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2[  sin (pi *t  ) + cos (pi * t)

From 1 to 2 seconds we have

[  2 sin (2pi) + 2cos  (2 pi)  - 2sin (pi) - 2cos(pi)] / (2 - 1) =

[ 2  - (-2) ] / 1      =     4 cm /sec

From 1 to 1.1 seconds

[  2 sin (1.1pi) + 2cos  (1.1 pi)  - 2sin (pi) - 2cos(pi)] / (1.1 - 1) ≈  -5.20 cm/sec

From 1 to 1.01 seconds

[  2 sin (1.01*pi) + 2cos  (1.01* pi)  - 2sin (pi) - 2cos(pi)] / (1.01 - 1) ≈ -6.18 cm/sec

From  1 to 1.001 seconds

[  2 sin (1.001*pi) + 2cos  (1.001* pi)  - 2sin (pi) - 2cos(pi)] / (1.001 - 1)  ≈ - 6.27 cm /sec

It appears that  the instantaneous velocity at 1 sec   =  -2pi  cm/sec  ≈  -6.28 cm/sec

BTW....we can verify that this is true using some Calculus   !!!

If y=C(1+r)^t, then just plug in the numbers.

Y=C(The original amount)*(1+r(The rate of intrest))^t(The amount of time)

y=900(1.08)^8

y=$1665.84

Hope this helped!

Factor the following:
4 x^3 + 8 x^2 - 96 x

Factor 4 x out of 4 x^3 + 8 x^2 - 96 x:
4 x (x^2 + 2 x - 24)

The factors of -24 that sum to 2 are 6 and -4. So, x^2 + 2 x - 24 = (x + 6) (x - 4):

 4x (x + 6) (x - 4)

% decrease =    (  [5200 - 3140] / 5200  x 100 ] %   =

( [2060 ] /5200   x 100 ) % =

[ .396   x 100 ] % =

≈ 39.6%  =     40% rounded

My science teacher did a back flip 

I asked it to simplify the RHS and it gave these:
log(1296/5) - 8 log(π) + 16 log(Γ(3/4))
log(331776/5) + 8 (log(π) - 2 log(Γ(1/4)))
-log(5) + 4 log(6) - 8 log(π) + 16 log(Γ(3/4))
= -0.34774389941635582079534116072897304565732004913773310141

This looks like a fun problem. I will try later, perhaps tomorrow if I remember .

Proving the integral to be LHS is more feasible. Thanks for the help, I haven't tried rewriting the right hand side.

Hi Max:  Mathematica 11  gives this and says LHS =RHS but no other explanation. Sorry about that:

8 log(Γ(3/4)/Γ(5/4)) - log(1280/81) = 4 log(2) + 4 log(3) - log(5) - 8 log(π) + 16 log(Γ(3/4))

Thank you Melody :)

Define 2 linear functions: \(L_1:Ax+By+C=0\\ L_2:Dx+Ey+F=0\).

From the question: The 2 straight lines intersect at the x-axis. \(\Rightarrow \dfrac{C}{A} = \dfrac{F}{D}\)

As C/A = F/D, C = kF and A = kD for some real k. We can rewrite L₂ as: \(L_2:Ax+ny+C=0\).

The equation representing the 2 straight lines is 

\(A^2x^2+(An+AB)xy+Bny^2+2ACx+(BC+Cn)y+C^2=0\).

Define \(n + B \stackrel{\text{def}}{=} P\).

Now the rewritten equation is \(A^2x^2+APxy+Bny^2+2ACx+CPy+C^2=0\).

WLOG, we can assume A > 0.

In this case, A = \(\sqrt6\).

2AC = -24, so C = \(-2\sqrt6\). => b = C² = 24

CP = 3, so P =  \(-\dfrac{\sqrt6}{4}\).

-a = AP = -3/2

So a = 3/2 :D

There we get (a,b) = (3/2, 24). :D

You can use the clock angle formula, which is \(|0.5\times (60\times H-11\times M)|\), where H is the number of hours and M is the number of minutes.

