All Answers 358181 Answers

$$\frac{4pages}{hour}=\frac{4pages}{60 minutes}=\frac{1}{15 }\;\;page/minute$$

$$\\\frac{1}{a-1}+\frac{3}{3a-1}=0\\\\
(a-1)(3a-1)\left[\frac{1}{a-1}+\frac{3}{3a-1}\right]=0(a-1)(3a-1)\\\\
\frac{(3a-1)}{1}+\frac{3(a-1)}{1}=0\\\\
3a-1+3a-3=0\\\\
6a-4=0\\\\
6a=4\\\\
a=2/3$$

I suppose it depends on how hot the pockets are.  :/

It could be 4/54  I suppose.

y=2x+12   (1)

x+y=90      (2)

sub 1 into 2 and solve simultaneously

P(sheep dog) = 4/7

the answer wil be in secs, the km cancel out

$${\frac{\left({\mathtt{8.368}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{8}}}\right)}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{5}}}\right)}} = {\frac{{\mathtt{8\,368}}}{{\mathtt{3}}}} = {\mathtt{2\,789.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

2789 seconds

$${\frac{{\frac{\left({\mathtt{8.368}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{8}}}\right)}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{5}}}\right)}}}{{\mathtt{60}}}} = {\frac{{\mathtt{2\,092}}}{{\mathtt{45}}}} = {\mathtt{46.488\: \!888\: \!888\: \!888\: \!888\: \!9}}$$

46.49 minutes

$$\\\frac{455}{6.5}=\frac{2100}{x}\\\\
\frac{6.5}{455}=\frac{x}{2100}\\\\
\frac{6.5}{455}*2100=x\\\\
x=\frac{6.5}{455}*2100\\\\
x=30$$

30 hours

How bad is the traffic?

These are both exactly the same line so there is an infinite number of simultaneous solutions.

Mmm this is a long one.

J+25-12  =  M+12+31  =  C-31-25

J +13       =     M+43     =    C-56

J+M+13+43 = 2(C-56)

J+M+56 = 2C-112

J+M = 2C-168     (1)

J+M=(2/7)(J+M+C)         (2)

sub (1) into (2)                                 

2C-168=(2/7)(2C-168+C)  

2C-168=(2/7)(3C-168)                                                                           

7(2C-168)=2(3C-168)                                                                                  

14C-7*168=6C-2*168

8C = 5*168

C = 5*21

C = 105

M+43     =    C-56

M+43     =    105-56

M = 105-56-43

M =  6

J +13  =  M+43

J +13  =  6+43

J = 6+43-13

J= 36

Original number of lollies

 J= 36,  M =  6,    C = 105   

I have double checked that this fits with all the original information.

So originally Jon had 36 lollies 

Hallo Omi67,

du hast vollkommen recht!

Alle Lösungen lauten: $$x = 239.032604916\ensurement{^{\circ}} \pm k\cdot 360 \ensurement{^{\circ}}$$ und $$x = 41.3562528996\ensurement{^{\circ}} \pm k\cdot 360 \ensurement{^{\circ}}$$

k=0,1,2,...

Viele Grüße

The perimeter of a rectangular traffic sign is 106 inches. Also its length is 15 inches longer than its width. What is the length and the width of this sign ?

Let L = length and W = width

perimeter :  106 = 2L + 2W     (1)

L = W + 15

L-W = 15  | *2

2L - 2W = 30      (2)

$$\\(1) \qquad 2L + 2W = 106 \\
(2) \qquad 2L - 2W = 30$$

$$\begin{array}{lrcl}
\hline
(1)+(2):& 4L &=& 136\\\\
& L &=& \dfrac{136}{4} \\\\
& \textcolor[rgb]{1,0,0}{L} & \textcolor[rgb]{1,0,0}{=} & \textcolor[rgb]{1,0,0}{34} } \\\\
\hline
& W &=& L - 15 \\\\
& W &=& 34 - 15 \\\\
& \textcolor[rgb]{1,0,0}{W} & \textcolor[rgb]{1,0,0}{=} & \textcolor[rgb]{1,0,0}{19} }
\end{array}$$

yes it is

Nice one TJM 

it is asin

Differentiate y with respect to x:

y=6x^2+5x-10

$$y = 6x^{\textcolor[rgb]{1,0,0}{2}} +5x^{\textcolor[rgb]{1,0,0}{1}} -10x^{\textcolor[rgb]{1,0,0}{0}}}\\\\
y' = 6\cdot \textcolor[rgb]{1,0,0}{2} \cdot x^{ \textcolor[rgb]{1,0,0}{2} - 1 } + 5 \cdot \textcolor[rgb]{1,0,0}{1} \cdot x^{ \textcolor[rgb]{1,0,0}{1} - 1 } - 10 \cdot \textcolor[rgb]{1,0,0}{0} \cdot x^{ \textcolor[rgb]{1,0,0}{0} - 1 }\\\\
y'= 12x^1+5x^0+0\\\\
y' = 12x+5$$

How much is a year ?

Year Definitions

The most common definition in the western world of the year is based on the revolution of the Earth around the Sun and is therefore called a `Solar Year'. However, there are several possibilites to define beginning and end of one revolution and thus also several kinds of solar years:

• A tropical year is the interval between two successive passages of the Mean Sun through the mean vernal equinox and lasts 365.242199 days UT. The name refers to the changes of seasons (greek `tropai', the turning points) which are fixed in this kind of year. It is for this reason that the tropical year is of great importance in the construction of calendars.
• A siderial year is the interval between two successive passages of the Mean Sun at the same (fixed) star. It lasts 365.256366 days UT.
• An anomalistic year is the interval between two successive passages of the Earth through the perihelion (the point closest to the Sun) of its orbit and it lasts 365.259636 days UT.

