All Answers 352880 Answers

First one

This is a little tricky.......r is always positive.....so....we need to add 180° to 120°

(-3, 120°)  =  ( 3, 300°)

So

cos 300 =  x / 3

3cos300 = x  

3 (1/2)  =  x  

1.5  = x

And

sin300 = y / 3

3sin300 = y

3 * -√3/2   = y  

-(3/2)√3   = y

-3√3 / 2   = y

So

(x , y )   =   ( 1.5, -3√3 / 2 )

2.Let $C$ be a point not on line $AE$ and $D$ a point on line $AE$ such that $CD \perp AE.$ Meanwhile, $B$ is a point on line $CE$ such that $AB \perp CE.$ If $AB = 4,$ $CD = 8,$ and $AE = 5,$ then what is the length of $CE?$

                    C

                           B

              4

                     8

  A                D                                   E

                       5

We have two triangles ...  ABE  and CED

Angle ABE = Angle CDE

Angle BEA = Angle CED

Therefore.....by AA congruency....

Triangle ABE  is similar to Triangle  CDE

So

AE / AB   =  CE / CD

5 / 4   =  CE / 8        cross-multiply

8 * 5 /  4   = CE

40 / 4   =  CE

10  = CE

\(BD = k \sqrt{2}\)

1. In triangle $ABC$, $AB = 3$, $BC = 4$, $AC = 5$, and $BD$ is the angle bisector. If $BD = k \sqrt{2}$, then find $k$.

A

3           5

B       4           C

The bisector will divide AC  into  a ratio  of 3 : 4   =  AD : CD

Therefore.....AD  =  (3/7)*5  =    15/7

The sine of BAC = 4/5

So  the cosine of BAC  =  √ [ 5^2 - 4^2 ] / 5  =   3/5

Using the Law of Cosines

BD^2   = AB^2 + AD^2  - 2 (AB * AD)cos BAC

BD^2 =  3^2  + ( 15/7)^2 - 2(3 * 15/7) (3/5)

BD^2  = 9 + 225/49 - 2(3 * 3 * 3) / 7

BD^2 =  9 +  225/49 - 54/ 7

BD^2  =   [ 441 + 225 - 378] / 49

BD^2  = 288 /49

BD = √288/ 7  =   (12/7) √2

So....   k  = (12/7)

sin (5x) / 2x    multiply top/bottom by 5

5sin (5x) / [ 5 * 2x]

(5/ 2)  sin (5x) / [ 5x]

As x  ⇒  0  the limit   of  sin (5x) / [5x}  = limit of  sin (0) / 0   =  1

So....5/2    is correct !!!!

tan x              sinx / cosx                      sin x * cos x            sin x

_____    =      ____________ =          ___________ =      _____

xsecx               x  [ 1 /  cos x]               x * cos x                   x

And the limit  of the last function as x ⇒ 0   =   1     !!!!

Correct on both counts !!!

(x^2-5x-24)/(x-8)  =

(x + 3) ( x - 8) / ( x - 8)  =

x + 3

This will be a line with a "hole" at x = 8

We can make this coninuous by a piecewise function

f(x)   =  (x^2-5x-24)/(x-8)    if x  not equal to 8

                  11                        if x   =  8

Wow, now I get it, thanks Melody (and everyone else too)!

Q

P    6√2         R

Since angle Q  = angle R.......this is a 45 - 45 - 90 right triangle

Angle P is right

And the sides opposite the other two angles are equal....so... PR  = QP

But these are the leg lengths

And the area of a right triangle  =

[ Product of leg lengths ] / 2   =  

[ 6√  2  * 6√2 ] / 2  =

[ 36 * 2] / 2  =

36 units^2

Since 7  is greater than pi....the absolute value   of   pi - 7  =   7 - pi

So  we have

[ pi  - (7 - pi ) l   =

l  2pi - 7 l

And 2pi < 7   ...... so the absolute value =

7 - 2pi

The distance from the beach to the island  will form the altitude, A, of a triangle

Let x represent the distance  that the person with the 30 degree line of sight is from where the altitude intersects the base

So...the distance that the person with the 40 degree line of sight to this intersection point is 50 -x

So   we have this

tan 30  = A/ x       

Rearrange as  [ x  =  A / tan 30  ] ⇒

[ x =    A / (1/√3)  =  √3A ]

And we have that

tan 40 = A / (50 - x)       sub in for x  and we get that

tan 40  = A / ( 50 - √3A)

(50 - √3A) * tan 40  = A

50sin 40 - √3A tan40  = A

50tan 40   =   √3Atan 40 + A          factor out A on the right

50tan 40  =  A ( √3 tan 40 + 1)      divide both sides by  (sin 40 + 1)

A = 50tan 40 / [√3 tan 40 + 1 ] ≈   17.1 (m)

EDIT TO CORRECT AN ERROR!!!

