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Your help:

Intepretation: \((a-1)\cdot x^2+(a-3)\cdot x-2>0\)

Solution:

\(1.a=0, x<-1 \)

\(2.a>0, x>\frac{2}{a-1}\)

\(3.a>1, x<-1\)

\(4.-1

\(5.a<-1,\space-1

(Source: http://www.wolframalpha.com/input/?i=(a-1)*x%5E2%2B(a-3)*x-2%3E0)

Input: b(5px-3c) = a(qx-4)

Intepretation: Solve for \(x\) in \(\)\(b(5px+3c) = a(qx-4)\)

Expand:

\(5bpx-3bc=aqx-4a\)

Move the \(x\) terms to the right, integer ones to the left:

\(5bpx-aqx=3bc-4a\)

Pull the x out from the left term:

\((5bp-aq)x=3bc-4a\)

Divide both sides by \(5bp-aq\):

\(x=\frac{3bc-4a}{5bp-aq}\)

Done :D

This is called a \("Repeating\space Decimal"\), just like how the results are in fractional loops, the cause of it is because the remainder always cease to disappear in divisions, like how grass never go away no matter how many times you cut it.

In your case, \(\frac{191}{7}=27.285714285714...\)

The number \(\frac{191}{7}\) can also be written as \(27+\frac{2}{7}\)

Since we know that \(\frac{1}{7}=0.142857142857...\)

The number  \(\frac{2}{7}=2\cdot\frac{1}{7}=0.285714285714...\)

Adding the previously deattached \(27\)

\(\frac{191}{7}\) can be written in decimal form as: \(27.285714285714...\)

(Source: https://en.wikipedia.org/wiki/Repeating_decimal)

Hope this helps :)

3y - y² = ?

\(3y-y^2=\color{blue}y\times (3-y)\)

  !

Why do you have to be so NEGATIVE?

Whats 2/9 divided by 1/5

One divides by a fraction,
multiplying by the reciprocal value.

\(\frac{2}{9}:\frac{1}{5}=\frac{2}{9}\times\frac{5}{1}=\frac{10}{9}\color{blue}=1\frac{1}{9}\)

  !

We can think of concentric circles, here....if R is the radius from the middle of the fan to the halway point, then 2R is the distance from the center to the edge....thus....in one revolution the center point travels 2piR  inches 

and the outside point travels 2pi * 2R =   4piR  inches  = twice as far

In 30 minutes....the center point travels twice as far as it does in 15 minutes =   97968 * 2  = 195,936 inches

And the outside edge will travel twice as far as this = 391,872 inches

15= 15/1, 15/1*5/9= 5/1*5/3 = 25/3.

True area  =   pi (20/2)^2  = 100 pi cm^2

If he underestimates the circumference by 20%....the measured diameter  is 20 * .80  = 16 cm

And the  area in this case   =     pi (16/2)^2  = pi * 8^2  =  64 pi

So.....the percent error in the area in this case   =   l  100 pi - 64pi  l / l 100 pi l  = 36 / 100  = 36%

If he overestimates the circumference by 20%....the measured diameter  = 20 * 1.20  = 24 cm

And the area in this case = pi (24/2)^2  = pi * 12^2  = 144 pi

So.....the percent error in this case  =  l  144 pi - 100 pi  l  / l  100 pi  l   =   44 / 100  =  44%

( √5) ^-2    is the same as    1  / (√5)²     =   1 / [ √5 * √5 ]   =   1  / 5

23. Number of tickets.    Total Cost

                   2.                            $7.  

                   4.                            $14

                   6.                            $21

These are all pretty much the same type of linear problem .......let x be the Number of Tickets and y be the Total Cost

Note that  as y changes by $7, x changes by 2.....so...the slope of our line  = 7/2

And letting one point  be  (2, 7)....we have the following equation :

y = (7/2)(x - 2) + 7

24, 25  and 27 are similar to 23

26.   Total cost.      Change from $10

            $10.00.                    $0

            $9.00.                      $1.00 

            $7.50.                      $2.50

This one is a little tricky......let the Change from $10 be y  and the Total Cost  = x

The function  that models this  is :  y  = 10 - x

28.   Miles traveled.    Miles remaining 

                 0.                           500

               125.                         375

               350.                         150

Let the Miles Remaining be  y and the  Miles Traveled

This is similar to 26.....the function that models this is   y = 500 - x

Jeff, this is an example of messy:  Complex mathematics presented in ASCII text.

