All Answers 358222 Answers

Remember: two lines that are perpendicular to each others' slopes multiply to -1

I think I got it.

I subtracted the total possibilities from the possibilities and the number of ways to choose officers without any of the past to get 4242.

Thank you for all of your help, and sorry if I was a pain :(

R =A(0) x 1/2^(elapsed time/half-life)
R =Remaining substance
A(0) =Initial substance
44.5 =356 * 1/2^(210/h), solve for h
h = 70 minutes - half-life of this "goo"!!

s divided by the square root of 2 will give you the radius of the circle.

s / sqrt(2)  =  10 / sqrt(2)  =  5·sqrt(2)

The area of the (full) circle will be:  pi · [ 5·sqrt(2) ]²  =  50·pi

The area of the square = 100   

     --->   the area of the green portion  =  [ ( 50·pi ) - ( 100 ) ] / 2  =  25·pi - 50

2y = x-1         y = 1/2 x - 1/2      slope , m   = 1/2     perpindicular slope = - 1/m  = -2

y = -2x + b    sub in the point given to find 'b'

3 = -2(-1) + b      b = 1

y = -2x+1

Let  x  =  ?

Parallel lines cut off proportional segments:

x / (x + 63)  =  (65 - 35) / 35

Finish this ...

Umm...

Pretty easy imo.

3x625=1875

4.25x880=3740

1875+3740=5615

Yay!

Let a, b, and c be length of the three sides of the brick.

From the ratios given, we can write these equations.
(1) (a^2+b^2)=(3)^2=9
(2) (b^2+c^2)=(2√3)^2=12
(3) (a^2+c^2)=(√15)^2=15

(1)+(2)+(3)
2*(a^2+b^2+c^2)=36
(a^2+b^2+c^2)=18
(9+c^2)=18
c^2=9

(2)
(b^2+c^2)=12
(b^2+9)=12
b^2=3

(1)
(a^2+b^2)=9
(a^2+3)=9
a^2=6

(I put too much effort into this... why didn't I just use sqrt(#) whatever)

The longest and shortest are c, b.

b^2:c^2=3:9=1:3

So B:C=1:√3

Yay!

Compute  
\(\large \sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ)\)

Formula:
\(\begin{array}{|lrcll|} \hline (1) & \cos(x-y) &=& \cos(x)\cos(y)+\sin(x)\sin(y) \\ (2) & \cos(x+y) &=& \cos(x)\cos(y)-\sin(x)\sin(y)\\ \hline (1)-(2): & \mathbf{2\sin(x)\sin(y)} &=& \mathbf{\cos(x-y)-\cos(x+y)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \sin(10^\circ) \sin(30^\circ) \sin(50^\circ) \sin(70^\circ) \quad | \quad \sin(30^\circ)=\dfrac{1}{2} \\ &=& \dfrac{1}{2}* \sin10(^\circ) \sin(70^\circ) \sin(50^\circ) \\ \hline \\ && \quad 2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y) \\ && \quad 2\sin(10^\circ)\sin(70^\circ)=\cos(60^\circ)-\cos(80^\circ) \quad | \quad \cos(60^\circ)=\dfrac{1}{2} \\ &&\quad 2\sin(10^\circ)\sin(70^\circ)=\dfrac{1}{2}-\cos(80^\circ) \quad | \quad \cos(80^\circ)=\cos(90^\circ-10^\circ)=\sin(10^\circ) \\ &&\quad 2\sin(10^\circ)\sin(70^\circ)=\dfrac{1}{2}-\sin(10^\circ) \\ &&\quad \mathbf{\sin(10^\circ)\sin(70^\circ)=\dfrac{1}{4}-\dfrac{1}{2}*\sin(10^\circ)} \\ \\ \hline &=& \dfrac{1}{2}\left( \dfrac{1}{4}-\dfrac{1}{2}*\sin(10^\circ) \right) \sin(50^\circ) \\ &=& \dfrac{1}{8}\sin(50^\circ)-\dfrac{1}{4}*\sin(10^\circ)\sin(50^\circ) \\ \hline \\ && \quad 2\sin(x)\sin(y)=\cos(x-y)-\cos(x+y) \\ && \quad 2\sin(10^\circ)\sin(50^\circ)=\cos(40^\circ)-\cos(60^\circ) \quad | \quad \cos(60^\circ)=\dfrac{1}{2} \\ && \quad 2\sin(10^\circ)\sin(50^\circ)=\cos(40^\circ)-\dfrac{1}{2} \quad | \quad \cos(40^\circ)=\cos(90^\circ-50^\circ)=\sin(50^\circ) \\ &&\quad 2\sin(10^\circ)\sin(50^\circ)=\sin(50^\circ)-\dfrac{1}{2} \\ &&\quad \mathbf{\sin(10^\circ)\sin(50^\circ)=\dfrac{1}{2}*\sin(50^\circ)-\dfrac{1}{4} } \\ \\ \hline &=& \dfrac{1}{8}\sin(50^\circ)-\dfrac{1}{4}*\left(\dfrac{1}{2}\sin(50^\circ)-\dfrac{1}{4}\right) \\ &=& \dfrac{1}{8}\sin(50^\circ)-\dfrac{1}{8}\sin(50^\circ)+\dfrac{1}{16} \\ &=& \mathbf{\dfrac{1}{16}} \\ \hline \end{array} \)

