All Answers 358180 Answers

just cut or divide 360 by 36 and what you get multiply by 150

$${\mathtt{2\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.04}}}{{\mathtt{4}}}}\right)}^{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{4}}\right)} = {\mathtt{2\,440.380\: \!079\: \!895\: \!933\: \!649}}$$

$$\\5^1-3x=10\\
5-3x=10\\
-3x=5\\
x=-5/3$$

did you forget some brackets?

You function is missing again anon  

there is a calc on the home page :)

Well, The 2 vertical ones will have to both be in rows 1 and 2      OR     both be in rows 2 and 3.

I am going to look at how many ways they can be in rows 1 and 2     (the number of positions in rows 2 and 3 will be the same)

There will be 25 horizonal tiles in each of rows 1 and 2 and these tiles must be exactly under each other.  they can not be interlaced like bricks , or they won't all fit.

There must be 2 vertical ones placed between them somewhere, anywhere.

------------------------

I am going to look at an easier problem to get an understanding of this.  Say there were just 3 horzontal bricks, how many places could the vertical bricks be placed?

xx [brick]  xx [brick] xx [brick] xx

the crosses are where the vertial ones could go.  There are 2(3+1) = 8 posibilities.  We need just 2 of these.

So I think that there are 8C2 posibilities

--------------------------

BACK to your problem       2(25+1)=100 possibilities    We need just 2 of these.

So I think that there are   100C2   possible places for the 2 vertical bricks.   This is just for rows 1 and 2.

---------------------------

Altogether I think that there are   2*100C2 arrangements possible.

$${\mathtt{2}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{100}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{100}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{9\,900}}$$

----------------------------

I think that is right but I won't stake my life on it.   LOL

What is the "net" of the pyramid???

Can you provide a pic of this, Melody??

CPhill has probably interpreted this as you intended but it really needed brackets because

$$x+1/x-3 = x+\frac{1}{x}-3$$

Remember that you do division BEFORE you do addition or subtraction!

$$(x+1)/(x-3) = \frac{x+1}{x-3}$$

5-(-1)

=> 5-1 (remove the brackets)

=> +4

so,5-(-1)= +4

It is NOT   log base 2 times 16

It is           log base 2 of 16

$$16=2^4$$

so

$$log_2\;16=4$$

you plan to invest $16000 in 4 years when you graduate. If you earn 8.5 per year, how much will you have 14 years from now?

FV=16000(1.085)^10

$${\mathtt{16\,000}}{\mathtt{\,\times\,}}{\left({\mathtt{1.085}}\right)}^{{\mathtt{10}}} = {\mathtt{36\,175.735\: \!070\: \!678\: \!940\: \!733\: \!5}}$$

Area of base = 25*22 = 550 ft square

  $$\frac{72.4-25}{2}=\frac{47.4}{2}=23.7$$

so 2 of the triangle sides have base 22 and height 23.7 Their area combined = 2* (1/2)*22*23.7

$${\mathtt{2}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{22}}{\mathtt{\,\times\,}}{\mathtt{23.7}} = {\frac{{\mathtt{2\,607}}}{{\mathtt{5}}}} = {\mathtt{521.4}}$$    square feet

$$\frac{67.8-22}{2}=\frac{45.8}{2}=22.9$$

so 2 of the triangle sides have base 25 and height 22.9 Their area combined = 2* (1/2)*25*22.9

$${\mathtt{2}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{22.9}} = {\frac{{\mathtt{1\,145}}}{{\mathtt{2}}}} = {\mathtt{572.5}}$$     square feet

total area =  $${\mathtt{550}}{\mathtt{\,\small\textbf+\,}}{\mathtt{521.4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{572.5}} = {\mathtt{1\,643.9}}$$     square feet

If you want me to draw a picture to explain what i  have done I can do that but try and understand it by yourself first.   

side = $${{\mathtt{85\,184}}}^{\left\{{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right\}} = {\mathtt{43.999\: \!999\: \!999\: \!999\: \!99}}$$      =  44cm

diagonal of base =   $${\sqrt{{{\mathtt{44}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{44}}}^{{\mathtt{2}}}}} = {\mathtt{62.225\: \!396\: \!744\: \!416\: \!182\: \!1}}$$

diagonal of cube =  $${\sqrt{\left({{\mathtt{44}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{44}}}^{{\mathtt{2}}}\right){\mathtt{\,\small\textbf+\,}}{{\mathtt{44}}}^{{\mathtt{2}}}}} = {\mathtt{76.210\: \!235\: \!533\: \!030\: \!600\: \!9}}$$

$$\\y=(5/3)cos(4x-pi)+4\\\\
y=(5/3)cos[4(x-\frac{pi}{4})]+4$$

Phase shift is pi/4 to the right

This old post goes into more detail

$$\\\frac{3}{5}\div 7\\\\
=\frac{3}{5}\div \frac{7}{1}\\\\
=\frac{3}{5}\times \frac{1}{7}\\\\
=\frac{3}{35}$$

$$\\(a-b)(4a-2b)\\\\
=a(4a-2b)\textcolor[rgb]{1,0,0}{-b}(4a-2b)\\\\
=4a^2-2ab-4ab--2b^2\\\\
=4a^2-2ab-4ab+2b^2\\\\
=4a^2-6ab+2b^2\\\\$$

$$\\1\frac{1}{8}\times 3\frac{1}{3}\\\\
=\frac{9}{8}\times \frac{10}{3}\\\\
=\frac{\not{9}^3}{\not{8}^4}\times \frac{\not{10}^5}{\not{3}^1}\\\\
=\frac{15}{4}\\\\
=3\frac{3}{4}$$

the world will end tommorow in a storm of doritoes and mountain dew

you are welcome

10√​x​​​+3=2√​x​​​+2

$$\\10\sqrt{x}+3=2\sqrt{x}+2\\\\
8\sqrt{x}+3=+2\\\\
8\sqrt{x}=-1\\\\
\sqrt{x}=\frac{-1}{8}\\\\$$

If the square root is in the question then it is defined to be positive.

So there is no solution to this.   

You look for a square number that goes into them.

square numbers that are helpful are 4,9,16,25,....

ex

$$\sqrt{75}=\squr{25*3}=\sqrt{25}*\sqrt{3}=5\sqrt{3}$$

The domain of this function is all x's except x =3......that would make the denominator 0

This function has a horizontal asymptote at y = 1

Thus, the range is (-∞, 1) u (1,∞)

It is already a decimal and it  is already in scientific notation.