All Answers 356600 Answers

Anchor Bob and Anna in any of two chairs......and they can be arranged in 2 ways.... [ Bob on Anna's left or right....]

And the other people can be arranged in 7!  = 5040 ways

So....the total arrangements are  2 * 7!  =   10080  ways

A(t)  =  3 - 2t² + 4^t

A(2)  =  3 - 2(2)² + 4²

A(1)  =  3 - 2(1)² + 4¹

A(2) - A(1)  =  [ 3 - 2(2)² + 4² ] - [ 3 - 2(1)² + 4¹ ]

A(2) - A(1)  =  [ 3 - 2(4) + 16 ] - [ 3 - 2(1) + 4 ]

A(2) - A(1)  =  [ 3 - 8 + 16 ] - [ 3 - 2 + 4 ]

A(2) - A(1)  =  [ 11 ] - [ 5 ]

A(2) - A(1)  =    6

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f( √[x + 1] )  =  1/x      for  x ≥ -1  and  x ≠ 0  

We want to find  f(2)  , so we need an x value that makes

√[ x + 1]  =  2      square both sides

x + 1  =  4           subtract  1  from both sides

x  =  3                 This is a valid  x  to plug in.

So....

f( √[3 + 1] )  =  1/3

f( 2 )  =  1/3                 

Oh my!
I miscounted the numbers! 

Oops!

Desolee! 

Yes, my answer is incorrect. Thank you for pointing it out. 

Good day!

There will be two points each for the range values 1 - 9   and one value for 0

So......the max number is  19

saarad0: What is the smallest number of four digits divisible by 2,3,10  ??

Since the LCM[2, 3, 10] =30, then the smallest FOUR-digit number would be:

34 x 30 =1,020

Graphing y=-2x+2 using xy chart method. 

For this, you can substitute any number you want in x in order to solve for y. However, I would recommend using values -1, 0, 1 and 2 as values of x. 

So your points are:

(-1,4)

(0,2)

(1,0)

(2,-2)

Use your ordered pairs and graph the graph!

y  =  5x - 4

a.  The x-intercept is the value of  x  when  y = 0 , so plug in  0  for  y  and solve for  x .

0  =  5x - 4      Add  4  to both sides of the equation.

4  =  5x           Divide both sides by  5 .

4/5  =  x

The x-intercept is  4/5  , or the point  (4/5, 0)  .

b.  Now we need to find the y-intercept. So plug in  0  for  x  and solve for  y .

y  =  5(0) - 4

y  =  -4

The y-intercept is  -4  , which is the point  (0, -4) .

To graph this, plot the points  (4/5, 0)  and  (0, -4)  and draw a line through them.

Substitute (6,-2) into the x and y of the equation to solve

(6, -2) so x= 6, y= -2

Y = 1/2X-5

-2 = 1/2(6) - 5

(1/2) x 6 = 3

-2 = 3-5

-2 = -2 ✓

(6,-2) satisfies the equation and therefore is a soloution of y=1/2X-5 

La response est 120! 

120/3 = 40, 120/2 = 60, 120/10= 12

Bonne chance! 

If h(x) = (−x + 1)2 − 5 and g(x) = −x − 2 , then (h ◦ g)(x) =?

h(x) = (−x + 1)2 − 5 = -2x+2-5 = -2x-3

(h ◦ g)(x) = -2[−x − 2] - 3 

(h ◦ g)(x) = [2x + 4] - 3 

(h ◦ g)(x) = 2x + 1

You are very welcome :)

Students, including me when I was young, find this all very mysterious and difficult to remember, but if you understand the diagram it is completely logical and not difficult to remember at all.   :)

Thank you very much Melody!

Don't know but you cold try making sense of this video

Using the definition of deriative (f(x+h)-f(x))/h could anyone show me how to calculate the deriative of 4/x^2? I know the answer is -8/x^3 but I don't know the steps to get there.

If you understand the diagram that goes with this, then it is easy to remember because it is totally logical.

Just replace the a with an x in this diagram and see if you can make sense of it.

\(f(x)=\frac{4}{x^2}\\ f(x+h)=\frac{4}{(x+h)^2}\\ \text{The gradient of the secant joining these points }\\ =\frac{rise }{run}\\ =\frac{f(x+h)-f(x)}{x+h-x}\\ =[f(x+h)-f(x)] \div h\\ =[\frac{4}{(x+h)^2}-\frac{4}{x^2}] \div h\\ =[\frac{4}{(x+h)^2}-\frac{4}{x^2}] \div h\\ =\frac{4x^2-4(x+h)^2}{hx^2(x+h)^2}\\ \)

\( =\frac{4x^2-4(x^2+2xh+h^2)}{hx^2(x^2+2xh+h^2)}\\ =\frac{-4(2xh+h^2)}{hx^2(x^2+2xh+h^2)}\\ =\frac{-4h(2x+h)}{hx^2(x^2+2xh+h^2)}\\\)

Now the gradient of the tangent will be the gradient ofthe secant as h tends ot 0

\(\text{gradient of tangent}\\ =\displaystyle\lim_{h\rightarrow 0}\;\;\frac{-4h(2x+h)}{hx^2(x^2+2xh+h^2)}\\ =\displaystyle\lim_{h\rightarrow 0}\;\;\frac{-4(2x+h)}{x^2(x^2+2xh+h^2)}\\ =\dfrac{-8x}{x^4}\\ =\dfrac{-8}{x^3}\)

I actually thought your recent off-topic post was cool...

