All Answers 356325 Answers

Perimeter of a rectangle =  2 l + 2 w = 48

  Now if we scale these values 1.5  we just multiply both sides of the above equation by 1.5

   1.5 (2l + 2w)  = 1.5 (48)  = 72 ft    

Volume of a cylinder is area x height = pi r^2 h       We want to find 'r'

v = pi r^2 h    solve for 'r'

v/(pi h) = r^2

sqrt (v/(pi h) = r

sqrt (18488/(pi*23) = r = 15.995 m = ~ 16 m

Second one:

Area of cylinder   pi r^2 = 36 pi

Volume she ate =  area x  5 cm   (the jar went from 11 cm to 6 cm = 5 cm)

                          =  36 pi x 5  = 565 cm^2     (and now Kelly has heartburn ! ) 

Sorry Melody, but the axis of symmetry is x = 1/2 not -1/2.

Also, it's quicker to say that the equation of the parabola must be 

\(y=(x+2)(x-3)=x^2-x-6\)

so b + c =  -7.

100/7 = 14 .28 ~ 14 numbers

100/10 =10 numbers

24 numbers total are divisible by 10 or 7

One number ( 70 ) is divisible by 10 AND 7

What is the value of   \(b+c \)   if     \(x^2+bx+c>0 \)     only when    \(x\in (-\infty, -2)\cup(3,\infty)\)?

Consider the concave up parabola   y=x^2+bx+c

\(x^2+bx+c>0 \)    will be true when y is positive, which is when the graph is above the x axis.

The roots of this graph are 

        \(x=\frac{-b\pm\sqrt{b^2-4c}}{2}\\ x=\frac{-b}{2}\pm\frac{\sqrt{b^2-4c}}{2}\\\)

the axis of symmetry of this graph will be     \(x=\frac{-b}{2}\)

No consider the domain where x is poistive.

It is less than -2 and greater then 3  SO the axis of symmetry must be half way between these points.  (-2+3)/2=-1/2

so

\(\frac{-b}{2}=\frac{-1}{2}\\ b=1\)

Also

\(\frac{-b}{2}+\frac{\sqrt{b^2-4c}}{2}=3\\ -b+\sqrt{b^2-4c}=6\\ sub\;in\;b=1\\ -1+\sqrt{1^2-4c}=6\\ \sqrt{1-4c}=7\\ 1-4c=49\\ -4c=48\\ c=-12\\ ~\\ b+c=1+-12=-11\)

2. Solution:

I asume:
Prove that if \(w,z\) are complex numbers such that \(|w|=|z|=1\)
and
\(wz\ne -1\), then
\(\frac{w+z}{1+wz}\) is a real number.

\(\text{Let $w = a+\imath b \qquad a^2+b^2 = 1\quad (|w|=1)$ } \\ \text{Let $z = c+\imath d \qquad c^2+d^2 = 1\quad (|z|=1) $ }\)

\(\begin{array}{|rcll|} \hline w+z &=& a+\imath b + c+\imath d \\ &=& (a+c) + \imath( b + d) \\\\ w\cdot z &=& (a+\imath b) + (c+\imath d) \\ &=& (ac-bd)+\imath(ad+bc) \\ \hline \end{array}\)

