.)

  So, \(3.5n=42m \longrightarrow n=12\)

We have solved for n, but we have to solve for 2.75n, which we can do: \(2.75*12=\boxed{33}\) miles.

Assuming that we cannot repeat any natural number....

10       ( 1 way)

9, 1     ( 2 ways)

8, 2     (2 ways)

7, 3     (2 ways)

6, 4     (2 ways)

7, 2 , 1   (6 ways)

6, 3 , 1   (6 ways)

5, 4, 1   (6 ways)

4, 3, 2 , 1    (24 ways)

So....   1 + 4(2)  + 3(6)  + 24    =     51 ways

I think the answer to the first question is  20 ways.....

To see this.....let Anne occupy one position in a circle.....and we will fill the sets in a  clockwise manner,.....  (we could also choose a counter-clockwise manner.....the results will be the same )

To Anne's right, we can choose any of the other 5 classmatres to fill this position.....and to this person's right, we can choose any 4 of the remaining classmates.....and these three comprise one set.....so, by default.....the other set will be composed of the classmates we haven't chosen

So......5 ways  X 4 ways  = 20 ways to divide the class