All Answers 358233 Answers

a)A triangle has vertices at A(1,-3),B(3,3) and C(-3,1)

i) Find coordinates of the midpoint (labeled P) of AB and the midpoint (labeled Q) of BC

\(\vec{P}=\frac{ \vec{A}+\vec{B}}{2} =\frac{ \begin{pmatrix}1\\-3\end{pmatrix} +\begin{pmatrix}3\\3\end{pmatrix} } {2} =\frac{ \begin{pmatrix}4\\0\end{pmatrix} }{2} =\begin{pmatrix}2\\0\end{pmatrix}\\\\ \vec{Q}=\frac{ \vec{B}+\vec{C}}{2} =\frac{ \begin{pmatrix}3\\3\end{pmatrix} +\begin{pmatrix}-3\\13\end{pmatrix} } {2} =\frac{ \begin{pmatrix}0\\4\end{pmatrix} }{2} =\begin{pmatrix}0\\2\end{pmatrix}\)

ii) show that PQ=½AC

\(\small{ \overline{PQ}=|\vec{P}-\vec{Q}|\\ =\sqrt{ ( \vec{P}-\vec{Q} )\cdot ( \vec{P}-\vec{Q} ) }\\ =\sqrt{ \left[ \begin{pmatrix}2\\0\end{pmatrix} -\begin{pmatrix}0\\2\end{pmatrix} \right] \cdot \left[\begin{pmatrix}2\\0\end{pmatrix} -\begin{pmatrix}0\\2\end{pmatrix} \right] }\\ =\sqrt{ \begin{pmatrix}2\\-2\end{pmatrix} \cdot \begin{pmatrix}2\\-2\end{pmatrix} }\\ =\sqrt{ 4+4 }\\ \overline{PQ}=\sqrt{ 8 }\\ \overline{AC}=|\vec{A}-\vec{C}|\\ =\sqrt{ (\vec{A}-\vec{C})\cdot (\vec{A}-\vec{C}) }\\ =\sqrt{ \left[ \begin{pmatrix}1\\-3\end{pmatrix} -\begin{pmatrix}-3\\1\end{pmatrix} \right] \cdot \left[\begin{pmatrix}1\\-3\end{pmatrix} -\begin{pmatrix}-3\\1\end{pmatrix} \right] }\\ =\sqrt{ \begin{pmatrix}4\\-4\end{pmatrix} \cdot \begin{pmatrix}1\\-4\end{pmatrix} }\\ =\sqrt{ 4\cdot8 }\\ \overline{AC}=2\cdot \sqrt{ 8 }\\ } \)

\(\dfrac{ \overline{PQ} }{\overline{AC} } = \dfrac{ \sqrt{ 8 } }{ 2\cdot \sqrt{ 8 } } = \dfrac{ 1 }{ 2 }\\ \overline{PQ} = \dfrac{ 1 }{ 2 }\cdot \overline{AC}\)

iii) show that PQ||AC

\((\vec{P}-\vec{Q} )\cdot ( \vec{A}-\vec{C} ) = \overline{PQ}\cdot \overline{AC} \cdot \cos{(\alpha)}\\ \cos{(\alpha)} = \frac{ (\vec{P}-\vec{Q} )\cdot ( \vec{A}-\vec{C} ) } {\overline{PQ}\cdot \overline{AC} }\\ \cos{(\alpha)}= \frac{ \left[ \begin{pmatrix}2\\0\end{pmatrix}-\begin{pmatrix}0\\2\end{pmatrix} \right]\cdot \left[ \begin{pmatrix}1\\-3\end{pmatrix}-\begin{pmatrix}-3\\1\end{pmatrix} \right] } { \sqrt{ 8 }\cdot 2\cdot \sqrt{ 8 } }\\ \cos{(\alpha)}= \frac{ \begin{pmatrix}2\\-2\end{pmatrix}\cdot \begin{pmatrix}4\\-4\end{pmatrix} } {16}\\ \cos{(\alpha)}= \frac{ 16 }{16}\\ \cos{(\alpha)}= 1 \qquad \Rightarrow \qquad \alpha = 0\qquad \Rightarrow \qquad PQ||AC \)