\(\small{ \begin{array}{|rcll|} \hline && \dfrac{w+z}{1+wz} \\\\ &=& \dfrac{ (a+c) + \imath( b + d) } { 1+(ac-bd)+\imath(ad+bc) } \\\\ &=& \dfrac{ (a+c) + \imath( b + d) } { (1+ac-bd)+\imath(ad+bc) } \\ && | \quad \small{\text{ Multiply top and bottom by the conjugate}} \\\\ &=& \left[ \dfrac{ (a+c) + \imath( b + d) } { (1+ac-bd)+\imath(ad+bc) } \right] \left[ \dfrac{(1+ac-bd)-\imath(ad+bc)} { (1+ac-bd)-\imath(ad+bc) } \right] \\\\ &=& \dfrac{ \left[(a+c) + \imath( b + d)\right] \left[(1+ac-bd)-\imath(ad+bc) \right] } { (1+ac-bd)^2 - \imath^2 (ad+bc)^2 } \\\\ && \quad | \quad \imath^2=-1 \\\\ &=& \dfrac{ (a+c)(1+ac-bd) -\imath(a+c)(ad+bc) +\imath( b + d)(1+ac-bd)-\imath( b + d)\imath(ad+bc) } { (1+ac-bd)^2 + (ad+bc)^2 } \\\\ &=& \dfrac{ (a+c)(1+ac-bd) -\imath(a+c)(ad+bc) +\imath2( b + d)(1+ac-bd)-\imath^2( b + d)(ad+bc) } { (1+ac-bd)^2 + (ad+bc)^2 } \\\\ && \quad | \quad \imath^2=-1 \\\\ &=& \dfrac{ (a+c)(1+ac-bd) -\imath(a+c)(ad+bc) +\imath2( b + d)(1+ac-bd)+( b + d)(ad+bc) } { (1+ac-bd)^2 + (ad+bc)^2 } \\\\ &=& \dfrac{ (a+c)(1+ac-bd)+( b + d)(ad+bc) + \imath \Big(( b + d)(1+ac-bd)-(a+c)(ad+bc) \Big) } { (1+ac-bd)^2 + (ad+bc)^2 } \\ \hline \end{array} } \)

The imaginary part of the top line, showing that it's equal to zero:

\(\begin{array}{|rcll|} \hline \imath \left[ \underbrace{ ( b + d)(1+ac-bd)-(a+c)(ad+bc) }_{= 0\ ?}\right] \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && (b + d)(1+ac-bd)-(a+c)(ad+bc) \\\\ &=& b+abc-b^2d+d+acd-d^2b -(a^2d+abc+acd+c^2b ) \\ &=& b+\not{ab}c-b^2d+d+\not{acd}-d^2b -a^2d-\not{abc}-\not{acd}-c^2b \\ &=& b-b^2d+d -d^2b -a^2d -c^2b \\ &=& b+d -b(\underbrace{c^2+d^2}_{=1})-d(\underbrace{a^2+b^2}_{=1}) \\ &=& b+d -b -d \\ &=& 0 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \imath \left[ ( b + d)(1+ac-bd)-(a+c)(ad+bc) \right] = 0 \\ \hline \end{array} \)

I assume:

Prove that if \(w,z \) are complex numbers such that \(|w|=|z|=1\)

and  \(wz\ne -1\), then

\(\frac{w+z}{1+wz}\) is a real number.

\(\text{Let $w = \cos(\theta)+ \imath \sin(\theta)$ } \\ \text{Let $z = \cos(\phi) + \imath\sin(\phi)$ }\)