The years so defined differ in length because of the precession of Earth's rotation and the tumbling of the Earth orbit.
Julian year (365.25 days UT) and Gregorian year (365.2425 days UT) as defined in the calendars of the respective name are solar years as well.

YES

in the top corner of the calc it says "scientific" Press that to get the drop down menu.

Matrices are second from the bottom.

pi is circumference divided by the diameter.

difference implies subtract, which is not correct.

The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s = 2 sin πt + 4 cos πt, where t is measured in seconds. (Round your answers to two decimal places.)

i. [1,2]

$$\\s(1)=2sin(\pi *1)+4cos(\pi *1)\\
s(1)=2sin(\pi)+4cos(\pi)\\
s(1)=0+4*-1\\
s(1)=-4\\\\
\\s(2)=2sin(\pi *2)+4cos(\pi *2)\\
s(2)=2sin(2\pi)+4cos(2\pi)\\
s(2)=0+4*+1\\
s(2)=4\\\\
Average\; velocity=\frac{distance}{time}\\
Average\; velocity=\frac{4--4}{2-1}\\
Average\; velocity=\frac{8}{1}\\
Average\; velocity=8 \;cm/sec$$

ii. [1,1.1]

iii. [1.1.01]

iv. [1.1.001]

(b) Find the instaneous velocity of the particle when t=1

$$\\s=2sin(\pi t)+4cos(\pi t)\\
v(t)=\frac{ds}{dt}=2\pi cos(\pi t)+-4\pi sin(\pi t)\\
v(t)=\frac{ds}{dt}=2\pi cos(\pi t)-4\pi sin(\pi t)\\
v(1)=2\pi cos(\pi*1 )-4\pi sin(\pi *1)\\
v(1)=2\pi *-1 \quad -4\pi *0\\
v(1)=-2\pi \;\;cm/sec\\$$

Hi Vraces - it is good to see you on the forum again :)

Thanks CPhill

I have gone through and given you both points for all your questions and answers.

Both were great.  We want people to be persistent around here.    

Vraces, I think that you may have a problem with factorising. 

If you want help with this you have only to log back on and ask for it. 

I'll show you some examples.

Please try to see that each time I have taken the common factor out the front and left the rest in the brackets.

$$\\\textcolor[rgb]{0,0,1}{6}-0.2*\textcolor[rgb]{0,0,1}{6}\\
=1*\textcolor[rgb]{0,0,1}{6}-0.2*\textcolor[rgb]{0,0,1}{6}\\
=\textcolor[rgb]{0,0,1}{6}(1-0.2)\\
=\textcolor[rgb]{0,0,1}{6}*0.8 \\
= 0.8*\textcolor[rgb]{0,0,1}{6}\\\\\\\\
\textcolor[rgb]{0,0,1}{y}-0.2\textcolor[rgb]{0,0,1}{y}\\
= 1*\textcolor[rgb]{0,0,1}{y}-0.2*\textcolor[rgb]{0,0,1}{y}\\
= \textcolor[rgb]{0,0,1}{y}(1-0.2)\\
=\textcolor[rgb]{0,0,1}{y}*0.8\\
=0.8\textcolor[rgb]{0,0,1}{y}\\\\\\\\
\textcolor[rgb]{0,0,1}{10^x}-0.2*\textcolor[rgb]{0,0,1}{10^x}\\
=1*\textcolor[rgb]{0,0,1}{10^x}-0.2*\textcolor[rgb]{0,0,1}{10^x}\\
=\textcolor[rgb]{0,0,1}{\textcolor[rgb]{0,0,1}{10^x}}(1-0.2)\\
=\textcolor[rgb]{0,0,1}{\textcolor[rgb]{0,0,1}{10^x}}*0.8 \\
= 0.8*\textcolor[rgb]{0,0,1}{10^x}$$

suppose a capacitor has XC =-8800 Ohms at f = 830 kHz. What is C ?

$$\boxed{X_C = \dfrac{ 1 }{ 2 \cdot \pi \cdot f \cdot C } \qquad
C = \dfrac{ 1 }{ 2 \cdot \pi \cdot f \cdot X_C }
} \\$$

Capacitive reactance, where: X_C = reactance in ohms (ohm), f = frequency in hertz (Hz), C = capacitance in farads (F)

We have:

reactance:  $$\small{\text{$X_C = 8800\ \Omega$}}$$

frequency: $$\small{\text{$f = 830000\ \rm{Hz} $}}$$

capacitance:  $$\small{\text{$C=\dfrac{1}{2\cdot\pi\cdot f\cdot X_C} $}}
\small{\text{$=\dfrac{1}{2\cdot\pi\cdot 830000\cdot (8800)} $}}
\small{\text{$=0.00000000002179011\ \rm{F} = 21.79011\ \rm{pF}$}}$$

Do you mean 

$$\\6log_3(x-3)=-24\\\\
log_3(x-3)=-4\\\\
3^{log_3(x-3)}=3^{-4}\\\\
x-3=3^{-4}\\\\
x-3=\frac{3^{-4}}{1}\\\\
x-3=\frac{1}{3^{+4}}\\\\
x-3=\frac{1}{81}\\\\
x=3\frac{1}{81}\\\\$$