Mr. BB presentations as the Answer Man are usually just a plop of slop –a vast sea of ASCII soup coupled with descriptions of mathematical symbols pasted and plastered from Wolfram Alpha’s automated solutions. I usually need a double dose of Dramamine to prevent seasickness when I try to decipher them. This particular presentation is a little better than his usual slop.  This still requires wading, but you just need wading boot instead of the usual hip boots.  I didn’t need the Dramamine this time. 

Actually, I’m impressed that Mr. BB can query Wolfram proficiently enough to obtain this answer. Aside from Wolfram’s claim for natural language query, higher-level math usually requires precise wording or symbolic code to prevent absurd interpretations.

I was going to teach my cat how to do this, but my dog volunteered to teach him for me. (Good dog, Mr. Peabody!)

\(\text {Sir Alan's presentation in LaTeX}\\ \tiny \text {Reproduced verbatim (as much as practicable) with corrections from listed errata. }\\ \text { }\\ \small \text {definition: } \hspace{25 mm} years : = 315000000 {{\mkern 1mu\cdot\mkern 1mu}} s\\ \small \text {half-life: } \hspace{28 mm} \tau: =3{{\mkern 1mu\cdot\mkern 1mu}} years\\ \small \text {initial isotope power: } \hspace{6 mm} Q_0=100{{\mkern 1mu\cdot\mkern 1mu}}W\\ \text{ }\\ \small \text {isotope power as } \\ \small \text {a function of time: } \hspace{10 mm} Q(t): = Q_0{{\mkern 1mu\cdot\mkern 1mu}}e^{-ln(2)\cdot(\frac{t}{\tau})}\\ \small \text {plate area: } \hspace{22 mm} A: = 9{{\mkern 1mu\cdot\mkern 1mu}} mm^2\\ \small \text {plate distance: }\hspace{16 mm} x: = 5 {{\mkern 1mu\cdot\mkern 1mu}} mm\\ \)

\(\small \text {power received at plate } \\ \small \text {as a function of time: } \hspace{9 mm} P(t): = \frac{A \cdot Q(t)}{x^2}\\ \small \text {beginning of third year: } \hspace{9 mm} t_1: = 3 {{\mkern 1mu\cdot\mkern 1mu}} years\\ \small \text {middle of eighth year: } \hspace{12 mm} t_2: = 7.5\; years\\ \small \text { }\\ \small \text {total ENERGY received: } E:= \int_{t_1}^{t_2} P(t) \; dt\\ \text{ }\\ \hspace{55 mm} E=\frac{A}{x^2}\cdot Q_0 \cdot \frac{\tau}{ln(2)} \cdot (e^{\frac{ln(2)\cdot t_2}{\tau}} -e^{\frac{ln(2)\cdot t_2}{\tau}}) \text{ }\\ \hspace{55 mm} E = 2.224 \cdot 10^9 J\\\)

-----------

By reproducing this, it seems I’ve demonstrated two proofs. Q.E.D.

1) If you prefer the format in LaTeX, you can do it yourself. It has the same messiness but with a different font. For an example of true messiness, see my troll post for the “Answer Man,” aka Mr. BB.

2) You are an idiot on multiple levels! You’ve risk alienating the only person (in regular attendance) who can answer this type of question.  You are well too far on the wrong side of the education gulf to be this arrogant. If you continue, you will only ever earn a dimwit degree in Stupid. 

Mr. BB has a dimwit degree in Stupid. He is about as educated as you are but he’s more arrogant. He uses his degree as a license to torture forum members with presentations so incompetent that it amuses my cat.  My cat likes him for some reason. I really don’t know why. Maybe because he makes him laugh or maybe because my cat is a (Monopoly) banker and embezzler, or maybe it’s because he likes rodents. It’s probably all three.     

sqrt(-36)^3 =  You can write it like so:

sqrt(-1)^3 x sqrt(36)^3     =

sqrt(-1)^3 x           6^3      = 

  i^3           x          216      =

[-1 x i]       x           216      =

-i                x           216     = - 216i

Thanks, Melody.....here's another way that is non-trig based.....