x^2 + 2x + 5  = 0

Make sure we are using the correct values  (and formula)

a = 1   b  = 2   and c  = 5

So we have

[ - b ±√ ( b^2 - 4ac) ]

________________      =

        2a 

[ -2 ±√ (2^2 - 4*1 * 5) ]

_________________        =

            2*1

 - 2 ±√  [4 - 20 ] 

_____________       =

          2

-2 ±√ (-16)

_________       =

       2

-2 ± 4i

______      divide everything by 2

    2

-1 ± 2i

Let's  check this in the quadratic  ...one root is 2i - 1

(2i - 1)^2  + 2 (2i - 1) + 5  =

(2i - 1) (2i - 1) + 2 (2i - 1) + 5

4i^2 - 4i + 1 + 4i - 2 + 5

- 4 + 1 - 2 + 5 =

-3 + 3  =   

0

We don't need to check the other root....if one complex root makes a quadratic = 0, then its conjugate will, too !!!

\((a + b)^n =\sum_{k=0}^{n} \begin{pmatrix} n\\ k \end{pmatrix}a^\left ( n -k \right )b^k\)

The powers of a and b are reversed in Rom's LaTeX.

The distance traveled by the plane forms the hypotenuse of a right triangle

This equals 300 * 10  = 3000ft

The altitude , A,  can be found as

sin (5°)  = A / 3000    multiply both sides by 3000

3000* sin(5°)  = A ≈  261.47 ft

And call the distance [ horizontally ] from the takeoff point , D

So  we have

cos ( 5°) = D / 3000

3000cos(5°) = D ≈  2988. ft

         A

         6

B  8   K     6       C

Since AK is an altitude....we can use the Pythagorean Theorem here

AB  = √ [ BK^2 + AK^2 ]  =  √ [ 8^2 + 6^2]  = √100 =  10

And  AC  = √ AK^2 + KC^2 ]   =  √ [ 6^2 + 6^2] = √ [ 2 * 6^2 ]  = 6√2

So....the perimeter is

AB + BK + KC + AC   = 

10 + 8 + 6 + 6√2  =

24 + 6√2   units  ≈  32.49 units

The triangle on the lft is a 30-60-90 right triangle

The altitude   = 8 *  √3/2  =  4√3  = √48

Since the triangle on the right is a 45 - 45 - 90 right triangle

The altitude  = the base  = √48 =  length of the legs

So..by the Pythagorean Theorem

x  = 

√ [ ( √48 )^2 + ( √48)^2 ]  = 

√96  =

√16 * √6  =

4√ 6

this is exactly what i was confused by!

thank you very much!

I don't really understand what you are saying but for example take k=3

\(\dbinom{4}{3}2^3 \left(\dfrac{3}{x^2}\right)^{4-3} = \\ 4\cdot 8 \cdot \dfrac{3}{x^2} = \\ 32\dfrac{3}{x^2} = \dfrac{96}{x^2}\\ b=96\)

yes i know how to do that but i am not sure on the part of after simplifying everything which leaves me with an interger to multiply k/x^2. this is where i struggle. thank you for helping

I think this is supposed to be a demonstration of the binomial theorem.

\((a+b)^n = \sum \limits_{k=0}^n~\dbinom{n}{k}a^k b^{n-k}\)

In this case

\(a = 2\\ b= \dfrac{3}{x^2}\)

\(\left(2 + \dfrac{3}{x^2}\right)^4 = \sum \limits_{k=0}^4~\dbinom{4}{k}2^k\left(\dfrac{3}{x^2}\right)^{4-k}\)

you can plug in k=4, 0 to get a-e

Yes, that is right :)

I guess he/she can keep the name :))

LOL!!....when you think about it....most of us are all just "Partial Mathematicians"  ...!!!!

Hi PartialMathematician, welcome to the forum.

In a couple of years time are you going to change your name to FullMathematician ?  LOL

On a serious note:

Please put some effort into learning to presenting your questions properly.

The stuff between the dollar signs is LaTex code.

If you delete the dollar signs you can past the rest into the LaTex box that will appear when you click [LaTex] in the ribbon.

After you hit ok it will present properly.

Note:

Just delete the stuff that is in the box when you open it.

o kiek butu -2,7 dvyliktuoju : -0,3 dvyliktuoju atsakymas      (Lithuanian detected)

and how many would be -2.7 twelfth: -0.3 the twelfth answer 

\(\frac{-2.7}{12}\;:\;\frac{-0.3}{12}\\ -2.7:-0.3\\ 27:3\\ 9:1\)

That is if I understand your question properly.

Tai yra, jei aš tinkamai suprantu jūsų klausimą.

So far I have not worked out how to derive those initial formulas that were given.

I have not looked at the rest of your logic yet, I'd really like to convince myself of the accuracy of the book extract first :/