\(\begin{array}{|rcll|} \hline w+z &=& \cos(\theta)+ \imath \sin(\theta) + \cos(\phi) + \imath\sin(\phi) \\ &=& \Big(\cos(\theta)+ \cos(\phi)\Big) + \imath \Big( \sin(\theta) + \sin(\phi) \Big) \\\\ w\cdot z &=& \Big(\cos(\theta)+ \imath \sin(\theta) \Big) \Big( \cos(\phi) + \imath\sin(\phi) \Big) \\ &=& \cos(\theta)\cos(\phi) + \imath\sin(\phi)\cos(\theta) + \imath \sin(\theta)\cos(\phi) + \imath^2 \sin(\theta)\sin(\phi) \\ &=& \cos(\theta)\cos(\phi)+ \imath^2 \sin(\theta)\sin(\phi) + \imath \Big(\sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta) \Big) \\ && | \quad \imath^2 = -1 \\ &=& \underbrace{\cos(\theta)\cos(\phi)- \sin(\theta)\sin(\phi)}_{=\cos(\theta+\phi)} + \imath \Big(\underbrace{\sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta)}_{=\sin(\theta+\phi)} \Big) \\ &=& \cos(\theta+\phi) + \imath \sin(\theta+\phi) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \dfrac{w+z}{1+wz} \\\\ &=& \dfrac{ \Big(\cos(\theta)+ \cos(\phi)\Big) + \imath \Big( \sin(\theta) + \sin(\phi) \Big) } { 1+\cos(\theta+\phi) + \imath \sin(\theta+\phi) } \\\\ &=& \dfrac{ \Big(\cos(\theta)+ \cos(\phi)\Big) + \imath \Big( \sin(\theta) + \sin(\phi) \Big) } { \Big(1+\cos(\theta+\phi)\Big) + \imath \sin(\theta+\phi) } \\ && | \quad \small{\text{ Multiply top and bottom by the conjugate}} \\\\ &=& \left[ \dfrac{\Big(\cos(\theta)+ \cos(\phi)\Big) + \imath \Big( \sin(\theta) + \sin(\phi) \Big) } { \Big(1+\cos(\theta+\phi)\Big) + \imath \sin(\theta+\phi) } \right] \left[ \dfrac{\Big(1+\cos(\theta+\phi)\Big) - \imath \sin(\theta+\phi) } { \Big(1+\cos(\theta+\phi)\Big) - \imath \sin(\theta+\phi) } \right] \\\\ &=& \dfrac{ \left[\Big(\cos(\theta)+ \cos(\phi)\Big) + \imath \Big( \sin(\theta) + \sin(\phi) \Big)\right] \left[\Big(1+\cos(\theta+\phi)\Big) - \imath \sin(\theta+\phi) \right] } { \Big(1+\cos(\theta+\phi)\Big)^2 - \imath^2 \sin^2(\theta+\phi) } \\\\ &=& \dfrac{ \left[\Big(\cos(\theta)+ \cos(\phi)\Big) + \imath \Big( \sin(\theta) + \sin(\phi) \Big)\right] \left[\Big(1+\cos(\theta+\phi)\Big) - \imath \sin(\theta+\phi) \right] } { \Big(1+\cos(\theta+\phi)\Big)^2 + \sin^2(\theta+\phi) } \\\\ &=& \dfrac{\Big(\cos(\theta)+ \cos(\phi)\Big)\Big(1+\cos(\theta+\phi)\Big) -\Big(\cos(\theta)+ \cos(\phi)\Big) \imath \sin(\theta+\phi) \\ +\imath \Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) -\imath \Big( \sin(\theta) + \sin(\phi) \Big) \imath \sin(\theta+\phi) } { \Big(1+\cos(\theta+\phi)\Big)^2 + \sin^2(\theta+\phi) } \\\\ &=& \dfrac{\Big(\cos(\theta)+ \cos(\phi)\Big)\Big(1+\cos(\theta+\phi)\Big) -\imath \Big(\cos(\theta)+ \cos(\phi)\Big) \sin(\theta+\phi) \\ +\imath \Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) -\imath^2 \Big( \sin(\theta) + \sin(\phi) \Big)\sin(\theta+\phi) } { \Big(1+\cos(\theta+\phi)\Big)^2 + \sin^2(\theta+\phi) } \\\\ &=& \dfrac{\Big(\cos(\theta)+ \cos(\phi)\Big)\Big(1+\cos(\theta+\phi)\Big) + \Big( \sin(\theta) + \sin(\phi) \Big)\sin(\theta+\phi) \\ -\imath \Big(\cos(\theta)+ \cos(\phi)\Big) \sin(\theta+\phi) +\imath \Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) } { \Big(1+\cos(\theta+\phi)\Big)^2 + \sin^2(\theta+\phi) } \\\\ &=& \dfrac{\Big(\cos(\theta)+ \cos(\phi)\Big)\Big(1+\cos(\theta+\phi)\Big) + \Big( \sin(\theta) + \sin(\phi) \Big)\sin(\theta+\phi) \\ +\imath \left[ \Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) - \Big(\cos(\theta)+ \cos(\phi)\Big) \sin(\theta+\phi) \right] } { \Big(1+\cos(\theta+\phi)\Big)^2 + \sin^2(\theta+\phi) } \\ \hline \end{array}\)

The imaginary part of the top line, showing that it's equal to zero:

