All Answers 358237 Answers

Use the fact that sin(angle) = opposite/hypotenuse

or angle = sin^-1(opposite/hypotenuse)

Like so:

.

There is no last digit in the decimal representation of pi.  There are an infinite number of such digits.

Thank you CPhill for your clarification.

forgot a zero, ups. 

the sum was 1 539 000 if i remember correct.

Use the law of cosines:

c^2 = a^2 + b^2 - 2ab cos C    where a, b, c are your sides, and C is your angle.

Your problem should then look like:

5.83^2 = 6.71^2 + 7.81^2 - 2(6.71)(7.81)cos C

(Use 5.83 on the left because that side is across from your angle.)

Now solve for cos C:

cos C = (5.83^2 - 6,71^2 - 7.81^2) / (-2 * 6.71 * 7.81)

cos C = 0.687255

Using arccos you then get approximately 46.6 degrees.

Let x = 0.99999....  where there are an infinite number of 9s

10x = 9.99999.....

Subtract x from 10 x:

10x - x = 9.99999..... - 0.999999.....

9x = 9

Divide both sides by 9

x = 1

so  0.99999..... = 1

.

Not zero; see below:

.

thanks much CPhill

Use the fact that  a*ln(b) =  ln(b^a)

2 ln 6 + 3 ln c → ln 6² + ln c³

Now use the fact that  ln a + ln b = ln(a*b)

2 ln 6 + 3 ln c → ln 6² + ln c³  →  ln 36 + ln c³ → ln(36c³)

16x^2 - 25   =  a difference of two squares =

(4x - 5) (4x + 5)

0.006hm x 100m/1hm   = 0.6m

0.6m  x 1000mm/1m  = 600mm

x^2+5x=84   subtract 84 from both sides

x^2 + 5x - 84  = 0      factor

(x + 12) (x - 7)  = 0 

And setting both factors to 0, x =  -12   or x = 7

I'm assuming that this is : 9 - 3 ÷ [1/3] + 1   remember to perform any  multiplications/divisions from left to right before any additions/subtractions.....

9 - [3 ÷ 1/3] + 1  =

9 - [9] + 1  =

0 + 1  =

1

Let Jack's distance from the tree = J1    and Jill's distance from the tree = J2

Then.....letting h be the height of the tree....we have....

tan 15  = h / J1    and  tan 25 = h / J2

Thus  h = J1*tan 15

And   tan25 = [J1*tan 15] / J2   →   tan 25/tan15  = J1/J2  = about  1.74......so Jack is about 1.74 times as far from the tree as Jill......if we let Jill's distance = m  ....then Jack's = 1.74m

And....using the Law of Cosines, we have :

100^2 = (m)^2 + (1.74m)^2 - 2(m)(1.74m)cos65

This is a little sticky to solve.....so I'll let WolframAlpha do the "heavy lifting".....solving for m, we have  m  ≈ 62.538m

So....Jill  is ≈ 62.538m  from the tree and Jack is ≈ 1.74(62.538)m    ≈ 108.816 m  from the tree

So

tan15 =  h/108.816

108.816*tan15 = h  = about 29.157m

tan25 = h/62.538

62.538*tan25 =h =  about 29.16 m

Close enough !!!....we'll call the height of the tree 29.16 m  ..........

95040

Here is a picture of a regular tetrahedron inside a cube.

A tetrahedron is just a triangular pyramid where all the sides are congruent.

Perhaps from the diagram, you can compare the two :)

LOL

Nicely done Chris,

I knew there was an easy way !!

But you know, why fly across Mt Everest in air conditioned comfort when you could just climb over it. with ropes and oxygen tanks.     

All solutions can be expressed as That is assuming it is a concave up or down parabola with an axis of symmetry parallel to the y axis.

\(y=a ( x^2 -4x - 45)   \qquad where \;\;a\in R\)

Easy

f(x)  = (x + 5) (x - 9)  =   x^2 -4x - 45

Here's the graph .....   https://www.desmos.com/calculator/o71jos4cwn

Write an equation for a quadratic function such that f(-5)=0 and f(9)=0

y=ax^2+bx+c

\(0=a*(-5)^2+b*-5+c \qquad (1)\\ 0=25a-5b+c \qquad (1b)\\ and\\ 0=a*(9)^2+b*9+c \qquad (2)\\ 0=81a+9b+c \qquad (2b)\\~\\ so\\ 25a-5b+c=81a+9b+c\\ 25a-5b=81a+9b\\ -14b=56a\\ b=-4a\\~\\ c=5b-25a=-20a-25a=-45a\\ c=-81a-9b=-81a+36a=-45a\\ Let\;a=1\\ then\;\;b=-4 \qquad and \qquad c=-45\\ \mbox{So a solution is}\\ y=x^2-4x-45 \)

3 + 4i / 8 - 3i       multiply top/bottom   by 8 + 3i

[3 + 4i]   [8 + 3i]          [ 24 + 32i + 9i + 12i^2]          [ 24 + 41i - 12]         [12 + 41i ]

---------  ----------    =  ------------------------------    =    --------------------    =    -----------

[8 - 3i]   [8  + 3i]               [64  - 9i^2]                              73                          73

What gets bigger the more u take away?

I come once in a minute, twice in a moment, but never in a thousand years. What am I?

What word is spelled incorrectly in the dictionary?

thats the point! trick question!