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Compute:  log_2^a2^b  in terms of a and b.

Using this definition:  log_rs = t   --->   r^t  =  s

                              log_2^a2^b = x   --->   (2^a)^x  =  2^b

                                                   --->    2^ax  =  2^b

                                                   --->     ax  =  b

                                                   --->       x  =  b/a

 Therefore:  log_2^a2^b =  b/a

The problem is to find the two zeros of the function and then find the distance between them.

First:  multiply out the equation:  y  =  -0.25(x - 2)(x - 22)  =  -0.25(x² - 24x + 44)

Second: set this equation to zero and solve:   0.25(x² - 24x + 44)  =  0

              Divide both sides by 0.25:                           x² - 24x + 44  =  0

              Factor:                                                         (x - 2)(x - 22)  =  0

              Find the zeros:                                             x = 2  or  x = 22

Third:  find the distance between these two answers:                                            (your turn)

You can follow these steps:

1) Use the Pythagorean Theorem to find the straight-line distance from A to C.

2) Subtract the diameter of the circle from the answer of step 1.

3) Find the circumference of the circle.

4) Find one-half of the answer of step 3.

5) Add the result of step 4 to the result of step 2.

​ I need Help

There are 38 boys, so \(38 - 18 = 20\) boys rest

There are 22 girls,  so \(22 - 9 = 13\) girls rest

There's only this one number.

(a) How many cards does Shrek have at first?

(b) How many cards does Fiona have in the end?

simplist method.

5 times 51 is 255

875-255 is 620.

do the rest. iz easy.

right?

so easy

csc(3π/4)

cot(π)  ???

Angle  BCD = 80º

63-19 is 54

54 divided by 2 is 27

27+19 is 46.

wait.what.46+27 is 73.@!##%^^$$!

4.17.this web is a calculator!

Let B, A, and D be three consecutive vertices of a regular 18-gon. A regular heptagon is constructed on AB with a vertex C next to A. Find angle BCD in degrees.

Angle  BCD = 80°   

Well you can make the necessary adjustments to the working for yourself. :)

a+b=13 b+c=24 c+a=19 what is the value of a+b+c.

\(\begin{array}{|lrcll|} \hline (1) & a+b=13 \\ (2) & b+c=24 \\ (3) & c+a=19 \\\\ (1)+(2)+(3): &(a+b)+(b+c)+(c+a) &=& 13+24+19 \\ & (a+b+c)+(a+b+c) &=& 13+24+19 \\ & 2(a+b+c) &=& 13+24+19 \\\\ & a+b+c &=& \dfrac{13+24+19}{2} \\\\ & a+b+c &=& \dfrac{56}{2} \\\\ & \mathbf{a+b+c} &=& \mathbf{28} \\ \hline \end{array}\)

what are the coordinates?

The above typo of 9 mph for the speed of the wind should read 4.5 mph instead, Sorry about that.

A regular dodecahedron \(P_1 P_2 P_3 \dotsb P_{12}\) is inscribed in a circle with radius \(1\).
Compute  \((P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2\).
(The sum includes all terms of the form \((P_i P_j)^2\), where \(1 \le i < j \le 12\).

\(\small{ \begin{array}{|lrcll|} \hline \mathbf{\text{Compute}} \\ \hline &s=(P_1 P_2)^2 + (P_1 P_3)^2+(P_1 P_4)^2+(P_1 P_5)^2+(P_1 P_6)^2+(P_1 P_7)^2+(P_1 P_8)^2+(P_1 P_9)^2+(P_1 P_{10})^2+(P_1 P_{11})^2+(P_1 P_{12})^2 \\ & +(P_2 P_3)^2+(P_2 P_4)^2+(P_2 P_5)^2+(P_2 P_6)^2+(P_2 P_7)^2+(P_2 P_8)^2+(P_2 P_9)^2+(P_2 P_{10})^2+(P_2 P_{11})^2+(P_2 P_{12})^2 \\ & +(P_3 P_4)^2+(P_3 P_5)^2+(P_3 P_6)^2+(P_3 P_7)^2+(P_3 P_8)^2+(P_3 P_9)^2+(P_3 P_{10})^2+(P_3 P_{11})^2+(P_3 P_{12})^2 \\ & +(P_4 P_5)^2+(P_4 P_6)^2+(P_4 P_7)^2+(P_4 P_8)^2+(P_4 P_9)^2+(P_4 P_{10})^2+(P_4 P_{11})^2+(P_4 P_{12})^2 \\ & +(P_5 P_6)^2+(P_5 P_7)^2+(P_5 P_8)^2+(P_5 P_9)^2+(P_5 P_{10})^2+(P_5 P_{11})^2+(P_5 P_{12})^2 \\ & +(P_6 P_7)^2+(P_6 P_8)^2+(P_6 P_9)^2+(P_6 P_{10})^2+(P_6 P_{11})^2+(P_6 P_{12})^2 \\ & +(P_7 P_8)^2+(P_7 P_9)^2+(P_7 P_{10})^2+(P_7 P_{11})^2+(P_7 P_{12})^2 \\ & +(P_8 P_9)^2+(P_8 P_{10})^2+(P_8 P_{11})^2+(P_8 P_{12})^2 \\ & +(P_9 P_{10})^2+(P_9 P_{11})^2+(P_9 P_{12})^2 \\ & +(P_{10} P_{11})^2+(P_{10} P_{12})^2 \\ & +(P_{11} P_{12})^2 \\ \hline \end{array} }\)

