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Thank you so much.

arccos  [ -√2 / 2 ] =       3pi/ 4  rads

Here's one more way :

60 mph  = 88 ft/sec

So

15 mph  = 22 ft/ sec

So...in 60 seconds we have    60 * 22  =  1320 ft

See my answer under  your earlier post.....

We can use the  Angle of  Intersecting Chords Theorem

We first need to find the measure  of  minor arc FB....this is given by :

m arc FB  = 360  - 100 - 154 - 40   = 66°

So...the measure of AED  = (1/2) ( m arc FB  + m arc AD)  =  (1/2) ( 66 + 40)  = (1/2)(106)  = 53°

x  = 7 ^ (log ₇ 2  + log ₇ 3)

x = 7^[ log ₇  (2 * 3) ]  

x  = 7^ [ log₇ 6 ]   ......using the log property   a^[log _a b]   = b  , we have 

x  = 6

We can use the Exterior Angle Theorem to  solve this

75  = 28  +  x       subtract 28 from both sides

47°   = x   =  the measure of the other  non-adjacent  interior  angle

log₂ x  + 5 log₂ y  - 3 log ₂ x  =

log₂ x  + log₂ y⁵  - log₂ x³ =

log₂  [ ( x y⁵ ) / x³ ] =

log₂ [ y⁵ / x² ]

ln ( a + b)  + 4 ln ( a - b) - 3 ln c  =

ln (a + b) + ln ( a - b)⁴  - ln c³ =

ln  ( [ (a + b) (a - b)⁴ ] / c³ )

5 log x - 4 log (x² + 1)  + 2 log (x - 1)  =

log x⁵  - log (x² + 1)⁴ + log ( x - 1)²   =

log   [   (x⁵ * (x - 1)² ) / (x² + 1)⁴  ]

Via dimensional analysis solution:

15  m/hr  x  1/60 hr/min  x  5280 ft/m  = 1320 ft/min       (as Tertre found)

Sum of interior angles of a polygon    (n-2)180

THIS polygon sum = 160 + (n-1)112         set these two equations =

(n-2)180 = 160 + (n-1)112     and solve for  n = 6

More Explanation:

Since your expanding and changing the 4in*5in from a 48in*60in, this is an example of expansion, (A). By 12 :)

Solution:

Since there are 5280ft in 1 mile, we can find how many feet are in 15 miles.

Converting miles into feet, we have \(5280*15=79,200 \)feet.

So, he travels 79,200 feet in 1 hour or 60 minutes.

Since they're asking us to find how many feet he can travel in 1 minute, we can easily do: \(\frac{79,200}{60}=\boxed{1,320} \)

feet.

EP probably gave you the point. The “floaters” on the forum annoy him too.

You now have two points.

Hahahaha that’s great, Ginger!

Did you give me the point? I don’t usually get points on my posts.

\(\text{Compute}:\displaystyle\lim_{x\to0}(\frac{\sin{x}}{x})^\frac{1}{1-\cos{x}}\)

We can approximate \(\sin{x} \text{ and } \cos{x}\) by their Taylor series.

After apply substitutions:

\(\displaystyle\lim_{x\to0}(\frac{\sin{x}}{x})^\frac{1}{1-\cos{x}}=\lim_{x\to0}(1-\frac{x^2}{6})^\frac{2}{x^2}=\lim_{x\to0}(1+\frac{x^2}{6})^{-\frac{2}{x^2}}\)

\(=\displaystyle\lim_{x\to\infty}(1+\frac{1}{6x^2})^{-2x^2}=\lim_{x\to\infty}(1+\frac{1}{x})^{-x/3}=\boxed{e^{-1/3}}\)

This is because \(e=\displaystyle\lim_{x\to\infty}(1+\frac1x)^x\)

I hope this helped,

Gavin

Transformation Rule  :  ( 6, -4)

Scale Factor :   3

I saw the solution but I didn't get the insight of the questions and how there were able to derive to formulae... I am new on momentum

Why would you expect such a descriptive reply to a simple, brain-dead question? This question requires only elementary math skills to answer. Posting a nondescript, basic question causes you to appear as a one of the simpleminded idiots, who want someone to do his homework.   

The guest has a point in asking why you are fucking around on here.  Because you are.

Generally, you learn the formulas and their applications before learning the methods of derivation.  The derivation of most physics formulas require advanced levels of calculus, linear algebra, and trigonometry.

If you want an advanced or intermediate level of derivation or the reasons why, then you need to ask pointed questions with contrasting examples.  Here’s an example of such a question.

\(\text{Find } a: \int_1^a(3x^2-6x+3)dx=27\)

\(\text{The solution you seek is } a=\boxed4\\ \)

V_cone   =  pi *( diameter/ 2)^2 * height  / 3   = 3.14 * (1.5 / 2)^2 * 4  = 7.065 in^3

multiplication property of equality

m ∠ F

m ∠ D

substitution property of equality

full

θ / [ 2 pi  ]

A  = θ r^2 / 2

Thank you!

The ENTIRE square is 30 x 30 = 900 units^2

Now subtract the TWO unshaded triangles

Area of EACH = 1/2 bh     there are TWO , so their combined area = bh

900 - (20 x 20) = 500 units^2

Sin 30 = opp/hyp = 14/x

x=14/.5 = 28

I saw the solution but I didn't get the insight of the questions and how there were able to derive to formulae...

I am new on momentum