The answer has 1000 zeros its called a googolplex 1000000000000000000000000000000000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000000000000000000000

0000000000000000000000000000000000000000000000000000000000000000000000000000000000

0000000000000000000

667 days (rounded up)

10000000/15000=666.666667

x^3^2  is something that I am never quite sure of but I think this is how it is supposed to be interpreted :// 

(4x^3^2)x^5+7x^3(5x^4)^2

\(4x^9x^5+7 x^3(5 x^4)^2\\ =4x^{14}+7 x^3(25 x^8)\\ =4x^{14}+175x^{11} \)

The pressure in Denver, Colorado (elevation 5280 ft), averages about 24.9 in mmHg.

c) Convert this pressure to Pa.

\(\boxed{~ 1\ \text{mmHg} = 133.322\ \text{Pa} ~}\)

\(24.9\ \text{mmHg} = 24.9\cdot 133.322\ \text{Pa}\\ \mathbf{24.9\ \text{mmHg} = 3319.7178\ \text{Pa} }\\\)

b) Convert this pressure to psi.

\(\boxed{~ 1\ \text{pascal} = 0.00014503773801\ \text{pound force/square inch [psi]} ~}\)

\(24.9\ \text{mmHg} = 24.9\cdot 133.322\cdot 0.00014503773801\ \text{pound force/square inch [psi]}\\ \mathbf{24.9\ \text{mmHg} = 0.481484360543533578\ \text{pound force/square inch [psi]} }\\\)

a) Convert this pressure to atm.

\(\boxed{~ 1\ \text{pascal} = 0.0000098692326671601283\ \text{atm} ~}\)

\(24.9\ \text{mmHg} = 24.9\cdot 133.322\cdot 0.0000098692326671601283\ \text{atm}\\ \mathbf{24.9\ \text{mmHg} = 0.03276306735751295336779374\ \text{atm} }\\\)

milli means 1000 times smaller

so

a milliamp is 1000 times smaller than an amp.

that means there are 1000 milliamps in an amp.

just the same as there is 1000 millilitres in a litre. :)

Thanks answering guest that is a good suggestion.

I just want to look at it with substitution.

the graph of every exponential function f(x) = a^x, a>0, a cannot = 1, passes through which 3 points? 

f(1)=a^1 = a    so (1,a) is on the graph.

 a cannot be 0 so (1,0) is not     so that means the last 3 three are not right.

That just leaves the middle three.

(1/a, -1) , (1,0) , and (a,1)

(-1, 1/a) , (0,1) , and (1,a)

(0,0) , (1,1) , and (a,a)

That just leaves the middle three.

(-1, 1/a) , (0,1) , and (1,a)

f(1)=a    check

f(0)=a^0=1     check

f(-1)=a^(-1)=1/a     check     :)))

0.350 A = 350 mA

Later on in maths you will find that there is often more than one answer to a question. But each answer needs to be examined to see if it is valid.  

This is an example of that.

yes  side squared =area

therefore

side = +/-sqrt( area)

BUT the length of a square must be positive so the negative answer is invalid.

Does that help?

