All Answers 358090 Answers

I don't remember ever doing one like this before   

I will give it a shot :)

$$\\(a+2b+3c)^6\\\\
=(a+[2b+3c])^6\\\\
=\displaystyle\sum_{r=0}^{6}^6C_r(a^r)(2b+3c)^{6-r}\\\\
$Now the degree of a must be 1 so I am only interested in terms where $ r=1\\\\
^6C_1(a^1)(2b+3c)^{5}\\\\
(6a)*[(2b+3c)^{5}]\\\\$$

$$\\$The general term for this is $\\\\
6a*[ ^5C_k*(2b)^k(3c)^{5-k}]\\\\
$The power of b must be 2 so k=2\\\\
6a*[ ^5C_2*(2b)^2(3c)^{3}]\\\\
6a*[ 10*4b^2*27c^3]\\\\
6a*[ 10*4b^2*27c^3]\\\\
6x*[1080*b^2c^3]\\\\
6480xb^2c^3\\\\
$So the coefficient of $ bc^2c^3 $ is 6480$$

I assume Chris got his answer from Wolfram|Alpha  ?   

Anyway they are the same so we are on a roll.       

Yes.  asin = arcsine = sin^-1

.

A sample of microbes weighs 50 mg. There are approximately 2.5 × 10⁶ microbes in the sample. What is the approximate mass of each microbe expressed using scientific notation in grams.

$$\small\text{$
\begin{array}{lcl}
\mathbf{\dfrac{50~mg}{2.5\cdot 10^6~microbes} \cdot \dfrac{1~g}{1000~mg} } \\\\
\mathbf{ =\dfrac{50}{2.5\cdot 10^6\cdot 1000} \cdot \dfrac{g}{ microbe } } \\\\
\mathbf{ =\dfrac{50}{2.5\cdot 10^6\cdot 10^3} \cdot \dfrac{g}{ microbe } } \\\\
\mathbf{ =\dfrac{50}{2.5\cdot 10^9 } \cdot \dfrac{g}{ microbe } } \\\\
\mathbf{ =\dfrac{50}{2.5 } \cdot 10^{-9} \cdot \dfrac{g}{ microbe } } \\\\
\mathbf{ =20 \cdot 10^{-9} \cdot \dfrac{g}{ microbe } } \\\\
\mathbf{ =20 \cdot \dfrac{ng}{ microbe } } \qquad ng = $ Nano gram$
\end{array}\\
$}}$$

the approximate mass of each microbe  is 20 ng (Nano gram) or $$\small{\text{$20 \cdot 10^{-9} ~g$}}$$

$$\\1.76846\qquad$round to 2 decimal places\\\\
1.76\textcolor[rgb]{1,0,0}{8}\not{4}\not{6}$$

The 8 is used for rounding        The 46 are just crossed out and not used at all

8 is a big number (5.6.7.8.9 are all big numbers ) so the 6 goes up to 7

Answer

1.77

It is all too much for me Everymornin17  

Thanks Headingnorth for that sterling effort.   

annon - I am pleased that you have persisted, that is a good thing, but do make sure you give yourself a little time to try and understand what you are being taught.

I shall try to help.    

√24 x^6 y^9 (Everything is under the square root sign)

The answer given was 2x^3 y^4 √6y  But I don't understand how the 3 and 4 was found

$$\\\sqrt{24*x^6*y^9}\\\\
=\sqrt{4*6}*\sqrt{x^3*x^3}*\sqrt{y^4*y^4*y}\\\\
=\sqrt{4}*\sqrt{6}*\sqrt{x^3*x^3}*\sqrt{y^4*y^4}*\sqrt{y}\\\\
=\sqrt{4}*\sqrt{x^3*x^3}*\sqrt{y^4*y^4}*\sqrt{6}*\sqrt{y}\\\\
=2*x^3*y^4*\sqrt{6y}\\\\
=2x^3y^4\sqrt{6y}\\$$

