All Answers 358108 Answers

Thank you so much. 

1  2  3  4 ==5 odd  x  5 even  x  5 odd  x  5 even==625 such 4-digit integers.

There is no numbers, try proof reading you questions!

John counts up from 1 to 13, and then immediately counts down again to 1, and then back up to 13, and so on, alternately counting up and down.  What is the 1000th integer in his list?

12up then 12 down makes 24

1000/24=41.666

24*41=984

1000-984=16

12+4

13-4=9

So I get 9,  could be wrong though.  It is up to you what logic you accept.

Here is the pic, drawn to scale.

Drawing a good enough pic was the most important part.

From the pic find

sin(alpha)

then

cos(alpha)

then

sin(2*alpha)

then

area of the triangle.

Ok thanks!

In general cases, you can treat complex numbers as vectors because there is a one-to-one correspondence \(\phi:\mathbb C \to \mathbb R^2\) defined by \(\phi (z) = \begin{bmatrix}\operatorname{Re}(z)\\\operatorname{Im}(z)\end{bmatrix}\).

You just have to be careful that if \(BC = 2AB\), that actually means \(\left\|\overrightarrow{BC}\right\| = 2 \left\|\overrightarrow{AB}\right\|\), not \(\overrightarrow{BC} =2 \overrightarrow{AB}\). This goes for any equation with lengths.

Thank you!!
One last question: Is it ok to treat complex numbers as vectors? (Because I usually did and got correct answer, but here it is incorrect, as you said, because the vectors are not parallel. So, Is treating complex numbers as vectors only works if they are parallel, and if they are not parallel then we have to take into account the direction? I.e. rotation of the vector by angle say \(\theta\)?)
I mean is the following general formula correct (I am just guessing): \(BC=2AB \iff BC_{vector}=r(cos\theta+isin\theta)AB_{vector}\), so in this case, \(\theta=90\) and r = 2, and it worked. But, is this formula correct in general or? (Also the case when the two lines are parallel, if we used \(\theta=180\), then \(cos\theta=-1\), but we want it 1 to be correct I think? But if we used \(\theta=0\) then I think it works. So, is the angle between two parallel lines 180 or 0? (Because with 0, the direction is maintained but with 180, the direction switched, which won't give the correct answer I think).

You treated BC and AB as vectors. If \(\overrightarrow{BC} = 2\overrightarrow{AB}\), that means \(\overrightarrow{BC}\) is parallel to \(\overrightarrow{AB}\), which is not what you want, since \(\angle ABC = 90^\circ\).

Using 2i instead of 2 gives the vectors a rotation of 90 degrees, which gives the correct answer.

The multiplication by -2i gives a rotation of 90 degrees clockwise, maybe the resulting point C is not in the first quadrant, which isn't what the question wanted.

1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 12
15 11

.

.

.

985  1

986 2
987 3
988 4
989 5
990 6
991 7
992 8
993 9
994 10
995 11
996 12
997 13
998 12
999 11
1000 10  [The 1000th integer is 10 - in counting down frm 13 to 1]

For the differentiation part: notice that r(t) and h(t) are functions of time and they are varying. So when we differentiate r, we do the normal differentiation (r -->1) but then applying the chain rule, we multiply by r' which is dr/dt, and similarly for h). So to differentiate (*) you need to know the product rule and chain rule.

Here is my attempt:

Since: \(V_{cone}=\frac{1}{3}\pi r^2h\)              (*)

and given: \(\dfrac{dr}{dt}=3\), and \(\dfrac{dh}{dt}=-4\)

Differentiate (*) with respect to t

\(\dfrac{d}{dt}V(t)=\frac{1}{3}\pi(2r\frac{dr}{dt}h+r^2\frac{dh}{dt})\)

We want to find the rate at \(t_0\) and we are given r=8 and h=10: 

\(V'(t)=\frac{1}{3}\pi(2(8)(3)+(8)^2(-4))=\frac{224}{3}\pi\)

Y

X        Z

tan Y  =  XZ / XY =   3/2   ⇒   XZ =  (3/2)XY

Therefore, by Pythagoras

XY^2  + XZ^2  = 80^2

XY^2  +  [ (3/2) XY ] ^2  =  6400

XY^2  + (9/4)XY^2  = 6400

(13/4)XY^2  = 6400

XY^2 =  6400 / (13/4)

XY^2  = 6400 * 4  / 13

XY  =   sqrt [   (6400 * 4 )  / 13  ]  =   160  / sqrt 13  ≈  44.38

THX, Guest....I realized later that mine was wrong  !!!!

Lets do this step by step.

f(-7) = -7^2 - 2(-7) = 49+14 = 63

f(63) = 63^2 - 2(63) = 3969-126 = 3843

f(3843) = 3843^2 - 2(3843) = 14,768,649 - 7686 = 14,760,963

:) :) :)

m = -3.

Draw CF ⊥ DA ..( F on DA)
Let z= x^2
Applying Pythagoras’s theorem in triangles ABD, BCE and ACF we get
BE^2=x^2-121=z-121
BD^2=z-169
AC^2=CF^2+AF^2 => (BE+BD)^2+4=z
Or 2.BE.BD=z-BE^2-BD^2-4= 286-z
4.BE^2.BD^2=4(z-121)(z-169)=(286-z)^2
Simplify this equation we get 3.z^2-588.z= 0
We get 2 solutions z= 0 and z= 196
Only z=196 is acceptable so x= 14

The roots are     -2  and 3 

So  the base of the triangle =  3  - -2 =   5

So

We need to  find  a    y intercept  such  that   (1/2)(5) (h) =  12 ....... so

(5/2) h = 12

h = 12 * 2/5  =    24 / 5

Expand the polynomial

a ( x + 2) (x -3)  =

a [x^2 -x - 6 ]

ax^2  - ax   -  6a

Now....when x  = 0   the y intercept =  -6a

And we need to have this  =   24/5

So

-6a =  24/5

a = -24/ 30 =  - 4/5

But.....we need   abs (a) =   4/5   since the parabola turns upward

So

a = 4/5

A = arctan (0.9681)   =  44°

Here's another way

Multiply all the constants  on top  and  bottom  =   7 * 2 * 27 * 6              42 *27          21 * 27         

                                                                               __________  =         ______  =    _______  =    

                                                                                4 * 2 * 1 * 8                 4 * 8              2 * 8 

567 / 16          

Now look at  the  rest

The first  fraction   gives us   x^(2014 -2013)  =  x^1

The second fraction gives us x^(2012 - 2007) = x^5

The third fraction gives us    x^(2006 - 2002)  = x^4

And the last one  gives us     x^(2008 -2010) =x^(-2)

So

x * x^5 * x^4 * x^(-2)  =  x^( 1 + 5 + 4 - 2)  =  x^8

So...our  answer is  just     (567 / 16) x^8

NVM .....Guest has the correct answer  !!!

We have similar triagles such that

BA / AC =  BD / DE

( BD + DA)  / AC =  BD /DE

(BD + 7) / 12  =  BD / 8       cross-multiply

8 (BD + 7)  = 12 BD

8BD + 56  = 12BD

56  = 12BD - 8BD

56  = 4BD

BD =  56  /   4    =     14

sum of 6 numbers is   6*14 =  84

Sum of 5 numbes is   5* 10  = 50

so ahat is the 6th number?