Thanks for that factorisation answering guest :)

I can take it one step further :))

\(k^3-k(a)^2=k^3 - ka^2=k(k^2 - a^2)\\=k(x-a)(k+a)\)

5x cubed divided by 10x to the power of 7

\(\begin{array}{rcl} \dfrac{(5x)^3}{(10x)^7} \\ &=&\dfrac{(5x)^3}{(2\cdot 5\cdot x)^7} \\ &=&\dfrac{(5x)^3}{2^7\cdot (5\cdot x)^7} \\ &=&\dfrac{1} {2^7\cdot (5x)^{7-3}} \\ &=&\dfrac{1} {2^7\cdot (5x)^{4}} \\ \dfrac{(5x)^3}{(10x)^7} &=&\dfrac{1} {2^7\cdot (5x)^{4}} \\ \end{array}\)

I'll just do some  examples

\(\sqrt2+\sqrt3\qquad \mbox{Cannot be simplified}\\ \sqrt2-\sqrt3\qquad \mbox{Cannot be simplified}\\ \sqrt2*\sqrt3=\sqrt{2*3}=\sqrt6\\ \sqrt2\div\sqrt3=\frac{\sqrt2}{\sqrt3}=\sqrt{\frac{2}{3}}\)

Although the first two cannot be simplified they are still added together.  :)

You can approximate a single number answer if you want to.

A milliamp is one 1000th of an amp.  

https://www.google.com.au/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=millliamp

(sin23)^2

It doesn't display the way I would like but it does give the correct answer.

(except for a little rounding error - that is normal)

\(PV=nRT\)

\(V_1=\dfrac 4 3 \pi \left(\dfrac {50}{2}\right)^3 \\ V_2 = \dfrac 4 3 \pi \left(\dfrac {51}{2}\right)^3 \\ \mbox{The rest of the quantities other than temperature stay constant.} \)

\(\dfrac{P V_1}{P V_2} = \dfrac{n R T_1}{n R T_2} \Rightarrow \dfrac {V_1}{V_2}=\dfrac {T_1}{T_2}\)

\(\dfrac {V_1}{V_2}=\dfrac{50^3}{51^3} = \dfrac {125000}{132651} \\ \mbox{so }\dfrac {T_1}{T_2} = \dfrac {125000}{132651} \\ T_2 = T_1 \dfrac{132651}{125000} = 19 \dfrac{132651}{125000} = 20.163 ^\circ C\)

Neglecting the effects of the atmosphere and the relativistic effects, we have:

.60 X 299,792,458=179,875,474.8m, the speed of the particle; the height of 500Km X 1,000=500,000m, hence 500,000/179,875,474.8=1/360 of 1 second. Time that the particle will take to reach the ground.

\(\mbox{20% of 4.2 }=\dfrac {20}{100} \times 4.2 = 0.2 \times 4.2 = 0.84\)

\(10C1 = \begin{pmatrix}10 \\ 1\end{pmatrix} = \dfrac {10!}{1!(10-1)!} = \dfrac {10!}{9!} = 10\)

Why don't you try and use this graphing calculator, so that you can learn and see it with your own eyes: https://www.desmos.com/calculator

Learn how to use it, because it is one of the better graphing calculators on the net.

-1-(-5x)-2x+2x2+7    

Simplify the following:
2 x^2--5 x-2 x-1+7

-(-5)  =  5:
2 x^2+5 x-2 x-1+7

Grouping like terms, 2 x^2+5 x-2 x-1+7 = 2 x^2+(5 x-2 x)+(-1+7):
2 x^2+(5 x-2 x)+(-1+7)

5 x-2 x = 3 x:
2 x^2+3 x+(-1+7)

7-1 = 6:
Answer: | 
| 2x^2+3x+6

(5/2)^x = (125/8)

x=3

Solve for x over the real numbers:
(5/2)^x = 125/8

Multiply both sides by 8:
2^(3-x) 5^x = 125

125 = 5^3:
2^(3-x)·5^x = 5^3

Express both sides with bases 2 and 5:
2^(3-x)·5^x = 2^0·5^3

Equate exponents of 2 and 5 on both sides:
3-x = 0 and x = 3

All equations give x = 3 as the solution:
Answer: | 
| x = 3

2x²⁺3x+6

In theory, yes! But, in real world that we live in, no.

It cannot be. Because the cube root of 100=4.6416....... is an IRRATIONAL number, which means you cannot write it as a fraction.