Simplify 12a³b² ÷ 4ab⁻².

$$\\\small{\text{
$
\begin{array}{rcl}
\dfrac{ 12\cdot a^3 \cdot b^2 }{ 4 \cdot a \cdot b^{-2} } \\\\
=\dfrac{ 12 } {4} \cdot \dfrac{ a^3 }{ a } \cdot \dfrac{ b^2 }{ b^{-2} } \\\\
=\dfrac{ 3\cdot 4 } {4} \cdot \dfrac{ a\cdot a^2 }{ a } \cdot \dfrac{ b^2 }{ b^{-2} } \\\\
=\dfrac{ 3\cdot \not{4} } {\not{4}} \cdot \dfrac{ \not{a}\cdot a^2 }{ \not{a} } \cdot \dfrac{ b^2 }{ b^{-2} } \\\\
= 3\cdot a^2 \cdot \dfrac{ b^2 }{ b^{-2} } \\\\
= 3\cdot a^2 \cdot b^{2-(-2) } \\\\
= 3\cdot a^2 \cdot b^{2+2} \\\\
= 3\cdot a^2 \cdot b^4
\end{array}
$}}$$

It is not help you want, you want someone to do your homework for you

just like you did here:

http://web2.0calc.com/questions/i-need-help-on-these-questions-please#r0

5(x-4) + 4 == 5x - 20 + 4

then-20 + 4 = -16

Final answer

5x - 16

Perhaps you mean WSW   ?

This is west of south west.

Which is South 67.5°West

or

157.5°

bottom left corner

For every 7  gallons of water in the original mixture, the orange juice concentrate comprises 2 gallons.

So, if this 7 gallons  is mixed with lemonade in a ratio of 3 : 4....there must be  7 x (4/3)  = 28/3  gallons of

lemonade in this mixture.

Then the ratio of orange juice concentrate to lemonade in this mixture is   2  : 28/3    =  6 : 28 =  3 : 14

1.

$${\mathtt{76.5}}{\mathtt{\,-\,}}{\mathtt{10}} = {\frac{{\mathtt{133}}}{{\mathtt{2}}}} = {\mathtt{66.5}}$$

$${\frac{{\mathtt{66.5}}}{{\mathtt{2}}}} = {\frac{{\mathtt{133}}}{{\mathtt{4}}}} = {\mathtt{33.25}}$$

The shoes are $33.25 each pair.

2.

$${\mathtt{300}}{\mathtt{\,-\,}}{\mathtt{50}} = {\mathtt{250}}$$

$${\frac{{\mathtt{250}}}{{\mathtt{5}}}} = {\mathtt{50}}$$

You need 50 weeks of saving.

3.

$${\mathtt{108}}{\mathtt{\,-\,}}{\mathtt{6}} = {\mathtt{102}}$$

$${\frac{{\mathtt{102}}}{{\mathtt{3}}}} = {\mathtt{34}}$$

The player bats in 34 times.

4.

$${\mathtt{35}}{\mathtt{\,-\,}}{\mathtt{8}} = {\mathtt{27}}$$

$${\frac{{\mathtt{27}}}{{\mathtt{5}}}} = {\mathtt{5.4}}$$

She can take 5 children with her (and keeps $2).

5.

$${\mathtt{5\,610}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,618}} = {\mathtt{7\,228}}$$

$${\frac{{\mathtt{7\,228}}}{{\mathtt{2}}}} = {\mathtt{3\,614}}$$

The depth of Kazumura Cave is 3614 feet.

I hope I could help you, even tough this is not the best way to learn it ;)

Use two unit multipliers to convert 1000 mm² to square centimeters.

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

First, write down known facts:>     1cm² = 100mm²

Second, using these  facts create a pair of unit multipliers:>      1cm² / 100mm²      or      100mm² / 1cm²

Third, go ahead with the unit conversion:>

                                     1000mm² / 1  x  1cm² /100mm²  = 1000cm² / 100 = 10cm²

As you can see,  1000mm²  has been turned into a fraction by dividing it by 1; then that fraction is multiplied  by adequate unit multiplier, the one that has   mm²   in its denominator.

