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\(\sqrt[6]{50} cis (5pi/4) \)
\(\sqrt[6]{50}cis (23pi/12)\)

\(\sqrt[6]{50}cis(7pi/12)\)
took the test

The domain of the function consists of all real numbers except those numbers that make the denominator equal to zero.

Let's try to find those numbers.

2·sqrt[  sqrt(x) - x  ] - 1  =  0

     2·sqrt[  sqrt(x) - x  ]  =  1

        sqrt[  sqrt(x) - x  ]  =  ½

                    sqrt(x) - x  =  ¼

            -x + sqrt(x) - ¼  =  0

             x - sqrt(x) + ¼  =  0

Using the quadratic equation:  sqrt(x)  =  [ -(-1) +/- sqrt( (1)² - 4·1·¼ ) ] / (2·1)

                                                 sqrt(x)  =  [ 1 + sqrt(0) ] / 2

                                                 sqrt(x)  =  ½

                                                          x  =  ¼

¼ is the number that cannot be included in the domain; so the domain consists of all real numbers except ¼.

d = distance      the same both ways

rate * time = d  

10 m/hr *x   =  9 m/hr * (x + 1/12 )                          5 min =  1/12 hr

10x = 9x + 9/12

x = 9/12 hr

in 9/12 hr     at 10m/hr    Terry covers the distance of 7.5 miles

Car A = x_a

Car B = x_a+10

rate x time = distance        time is 1/2 hr

(x_a +   xa+10)* 1/2 = 60

x_a = 55 k/hr           in 1/2 hr    car A  travels  27.5 km

x_a+10 = 65 k/hr     in 1/2 hr    car B travels  32.5 km              Check:    27.5 + 32.5 = 60 km 

Because both square roots and natural logs are not defined for negative values, let's find where the

function is zero.

sqrt[  ln[  (4x - x²) / 3 ] ]  =  0

--->   ln[  (4x - x²) / 3 ] ]  =  0                  (squaring both sides)

--->             (4x - x²) / 3  =  e⁰                (writing as an exponent)

--->              (4x - x²) / 3  =  1

--->                   (4x - x²)  =  3

--->              -x² + 4x - 3  =  0

--->               x² - 4x + 3  =  0

--->             (x - 1)(x - 3)  =  0

--->                      x = 1   or   x = 3

Checking values below 1 results in an undefined situation.

Checking values above 1 results in an undefined situation.

But, values  1 <= x <= 3  work so this is the domain.

The probability is 13/52*12/51*11/50 = 11/850.

\(z=\sqrt{5+12i},-\sqrt{5+12i}\)

The volume of a cylinder of height 11 inches and radius r inches is given by the formula V=11πr2.

Suppose that the radius is expanding at a rate of 0.6 inches per second. How fast is the volume changing when the radius is 2.3 inches? Use at least 5 decimal places in your answer.  ____???cubic inches per second

\(V=11\pi r^2\\ \frac{dV}{dR}=22\pi r\)

\(\frac{dr}{dt}=0.6\)

\(\frac{dV}{dt}=\frac{dV}{dr}\times \frac{dr}{dt}\\ \frac{dV}{dt}=22\pi r\times 0.6\\ \frac{dV}{dt}=13.2\pi r\\ When\;\;r=2.3\; inches\\ \frac{dV}{dt}=13.2\pi *2.3\\ etc \)

LaTex:

V=11\pi r^2\\
\frac{dV}{dR}=22\pi r

\frac{dr}{dt}=0.6

\frac{dV}{dt}=\frac{dV}{dr}\times \frac{dr}{dt}\\
\frac{dV}{dt}=22\pi r\times 0.6\\
\frac{dV}{dt}=13.2\pi r\\
When\;\;r=2.3\; inches\\
\frac{dV}{dt}=13.2\pi *2.3\\
etc

Try proof reading your question and not just posting garbage.

With a short computer code I can verify geno3141's answer. You must have the patience of Job !

n=1;a=30000;s=0;m=n;p=0;n=m;cycle:if(n%2==0 and int(n/10)%2==0 and int(n/100)%2==0 and int(n/1000)%2==0 and int(n/10000)%2==0, goto loop, goto next);loop:print", ",n,;p=p+1;next:n++;if(n<=a, goto cycle,discard=0;print">Total Num =", p

OUTPUT: =26,000 as being the 1000th integer with all of its digits EVEN.

Thanks Kadjaye   

No, but I will give you a good hint.

Multiply both sides by the lowest common denominator (which is 12)

I do not know if I understand the question but if the biggest difference between the pairs is 3 then the pairs could be

1,4

2,5

3,6

7,10

8,11

9,12

13,16

14,17

15,18

19,22

20,23

21,24

25,28

26,29

27,30

31,34

32,35

33,36

37,40

38,39

19*3+1 = 58

So I get the same 'guess' answer as guest does.   