\(\begin{array}{|rcll|} \hline \imath \left[ \underbrace{\Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) - \Big(\cos(\theta)+ \cos(\phi)\Big) \sin(\theta+\phi) }_{= 0\ ?}\right] \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) -\Big(\cos(\theta)+ \cos(\phi)\Big) \sin(\theta+\phi) \\\\ &=& \sin(\theta) + \sin(\phi) \\ &\qquad +& \sin(\theta)\cos(\theta+\phi) \\ &\qquad +& \sin(\phi)\cos(\theta+\phi) \\ &\qquad -& \cos(\theta)\sin(\theta+\phi) \\ &\qquad -& \cos(\phi)\sin(\theta+\phi) \\\\ &\qquad & \boxed{\cos(\theta+\phi) = \cos(\theta)\cos(\phi)- \sin(\theta)\sin(\phi) \\ \sin(\theta+\phi) = \sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta)} \\\\ &=& \sin(\theta) + \sin(\phi) \\ &\qquad +& \sin(\theta)\Big( \cos(\theta)\cos(\phi)- \sin(\theta)\sin(\phi) \Big) \\ &\qquad +& \sin(\phi)\Big( \cos(\theta)\cos(\phi)- \sin(\theta)\sin(\phi) \Big) \\ &\qquad -& \cos(\theta)\Big( \sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta) \Big) \\ &\qquad -& \cos(\phi)\Big( \sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta) \Big) \\\\ &=& \sin(\theta) + \sin(\phi) \\ &\qquad +& \sin(\theta)\cos(\theta)\cos(\phi)- \sin(\theta)\sin(\theta)\sin(\phi) \\ &\qquad +& \sin(\phi)\cos(\theta)\cos(\phi)- \sin(\phi)\sin(\theta)\sin(\phi) \\ &\qquad -& \cos(\theta)\sin(\theta)\cos(\phi) - \cos(\theta)\sin(\phi)\cos(\theta) \\ &\qquad -& \cos(\phi)\sin(\theta)\cos(\phi) - \cos(\phi)\sin(\phi)\cos(\theta) \\\\ &=& \sin(\theta) + \sin(\phi) \\ &\qquad +& \sin(\theta)\cos(\theta)\cos(\phi)- \sin^2(\theta)\sin(\phi) \\ &\qquad +& \sin(\phi)\cos(\theta)\cos(\phi)- \sin^2(\phi)\sin(\theta) \\ &\qquad -& \cos(\theta)\sin(\theta)\cos(\phi) - \cos^2(\theta)\sin(\phi) \\ &\qquad -& \cos(\phi)\sin(\phi)\cos(\theta) - \cos^2(\phi)\sin(\theta) \\\\ &=& \sin(\theta) + \sin(\phi) \\ &\qquad +& -\sin(\theta)\Big( \underbrace{\sin^2(\phi)+ \cos^2(\phi)}_{=1} \Big) \\ &\qquad +& - \sin(\phi)\Big( \underbrace{\sin^2(\theta)+\cos^2(\theta)}_{=1} \Big) \\ &\qquad +& \underbrace{\sin(\theta)\cos(\theta)\cos(\phi) - \sin(\theta)\cos(\theta)\cos(\phi)}_{=0} \\ &\qquad +& \underbrace{\sin(\phi)\cos(\phi)\cos(\theta) - \sin(\phi)\cos(\phi)\cos(\theta)}_{=0} \\\\ &=& \sin(\theta) + \sin(\phi) -\sin(\theta) - \sin(\phi) \\ &=& 0 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \imath \left[ Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) - \Big(\cos(\theta)+ \cos(\phi)\Big) \sin(\theta+\phi) \right] = 0 \\ \hline \end{array}\)

Apparently I've been up to long. I didn't put the 10 in when I answered the question! Thanks again for your help. 

Here is the diagram of the info given:

Break the two speeds into their x and y components...then add them together

Airplane  ax=0  ay=700

Wind      wx= -40 cos45    wy= -40sin45     (see diagram)

                     = -28.28             = -28.28

Add together for the RESULTANT

x=-28.28   y = 700-28.28= 671.71

Degrees off course will be arctan (-28.28/671.71) = 2.41 degrees off course to the west

Speed will be   sqrt ( x^2 + y^2) = 672.3 km/hr

Yes, the SECOND question is right. After I converted it ends up being 32.24kcal. 