Similar triangles

54 is to 30     as     ? is to 5

54/30 = ?/5      solve for ' ? '

x = log64/log81

27^{(log64/log81 +1)} =610.94026

The kitchen tiles will fit evenly because 20 cm fits evenly to multiples of meters

3 m = 300 cm     5 m = 500 cm

area to be tiled =   300 x 500 = 150 000 cm^2

   area of each tile   20 x 20 = 400 cm^2

       # tiles required   150 000 / 400 = 375 tiles

Tiles come 10/box      375 tiles / 10 tiles/box = ~ 38 boxes

38 boxes * L6 / box =   L 228 total tile cost

81^x = 64. What is 27^{x+1}?

For (a) use law of sines to find angle at 'b'   then use the fact that triangle angles sum to 180^o

sin 110 / 46  =  sin b / 35

for area , you have two irregular triangles

   if you have two sides (a and b ) and the included angle (c)     area = 1/2 ab sin c

INn FOUR hours Jennifer will travel 12 m/h x4 = 48 miles

  it will take Thomas  48 miles / 4 m/hr = 12 hours to reach the bike

    in the meantime Jennifer has EIGHT hours of walking   3 m/hr x 8 =  24 miles form the bike

        Thomas now sets out at   10 m/h    this is     10 - 3 = 7 m/h faster than Jennifer is walking

             he will close the 24 miles and catch Jennifer in  7 m/hr *x =  24 m

                     x =  24/7hours

So Thomas walks for   12 hours then bikes for  24/7 hr   =  15  3/7 hrs  = ~ 15 hr 26 min

tahnk you!

To start, you have the difference of two squares, so you can write \((3^{1001}+4^{1002})^2 - (3^{1001}-4^{1002})^2=2\cdot3^{1001}\cdot2\cdot4^{1002}\)

This can be written as \(2\cdot3^{1001}\cdot2\cdot4^{1002}=4\cdot(3^{1001}\cdot4^{1001})\cdot4^1=16\cdot12^{1001}\)

So k = 16.

A question to asker guest.

How can you tell if a number is divisible by 3?

If you do not know then google it and get back to us with the answer.

Please no one else say more until we get a proper response from asker guest.

See https://web2.0calc.com/questions/help_50204#r2 

Hmm.  Check:  If   w = -1, x = -9, y = 6, z = 12,  then w+x+y = -1 - 9 + 6 = 4 not -2.  I think Guest made a mistake in inverting the matrix.

Here's the correct matrix inversion:

Here's a graphical solution:

80% of the tank holds 10litres, so 20% of the tank will hold 10/4 = 2.5 litres; hence a full tank holds 12.5 litres.

Your 5-digit cryptogram =21,978

21, 978 x 4 =87,912 - which are same 5 digits in reverse.

Yeah, you're right, Melody.

Sorry about that.

When HB sees this, I hope he/she'll be able to figure it out!