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Thanks, CC.......welcome aboard!!!

Thanks so much!

Yeah, I do.

5/9 = 0.5555555555. . .   (Tbh I just googled what is 0.5 repeating)

5/900 = 0.0055555555. . .

81/100 + 5/900 = 0.815555555. . .

729/900 + 5/900 = 0.815555555. . .

734/900 = 0.815555555. . .

367/450 = 0.81555555. . .

2 3/4 * 2 9/5 = 209/20= 10.45 

The edge of a cube with a volume  of 1728mm^3   =   12mm

So....the allowable error, E, would be

(12 - .04)^3 < E < (12 +  .04)^3

1710.77mm^3 < E < 1745.34mm^3

Thank you so much 

I'm particularly fond of pumpkin pi:

www.piday.org/million/

21. Using the Law of Sines, we have

sin A / 8  = sin (120)/ 10

sin A  = (8/10)sqrt(3)/2

sinA  = .4sqrt(3)

arcsin(.4sqrt(3))  = A  

So m< C  = 180 - B - A  =  180 - 120 - arcsin(.4sqrt(3)) ≈ 16.1°

I cannot see any DIRECT solution for N. There maybe a way but I cannot see it. However, using the method of iteration and interpolation, it gives a value for N =5.36064361355.

I have checked it by substituting it into the equation, and it balances the equation to the penny.

Good luck to you.

11b - 15.

Simplifying equations like those usually just involves combining like terms:

7b + 4b = 11b

-10 - 5 = -15

so 7b - 10 - 5 + 4b = 11b - 15

20. Using the Law of Cosines, we have

13^2 = 10^2 + 8^2  - 2(10)(8)cos(K)

5 =   -160 cos (K)     divide by - 160

-1/32  = cos(K)

arccos (-1/32) = K  ≈ 91.8°

2x - 1  is the first term

And we have that  the first term + d  = the second term

2x - 1 + d    = 4x   →   d  = 2x + 1   (1)

And   the first term + 2d  = the third term

2x - 1 + 2d  = 5x + 3  →  2d  = 3x + 4    (2)

Sub (1)  into (2)   and we have that

2 [2x + 1]  = 3x + 4

4x + 2  = 3x + 4       subtract 3x, 2 from both sides

x = 2     and using (1) →  d  = 2(2) + 1   = 5

So

The first term  is 2(2) - 1  = 3

The second term is 4(2)  = 8

And the third term is 5(2) + 3  = 13

Note that the difference between terms = d = 5 

Let a1 be the first term

Then the second term is  a1 + d

And the third term is  a1 + 2d

So we have

a1 + [ a1 + d ] = 16    →    2a1 + d  = 16   →   d  =  16 - 2a1      (1)

[a1 + d] + [a1 + 2d]  =  28    →  2a1 + 3d =  28      (2)

Sub  (1)  into (2)   and we have

2a1 + 3 [16 - 2a1]   = 28

-4a1  + 48   =  28       subtract 48 from both sides

-4a1  = -20     divide both sides by -4

a1  = 5

And using (1).....d = 16 - 2(5)   =  16 - 10  = 6

So  the first term  =  5

The second term is 11

And the third term = 17

This is a Law of Cosines problem.

Law of Cosines:

\(c^2 = a^2 + b^2 - 2ab\cos C \\ \\ c = TU, a=24, b=12, C=38^{\circ} \\ \text{ (a and b are interchangeable, just pick one and stick with it.)} \\ TU^2 =24^2 +12^2-2(24)(12)\cos38 \\ TU^2 = 576 + 144 - 576\cos38 \\ TU^2 = 720 - 576\cos38 \\ TU = \sqrt{720 - 576\cos38} \\ TU \approx 16.3 \text{ mm}\)

This is not an identity for  either

[1-sec 8x] /[1- sec4x]  = tan8x / tan4x      nor

(1 - sec^8(x)) / (1 - sec^4(x))  = tan^8(x) / tan^4(x)

See here : https://www.desmos.com/calculator/figbjj45d8 

And  here : https://www.desmos.com/calculator/mbarrthbsp

Using cos Law, we can easily find TU.

\(TU = \sqrt{24^2+12^2-2*24*12*cos(38)}\)

TU ≈ 16.3127497964741677

Note: in cos(38), the 38 is in degrees, not radians.

We can use Law of Sines for this.

\(\frac{\sin A}{a} = \frac{\sin B}{b} \\ \frac{\sin A}{9} = \frac{\sin72}{38} \\ \sin A = (9)(\frac{\sin72}{38}) \\ A = \arcsin(\frac{9\sin72}{38}) \\ A \approx 13.018 \text{ degrees}\)

(To the nearest tenth it's 13.0º )

sqrt(160)   =

sqrt (16 * 10) =

sqrt(16) * sqrt(10) =

4 * sqrt(10)  ≈   12.65

\(\sqrt{160} \\ = \sqrt{(2)(2)(2)(2)(2)(5)} \\ = (2)(2)\sqrt{(2)(5)} \\ = 4\sqrt{10}\)

≈ 12.650

First we will find the distance to the TOP of the window:

x=xo + vot + 1/2 at^2      Where xo = 0 (top of window)  and x = window height = 1.2 m

1.2 = vo (.125) + 1/2 (9.8)(.125^2)

vo= 8.9875 m/s   at top of window

NOW:  v = vo + at     where  vo = 0 (at start)

           8.9875 = 0 + 9.8 (t)

            t = time to reach window top = .91709 seconds

NOW  Distance to TOP of window :

            x = xo + vot + 1/2 a t^2

               = 0   + (0)t + 1/2 (9.8) (.91709)^2

            x = 4.12 m  to top of window

NOW it takes 1 second + .125 second to travel back to the top of the window = 1.125 seconds

  again:  x = xo + vot + 1/2 at^2

                 = 0 + 0(t) + 1/2(9.8)(1.125^2) = 6.20 meters

Finally:

6.20 meters + 4.12 meters = 10.32 meters tall

I don't really know how to do this but I was just thinking about it and if we just knew the total time of the fall then we could figure out how high it fell from because we know gravity makes stuff fall at about 9.8 m/s² . (But you have to forget about air resistance.)

So wouldn't the time of the fall be 2.0 + 0.125 + 0.125? Because if the bounce up was identical to the fall then there is exactly as much height above the window as below it.

So the total time is 2.25 seconds...and gravity makes stuff fall at 9.8 m/s/s.

So height divided by time divided by time is 9.8

height divided by 2.25 divided by 2.25 is 9.8

height is 9.8*2.25*2.25

height is about 49.6125 meters.

That's probably wrong though because I'm just guessing at how to do it.

I get an answer of 10.32 m   .....anyone else get the same?

No, she doesn't. She needs:

2,500/300 =8 1/3 packs of sugar.

8 1/3 - 6 = 2 1/3 packs of sugar she has to buy.

Nie przekłada się doskonale      

Być może można spróbować użyć kilku różnych słów.