I hope this can help...

cos(13 pi/4)  = cos (5 pi / 4)   =  -1 /√2

could you explain WHY this works ?

$$\vec{P_1}=\begin{pmatrix} 1 \\ 0 \end{pmatrix} \qquad\vec{P_2}=\begin{pmatrix} 4 \\ 3 \end{pmatrix} \qquad\vec{P_3}=\begin{pmatrix} 5 \\ -2 \end{pmatrix} \\\\
\small{\text{direction vectors: }}\\
\begin{array}{crcl}
& \vec{d_1} &=& \vec{P_2} - \vec{P_1}\\
& \vec{d_2} &=& \vec{P_3} - \vec{P_2} \\
& \vec{d_3} &=& \vec{P_1} - \vec{P_3}\\
\end{array}\\\\\\
\small{\text{$\begin{array}{lclclclcl}
\vec{P_4} = \vec{P_1}+\vec{d_1}-\vec{d_3}
&=& &-&\vec{P_1} &+& \vec{P_2} &+& \vec{P_3} =\begin{pmatrix} -1+4+5 \\ 0+3-2 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}\\\\
\vec{P_4} = \vec{P_2}+\vec{d_2}-\vec{d_1}
&=& &&\vec{P_1} &-& \vec{P_2} &+& \vec{P_3}=\begin{pmatrix} 1-4+5 \\ 0-3-2 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}\\\\
\vec{P_4} = \vec{P_3}+\vec{d_3}-\vec{d_2}
&=& &&\vec{P_1} &+& \vec{P_2} &-& \vec{P_3}=\begin{pmatrix} 1+4-5 \\ 0+3+2 \end{pmatrix} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\\\\\end{array}$}}$$

what is 0.4441 as a fraction

 It's  4441/10000

(a+2b)²-(a-2b)*(a+2b)-(3a-b)² = ???

$$\small{\text{
$
(a+2b)^2-(a-2b)*(a+2b)-(3a-b)^2 = 7b^2+10ab-9a^2
$}}$$

Learning sites for MG and other keen students  

This site looks good and it appears to be free.  If you join up they track your progress  :)

Amended answer...sorry, Mellie....I realized the answer I gave before was incorrect....

(a) It appears that a constant term will occur when n is a multiple of 3

For instance  

When n = 3, the constant term is 3  =  C(n, [2/3]n) = C(3,2)

When n = 6,  the constant term is 15  = C(6,4)  

When n = 9, the constant term is 84  = C(9,6)

(b) So  ...it appears that the constant term is produced by C(n,[2/3]n)    

This happens because at C(n, [2/3]n ).......x^1  is being raised to the [n - (2/3)n]  = (1/3)n power...and...

x^-1 is being raised to the (2/3)n power....

So   (x^2)^[(1/3)n]  * (x^-1)^[(2/3)n]   =   (x)^(2/3)n  * (x)^(-2/3)n  = 1

And  C(n, [2/3]n) * (1)   =  C(n, [2/3]n )   = a constant

///////////////////////////////////

Whats the area of a regular hexagon with a height of 2 times the square root of 3 ?

$$\boxed{
\mathbf{
A = 2\cdot h^2\cdot\sqrt{3}
}
}\\\\
\small{\text{
$
\begin{array}{lrcl}
A = 2\cdot \textcolor[rgb]{150,0,0}{h^2}\cdot\sqrt{3} \qquad & h &=& 2\cdot\sqrt{3}\\
& h^2 &=& 4\cdot 3 \\
& h^2 &=& 12 \\
A = 2\cdot \textcolor[rgb]{150,0,0}{12}\cdot\sqrt{3}\\
A = 24 \cdot \sqrt{3}\\
A = 41.5692193817 \\
A \approx 41.569
\end{array}
$}}$$

Each variable stands for a different thing :)