Here is my attempt:

14, 25, 36, 710, 811, 912, 1316, 1417, 1518, 1922, 2023, 2124, 2528, 2629, 2730, 3134, 3235, 3336, 3740, 3839

The 3-digit numbers should be read as: 7 and 10 or 8 and 11.....etc. The 4-digit numbers as 13 and 16 or 14 and 17...etc.

There is an "absolute" difference of 3 between all the pairs, except the last one. Therefore, the maximum sum I get is:

19 x 3 + 1 =58

Two circles are drawn, with the same center. A chord of the large circle is drawn, so that it is tangent to the small circle.

If the chord has a length of 12 then find the area of the ring-shaped region that is inside the large circle but outside the small circle.

\(\begin{array}{|rcll|} \hline \mathbf{R^2} &=& \mathbf{6^2 + r^2} \\ \mathbf{R^2 - r^2} &=& \mathbf{36} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\text{area of the ring}} &=& \mathbf{\pi R^2 - \pi r^2} \\ \text{area of the ring} &=& \pi (R^2 - r^2) \quad | \quad \mathbf{R^2 - r^2=36} \\ \mathbf{\text{area of the ring}} &=& \mathbf{36\pi} \\ \hline \end{array} \)

Terry, a marathon runner ran from her house to the high school then back.

The return trip took 5 minutes longer. If her speed was 10 mph to the high school and 9 mph to her house,

How far is Terry's house from the high school?

\(\begin{array}{|rcl|c|rcl|} \hline &(1)& && &(2) \\ \hline \mathbf{v_1} &=& \mathbf{\dfrac{s}{t}} && \mathbf{v_2} &=& \mathbf{\dfrac{s}{t+\dfrac{5}{60} }} \\ && & \boxed{v_1 = 10,\ v_2 = 9 }\\ 10 &=& \dfrac{s}{t} && 9 &=& \dfrac{s}{t+\dfrac{5}{60} } \\\\ \mathbf{t} &=& \mathbf{\dfrac{s}{10}} && 9 &=& \dfrac{s}{t+\dfrac{5}{60} } \\\\ & & && 9\Big(t+\dfrac{5}{60}\Big) &=& s \\\\ & & && 9t+\dfrac{9*5}{60} &=& s \\\\ & & && 9t+0.75 &=& s & \mathbf{t=\dfrac{s}{10}} \\\\ & & && \dfrac{9s}{10}+0.75 &=& s & | \quad * 10 \\\\ & & && 9s + 7.5 &=& 10s \\ & & && \mathbf{s} &=& \mathbf{7.5} \\ \hline \end{array}\)

Terry's house from the high school is 7.5 miles far away.

A wooden rectangular prism has dimensions 4 by 5 by 6. This solid is painted green and then cut into 1 by 1 by 1 cubes. The ratio of the number of cubes with exactly two green faces to the number of cubes with three green faces is

At 2:00 PM, a car leaves town A and another leaves town B.
The distance between the two towns is 60 km. One car travels 10 kph faster than the other.
The car pass each other at 2:30 PM. How far does each travel?

\(\begin{array}{|rcl|c|rcl|} \hline &(1)& && &(2) \\ \hline v_1 &=& \dfrac{s_1}{t} && v_2 &=& \dfrac{s_2}{t} \\ && & \boxed{v_1 = v_2+10,\\ s_1+s_2=60,\ s_2 = 60-s_1 }\\ v_2+10 &=& \dfrac{s_1}{t} && v_2 &=& \dfrac{60-s_1}{t} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)-(2): & v_2+10-v_2 &=& \dfrac{s_1}{t}-\dfrac{60-s_1}{t} \\\\ & 10 &=& \dfrac{s_1}{t}-\dfrac{60-s_1}{t} \quad | \quad *t \\\\ & 10t &=& s_1 - (60-s_1) \\ & 10t &=& 2s_1 - 60 \\ & 2s_1 &=& 10t+60 \quad | \quad :2 \\ & \mathbf{s_1} &=& \mathbf{5t+30} \\ \hline & s_2 &=& 60-s_1 \\ & s_2 &=& 60-(5t+30) \\ & \mathbf{s_2} &=& \mathbf{30-5t} \\ \hline \end{array}\)

\(\text{Let $t=2:30\ PM-2:00\ PM=0.5$ hours}\)

\(\begin{array}{|rcll|} \hline \mathbf{s_1} &=& \mathbf{5t+30} \quad | \quad t=0.5 \\ s_1 &=& 5*0.5+30 \\ \mathbf{s_1} &=& \mathbf{32.5\ \text{km} } \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{s_2} &=& \mathbf{30-5t} \quad | \quad t=0.5 \\ s_2 &=& 30-5*0.5 \\ \mathbf{s_2} &=& \mathbf{27.5\ \text{km} } \\ \hline \end{array}\)