\(|0.5\times (60\times 6-11\times 48)| = |0.5\times (360-528)| = |0.5\times (-168)|=|-84|=84\)

Therefore, the angle between the hands is 84 degrees.

You are very welcome!

:P

Let \(P(z) = z^8 + \left(4\sqrt{3} + 6\right)z^4 - \left(4\sqrt{3} + 7\right)\).

What is the minimum perimeter among all the 8-sided polygons in the complex plane whose vertices are precisely the zeros of P(z)?

Try [3, 8)   i.e. open interval on the right.

Это будет зависеть от того, какие месяцы.
Но 60 месяцев - это 5 лет

5*365 = 1825

У февраля есть дополнительный день, поэтому вы должны добавить 1 или 2 к этому числу.

1826     или же      1827

Solve !

NVM. Answer is 160.

Rectangle ABCD is the base of pyramid PABCD.

If AB = 8, BC = 4,\(\overline{PA}\perp \overline{AD}\), \( \overline{PA}\perp \overline{AB}\),  and PB = 17,

then what is the volume of PABCD?

Let h = height of the pyramid

Let B = Area of the Base of the pyramid = Area of the rectangle ABCD = \(4\cdot 8\) = \(32\)

\(\mathbf{h=\ ? }\)

\(\begin{array}{|rcll|} \hline h^2 + AB^2 &=& PB^2 \\ h^2 + 8^2 &=& 17^2 \\ h^2 &=& 17^2-8^2 \\ h^2 &=& 289-64 \\ h^2 &=& 225 \\ \mathbf{h} & \mathbf{=}& \mathbf{15} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{Volume of PABCD} &=& \dfrac{1}{3}\cdot B \cdot h \\\\ &=& \dfrac{1}{3}\cdot 32 \cdot 15 \\\\ &=& 5\cdot 32 \\\\ \mathbf{\text{Volume of PABCD}} & \mathbf{=}& \mathbf{160} \\ \hline \end{array}\)

Set up table of values and then graph each line from its parametric form.

x=-3+5t

y=2-4t

I know how to get the points by substituting t with 0 and 1, getting (-3,2) and (2,-2). Do I plot the points right away or should I find the slope first.

Thanks Chris and Alan. 

Thank you very much!!!

I think you mean 5.3912e+35 = 5.3912 x 10³⁵ Planck lengths

As a comparison it is estimated that there are between 10⁷⁸ and 10⁸² atoms in the universe

\(\text{let's use the navigational assignment of degrees to the clock, i.e. }\\ \text{12 is 0 degrees increasing as we move clockwise back to 360 degrees again at 12}\\ \phi_m = 360\dfrac{min}{60} = 360\dfrac{48}{60} = 360 \dfrac 4 5 = 72\cdot 4 = 288^\circ\\ \phi_h = 360 \dfrac{60hr + min}{720} = 360 \dfrac{60\cdot 6 + 48}{720} = \dfrac 1 2(360+48) = 204^\circ\\ |288^\circ - 204^\circ| = 84^\circ\)

\(\text{I assume this is one question and that both must be satisfied}\\ x^2 + (2a+1)x + a^2 \text{ has solutions of }\\ x = \dfrac{-(2a+1)\pm \sqrt{(2a+1)^2 -4a^2}}{2}\\ \text{in order for these to be integers}\\ (2a+1)^2 -a^2 \text{ must the be the perfect square of an odd number}\)

\((2a+1)^2 - 4a^2 = \\ 4a^2 + 4a+1-4a^2 = 4a+1 \\ \text{and this must be the perfect square of an odd number if }x \text{ is to be an integer}\)

\(\text{The values of }a \text{ that satisfy this are }\\ (2,6,12,20,30,42) \\ \text{and all of these lead to 2 unique integer solutions of the given quadratic}\)