Look at the following image :

Let's forget about the circle for a moment -  we'll return to this in a second.....

Let us place the side with length 6 along the x axis such that its endpoints lie at (-3,0)  and (3,0)...let's label these as  A and C

Let the altitude of the triangle lie along the y axis......this will actually form two right triangles.....each with a hypotenuse of 5

So....using the Pythagorean Theorem the altitude of our triangle will be sqrt [ 5^2 - 3^2]  = sqrt [ 16]  = 4

And let us call the altitude BD

Now....the center of our circle will lie on  AD....and let us call this point (0, y)

And.......the radius of our circle will pass through all three vertex points of triangle ABC.....this means that the distance from (0, y) to B    will equal the distance from  (0,y)  to C...and  B has the coordinates (0, 4)  and C has the coordinates (3,0)

So...using the distance formula,  we have that

sqrt  [ ( 0 - 0)^2  + ( y - 4)^2 ]  = sqrt [ (0-3)^2 + (y - 0)^2]     squaring both sides and simplifying, we have that

(y - 4)^2  =  9 + y^2

y^2 - 8y +  16  = 9 + y^2       subtract y^2 from both sides

-8y + 16 =  9                       add 8y to both sides, subtract 9 from both sides

7  = 8y      divide both sides by 8

7/8  = y....  so the center of our circle will lie at (0, 7/8)

And the distance from this point to D  is the radius   =  4 - 7/8 =  32/8 - 7/8 =  25 / 8   = 3.125 =  3 + 1/8  

And the equation of the circle will be  x^2 + (y - 7/8)^2  = (25/8)^2

pi radians  = 180 degrees

(pi/6)  radians  = 30 degrees

cos 30  degrees  = (sqrt3)/ 2

Yes  of course I made a typo.    

Thanks for correcting it   

I would assume that this is a careless typo by Melody, but the final arithmetic calculation is incorrect.

8^2 =    64

            X

10^2 = 100

____________

             6400

Hi Chris :D

Yes that is very understandable :)

But then who'd want to quit eating dinner!    Unless it include brussel sprouts of course :)

what 512^2/3 * 100,000^2/5

\((512^{1/3})^2 * (100000^{1/5})^2\\ =8^2*10^2\\ =6200\)

An isosceles triangle with equal sides of 5 inches and a base of 6 inches is inscribed in a circle. What is the radius, in inches, of the circle? Express your answer as a mixed number.

Referring to my diagram.

\(\theta+\alpha =\pi\;\;radians\\ \alpha = \pi-\theta\\ so\\ sin\alpha=sin\theta\\ cos\alpha = -cos\theta\\ \text{I am going to use the identity }cos^2\theta+\sin^2\theta=1\\\)

\(\text{Using Cosine Rule}\\ 5^2=r^2+r^2-2*r*r*cos\theta\\ 25=2r^2-2r^2cos\theta\\ \frac{25-2r^2}{-2r^2}=cos\theta\\ cos\theta=\frac{25-2r^2}{-2r^2}\\ cos^2\theta=\frac{4r^4-100r^2+625}{4r^4}\\~\\ sin\alpha=\frac{3}{r}=sin\theta\\ sin^2\theta=\frac{9}{r^2}=\frac{36r^2}{4r^4}\\~\\ Sin^2\theta+cos^2\theta=\frac{36r^2+4r^4-100r^2+625}{4r^4}=1\\ 36r^2+4r^4-100r^2+625=4r^4\\ -64r^2+625=0\\ 64r^2=625\\ r=\frac{25}{8}=3\frac{1}{8} \)

The radius of the circle is 3 and 1/8 inches. 

Sorry, but can you tell us the meaning and the formula of calculating these points? I can't find any translation on the web of what those abbreviations mean besides it is some kind of a special word for courts in India, any help and clearance will help us answer your question more easily, cheers.

Source: https://forum.wordreference.com/threads/obc-general-sc-st-pc-university.1806282/

Oh yes XD I messed up the numbers and my vocabulary

(Corrected :P)

A box has 3 dimensions SO I assume you are talking about a rectangle.

Draw the rectangle and put the diagonal and the angle that is spoken of on it. 

Find the length of the diagonal using pythagoras's theorum.

Sine of the angle is the opposite side over the hypotenuse. Just read it off your diagram.