How far does each travel:
One car travel \(\mathbf{32.5\ \text{km}}\) and the other car travel \(\mathbf{27.5\ \text{km}}\)

Simplify

\( \dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k - 1}} &=& \dfrac{n!}{(k-1)!(n-k+1)!} \quad &| \quad (k-1)!k=k!~ \text{or}~(k-1)!= \dfrac{k!}{k} \\\\ &=& k *\dfrac{n!}{k!(n-k+1)!} \quad &| \quad (n-k)!(n-k+1)=(n-k+1)! \\\\ &=& \dfrac{k}{n-k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{k}{n-k+1} \dbinom{n}{k} } \\ \hline \end{array}\)

\\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}} \\\\ &=& \dfrac{\dbinom{n}{k}}{\dfrac{k}{n-k+1} \dbinom{n}{k} } \\\\ &=& \dfrac{n-k+1}{k} * \dfrac{\dbinom{n}{k}}{\dbinom{n}{k} } \\\\ &=& \mathbf{\dfrac{n-k+1}{k}} \\ \hline \end{array}\)

In the figure, each of the three circles is tangent to the other two

and each side of the equilateral triangle is tangent to two of the circles.

If the length of one side of the triangle is 4, what is the radius of each circle?

\(\begin{array}{|rcll|} \hline \mathbf{\tan(30^\circ)} &=& \mathbf{\dfrac{r}{2-r}} \\\\ (2-r)\tan(30^\circ) &=& r \\\\ 2\tan(30^\circ)-r\tan(30^\circ) &=& r \\\\ 2\tan(30^\circ) &=& r+r\tan(30^\circ) \\\\ 2\tan(30^\circ) &=& r\Big(1+\tan(30^\circ)\Big) \\\\ r &=& \dfrac{2\tan(30^\circ)}{1+\tan(30^\circ)} & \boxed{\tan(30^\circ)=\dfrac{ \sqrt{3} } { 3 } }\\\\ r &=& \dfrac{2*\dfrac{ \sqrt{3} } { 3 }}{1+\dfrac{ \sqrt{3} } { 3 }} \\\\ r &=& \dfrac{2\sqrt{3}}{3\Big( 1+\dfrac{ \sqrt{3} } { 3 } \Big) } \\\\ r &=& \dfrac{2\sqrt{3}}{3+ \sqrt{3} } \\\\ r &=& \dfrac{2\sqrt{3}}{\Big(3+ \sqrt{3}\Big) } * \dfrac{ \Big(3- \sqrt{3}\Big) } { \Big(3+ \sqrt{3}\Big) } \\\\ r &=& \dfrac{2\sqrt{3}\Big(3- \sqrt{3}\Big)} { \Big(3+ \sqrt{3}\Big)\Big(3- \sqrt{3}\Big) } \\\\ r &=& \dfrac{2\sqrt{3}\Big(3- \sqrt{3}\Big)} {9-3 } \\\\ r &=& \dfrac{2\sqrt{3}\Big(3- \sqrt{3}\Big)} {6} \\\\ r &=&\dfrac{2}{6}* \sqrt{3}\Big(3- \sqrt{3}\Big) \\\\ r &=&\dfrac{1}{3}* \sqrt{3}\Big(3- \sqrt{3}\Big) \\\\ r &=&\dfrac{1}{3}* \sqrt{3}*3- \dfrac{1}{3}* \sqrt{3}\sqrt{3} \\\\ r &=& \sqrt{3} - \dfrac{1}{3}*3 \\\\ \mathbf{r} &=& \mathbf{\sqrt{3} - 1}& \sqrt{3}=1.73205080757 \\\\ r &=& 1.73205080757 - 1 \\\\ \mathbf{r} &=& \mathbf{0.73205080757} \\ \hline \end{array}\)

The radius of each circle is \(\approx \mathbf{0.732}\)

Let A and C be points on a circle of radius r, such that AC = r.  B lies on minor arc AC.  Find angle ABC.

No matter where point B is located on a minor arc, the angle ABC is 150 degrees. 

Below 100, we have: 

2, 4, 6, 8, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88   =  [24]

From 200 - 300, we get 200 plus all the above numbers (prefixed with a 2)                        [25]

From 400 - 500                                                                                                                     [25]

From 600 - 700                                                                                                                     [25]

From 800 - 900                                                                                                                     [25]     =  124

From 2000 - 3000                                                                                                               [125]

From 4000 - 5000                                                                                                               [125]

From 6000 - 7000                                                                                                               [125]

From 8000 - 9000                                                                                                               [125]  =  624

From 20000 - 21000                                                                                                            [125]

From 22000 - 23000                                                                                                            [125]     

From 24000 - 25000                                                                                                            [125]  =  999

26000                                                                        

You can use a Venn diagram!  The answer is 2a - 2b.