\(\begin{array}{|rcll|} \hline (P_1 P_2)=(P_2 P_3)=(P_3 P_4)=(P_4 P_5)=(P_5 P_6)=(P_6 P_7)=(P_7 P_8) \\ =(P_8 P_9)=(P_9 P_{10})=(P_{10} P_{11})=(P_{11} P_{12})=(P_1 P_{12})= 2*\sin(15^\circ) \\\\ (P_1 P_2)=(P_2 P_4)=(P_3 P_5)=(P_4 P_6)=(P_5 P_7)=(P_6 P_8)=(P_7 P_9) \\ =(P_8 P_{10})=(P_9 P_{11})=(P_{10} P_{12})=(P_{1} P_{11})=(P_2 P_{12})= 2*\sin(30^\circ) \\\\ (P_1 P_4)=(P_2 P_5)=(P_3 P_6)=(P_4 P_7)=(P_5 P_8)=(P_6 P_9)=(P_7 P_{10}) \\ =(P_8 P_{11})=(P_9 P_{12})=(P_{1} P_{10})=(P_{2} P_{11})=(P_3 P_{12})= 2*\sin(45^\circ) \\\\ (P_1 P_5)=(P_2 P_6)=(P_3 P_7)=(P_4 P_8)=(P_5 P_9)=(P_6 P_{10})=(P_7 P_{11}) \\ =(P_8 P_{12})=(P_1 P_{9})=(P_{2} P_{10})=(P_{3} P_{11})=(P_5 P_{12})= 2*\sin(60^\circ) \\\\ (P_1 P_6)=(P_2 P_7)=(P_3 P_8)=(P_4 P_9)=(P_5 P_{10})=(P_6 P_{11})=(P_7 P_{12}) \\ =(P_1 P_{8})=(P_2 P_{9})=(P_{3} P_{10})=(P_{4} P_{11})=(P_5 P_{12})= 2*\sin(75^\circ) \\\\ (P_1 P_7)=(P_2 P_8)=(P_3 P_9)=(P_4 P_{10})=(P_5 P_{11})=(P_6 P_{12})= 2*\sin(90^\circ)=2 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline s&=& 12*(2*\sin(15^\circ))^2 + 12*(2*\sin(30^\circ))^2+12*(2*\sin(45^\circ))^2 \\ && +12*(2*\sin(60^\circ))^2 +12*(2*\sin(75^\circ))^2+6*(2*\sin(90^\circ))^2 \\\\ s&=& 48\sin^2(15^\circ) + 48\sin^2(30^\circ)+48\sin^2(45^\circ) \\ && +48\sin^2(60^\circ) + 48\sin^2(75^\circ)+24 \\\\ s&=& 48\left( \sin^2(15^\circ) + \sin^2(30^\circ)+\sin^2(45^\circ)+\sin^2(60^\circ) + \sin^2(75^\circ) \right) +24 \\\\ s&=& 48\left( \sin^2(15^\circ) + \sin^2(30^\circ)+\sin^2(45^\circ)+\cos^2(30^\circ) + \cos^2(15^\circ) \right) +24 \\\\ s&=& 48\left( \sin^2(15^\circ) + \cos^2(15^\circ) + \sin^2(30^\circ)+\cos^2(30^\circ)+\sin^2(45^\circ)\right) +24 \\\\ s&=& 48\left( 2+\sin^2(45^\circ)\right) +24 \quad | \quad \sin(45^\circ)=\dfrac{\sqrt{2}}{2} \\\\ s&=& 48\left( 2+\dfrac{1}{2}\right) +24 \\\\ s&=& \dfrac{48*5}{2} +24 \\\\ s&=& 24*5 +24 \\\\ s&=& 24*6 \\\\ \mathbf{s}&=& \mathbf{144} \\\\ \mathbf{s} &=& \mathbf{12^2} \\ \hline \end{array}\)

The sum is \(\mathbf{12^2 = 144}\)

To make it readable:
 Let \(A=(4,-1)\)\(,B=(6,2),\) and \(C=(-1,2).\)There exists a point \(X\) and a constant \(k\) such that, for any point \(P\) \(PA^2 + PB^2 + PC^2 = 3PX^2 + k\). Find (the value of) constant \(k\)

Find the ordered pair (a, b) of real numbers for which
\((ax + b)(x^5 + 1) - (5x + 1)\)
is divisible by \(x^{2}+1\).

0.6 * 6.95 = ???

I;m not sure if thats what you're looking for

Let IJKLMN be a hexagon with side lengths IJ = LM = 2, JK = MN = 3, and KL = NI = 8. Also, all the interior angles of the hexagon are equal. Find the area of hexagon IJKLMN.

The area of hexagon is:   A = 39.83716857 u²   

Thank you, CPhill !