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Here is how to do 1 and 4.  See if you can use similar approaches on the others:

.

Hi Asad,

I  do not think you can download web2.0... You have to use it when you are online.

                                                       Origin of the Knot

As GA points out, Mr. Gumpach's tome, while illustrative, does not quite cover all the bases. For instance....sailing an equatorial path is always tricky.......crossing the Dateline insures problems because one never knows whether he/she is measuring speed in current terms - or in yesterday's terms. What might appear to be great speed yesterday might only appear to be a cipher, today.

Furthermore, historically......."knots"   were measured somewhat erratically. The once-accepted method was accomplished  by the "chip log."  Of course, finding logs at sea was sometimes as difficult as finding logs in Mathematics. So....if the ship carried livestock for provision of meat and/or dairy, the "cow chip" was sometimes an adequate substitute.  In the "Horse Latitudes,"  the whole cow was sometimes thrown overboard.  Lacking that, Mr. Newton himself was often shanghaied in port and thrown overboard later. (Not for any particular measurement purposes, but for the fact that few personages - if any - can abide a windbag scientist.)

So....the present dilemma remains.....the "knot"  is an unspecified quantity. In fact, it is the inspiration for the term "naught"  meaning  "nothingness"  or "nonexistence" or "insignificant." Therefore, I urge that we forget about chasing the quixotic and focus on the more attainable. For instance, the search for the Yeti immediately comes to mind.

I apologize if some of the facts I have laid out are errrant. However, mostly I told the truth....mostly........

Your Humble Servant and WebCalc Historian,

Sir CPhill, Esq 

Rewrite as

x^2 + 3x + y^2 - 2y = 1

Complete the square on x and y

x^2 + 3x + 9/4 + y^2 - 2y + 1  =  1  + 9/4  + 1

( x + 3/2)^2  +  ( y - 1)^2  =  2 + 9/4

( x + 3/2)^2  +  ( y - 1)^2  =  8/4 + 9/4

( x + 3/2)^2  +  ( y - 1)^2  =  17/4 

Look at the following :

I think you want to find the height of the triangle containing the isosceles triangle

AB = 2sqrt (2)   

Let  AEB  be isosceles......with  angles ABE, BAE  equal....so  AE  = BE

Bisect  AEB.......by SAS......BH = AH = EH

And ABE  = 90......so  ABE, BAE  = 45

And BH =  sqrt (2).......so EH = sqrt (2)

Let E be the intersection of medians BG and AF

Draw GF through D.....drop  perpendiculars from F, G to interesect AB at I and J  and by AAS,  triangle FIB is congruent to  triangle GJA.  ...then FI  = GJ ...so.. GF  is parallel to AB

And triangle AGF  is similar to triangle ABC.....so  DF  = 1/2  of HB  =  sqrt(2)/2

So  CD  =   1/2 CH   

And EDF  is right  and  angle DFE  = BAE  =  45

So DEF  = 45...so triangle EDF  is 45 - 45 - 90  with DF = DE  = sqrt (2) / 2

So.....HD  = EH + DE  =  sqrt (2) +  sqrt (2) / 2

And..........By  similar triangles......  2 CD  = 2 HD  =  HC

So  .....HC  =    2 [  sqrt (2) +  sqrt (2) / 2 ]   =  3sqrt (2)  = height of ABC

GOOD JOB!! Mr. BB. You sourced your quote, though Google isn’t the source. The quote is from: https://en.wikipedia.org/wiki/Knot_(unit)  

So . . .  no banana for you.

Also, your post is knot really an answer to his question.  From the context, his question should be read as “what is the linear value of one nautical mile at the equator?” 

Johannes von Gumpach makes the same error, using “knot” when he means “nautical mile” on page 253 of his book The True Figure and Dimensions of the Earth (1862) 

I use a quote from Gumpach’s book to partly answer the question

The explanation however will appear as simple, when as it is remembered that the nautical mile is an angular, rather than a linear measure, being one of 360x60=21,600 parts of the Earth's equatorial circumference whatever be the true linear value of that circumference. Hence considered as a linear measure it has as yet no definite value and its correctness depends absolutely on the correct linear measurement of an equatorial degree. If therefore the circumference of the Earth is taken too great by 166 or 167 miles, the nautical mile being one of its equal parts, and the subdivisions of the nautical mile or knots of the log-line__by which the distance sailed by a vessel is actually measured—are  likewise taken too great; and consequently, the linear distance sailed by a vessel when reduced to angular distance, is reduced by means of too  great a unit of measure; whence the number of nautical miles sailed both by computation and by the log-line, falls short of the true number.

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This book is in the public domain. Here’s a link for a high-quality PDF image copy of an original book in the NY Public Library archive collection.  The next time I’m there, I may “check it out.”

Note that   264  /  495  =    8  / 15

So......the ratio of the legs  is   8  :  15

And one of the Pythagorean "Triples"  is     8  : 15  : 17

So.........

264 / 8  =  33

And the hypotenuse must be......17 * 33  =   561

So......the  sides are    264 -  495  - 561

2nd quadrant.......

Just replace "x" in the equation with "6"

y  =  -2 (6) + 16  =  4   ⇒   (6, 4 )  

y  =  ax + b 

y - b =  ax

[ y - b]  /  a  =  x        swap x and y and for y write f^-1 (x)

So 

f^-1 (x)  = [ x - b ] / a  =  0         when     b  =  x