Yeah I've redone it a few times and I keep getting 2.28e+06 J, and it's still saying it's wrong.

Thank you for all of your help. I really appreciate it. 

GA  pointed out that in j the heat of fusion is 333.55 j/g  ~ 79.72 cal/g   

    I used 80 cal/gm   (which is what we used when I was in H.S. Physics many years ago)  in my calc....so you may want to change that constant to get a slightly different answer to the SECOND question.....  OK ????   Instead of 24000 it would be 23916 calories for that part of the solution.

Yes, if I use your EXAMPLE

How much heat in joules is needed to raise the temperature of 7.5 L of water from 0°C to 87.0°C? (Hint: Recall the original definition of the liter.)
 2.73e+06 J

7.5 L x 1000 gm/l   1 cal/gm-c x 87c = 652500 cal = 2.73 e +6    checks out fine

    I think the answer they are giving you is incorrect....

        either the temp change is wrong or the amount of water is incorrect.  

This problem was just an example I had. 

How much heat in joules is needed to raise the temperature of 7.5 L of water from 0°C to 87.0°C? (Hint: Recall the original definition of the liter.)
 2.73e+06 J

The one I need to answer is this. 

How much heat in joules is needed to raise the temperature of 7.0 L of water from 0°C to 78.0°C? (Hint: Recall the original definition of the liter.)

I had already attempted to enter the answer in as you just said and it didn't work. So I'm really not sure what is going wrong.

Well then the first answer becomes

2284.5 kj = 2284500j = 2.28 e+06

I am thinking there may be a slight conversion factor error in there....or an error in your answer.....

    I see no other way to get other than the answer I calculated.

What numbers are you using for specific heat of water?

Thanx GA...I have always used 80  

Thank you! I tried to redo it, but I just could not figure it out.

The first answer was wrong as well, the answer needs to be in this format. 2.73e+06 J
Any ideas?

EP, the heat of fusion for 1g of water at 0 °C is approximately 334 joules or 79.7 calories. 

GA

Answer already posted to previous submission.

I apologize.....I used the wrong specific heat for ice   I use 1     it is .5

so to heat the ice to zero degrees will require

300 x .5 x 15 = 2250 cal    (not 4500)

NOW add them up   2250 + 24000 + 6000 = 32250 cal = 134934 j = 134.9 kj

Sorry!

Must be, thanks regardless!

Maybe a little easier way

16 ^2t  = n ^4t

(4^2)^2t = n^4t

4^4t = n^4t         so n = 4

Miriam starts with   16^2t      And J starts with n^4t    and we want these to be equal.....

Sooooo let's set them equal and solve for n    the amount of bacteria Jessica needs

16^2t  = n ^4t         Log both sides to get

2t log16  = 4t log n  divide by 2t

log16 = 2 log n        divide by 2

(log 16 )/2 = log n    10 power both sides

10^((log16)/2) = n = 4   the number J needs to start with to keep things equal

Unfourtanetly that wasn't correct.

I have been given this information, but I still can't seem to get the correct answer.

1. How much heat in joules is needed to raise the temperature of 7.5 L of water from 0°C to 87.0°C? (Hint: Recall the original definition of the liter.)
 2.73e+06 J

2.  How much heat is necessary to change 325 g of ice at -20°C to water at 20°C?

change g to kilograms; and to figure the heat here because it is original ice and will need to be melted you will do

The specific heat of ice calculation to get it from -20 to 0 °C + the heat of fusion to melt the ice +  the specific heat calculation to raise it to 20°C :

The specific heat of ice is .5kcal/kg*°C

H = m c ice ΔT + mLf  + m c ΔT =

 .325 kg (.5kcal   ) (20°C) + .325 kg (80 kcal)  + .325 kg ( 1 kcal)  (20°C)  = 35.75 kcal

                kg *°C                                        Kg                      kg * °C