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If j, k, and l are positive with jk = 24, jl = 48, and kl = 18,

find j + k + l.

1.

\(\begin{array}{|rcll|} \hline jk\cdot jl \cdot kl = (jkl)^2 &=& 24\cdot 48\cdot 18 \\ &=& 20736 \\ jkl &=& \sqrt{20736} \\ \mathbf{jkl} &\mathbf{=}& \mathbf{ 144} \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline j+k+l &=& \dfrac{jk\cdot jl}{jkl} + \dfrac{jk\cdot kl}{jkl} + \dfrac{jl\cdot kl}{jkl} \\ &=& \dfrac{jk\cdot jl+jk\cdot kl+jl\cdot kl}{jkl} \\ &=& \dfrac{24\cdot 48+24\cdot 18+48\cdot 18}{144} \\ \mathbf{j+k+l} &\mathbf{=}& \mathbf{ 17} \\ \hline \end{array}\)

Positive real numbers r, s satisfy the equations

r^2+s^2 = 1 and

r^4 + s^4 = 7/8.

Find rs.

\(\begin{array}{|rcll|} \hline (r^2+s^2)^2 &=& r^4+2r^2s^2+s^4 \\ (r^2+s^2)^2 &=& r^4+s^4+2r^2s^2 \\ 2r^2s^2 &=& (r^2+s^2)^2-(r^4+s^4) \quad & | \quad r^2+s^2 = 1 \qquad r^4 + s^4 = \dfrac{7}{8} \\ 2r^2s^2 &=& 1^2-\dfrac{7}{8} \\ 2r^2s^2 &=& \dfrac{1}{8} \\ r^2s^2 &=& \dfrac{1}{16} \\ \mathbf{rs} &\mathbf{=}& \mathbf{\dfrac{1}{4}} \\ \hline \end{array}\)

This picture may help:

Remember that we want to find the area of the walk, that is the shaded area in the picture.

You found the area of a circle with a radius of 8.3 .

Below is the full answer......

area of walk   =   area of outer circle - area of inner circle

radius of inner circle in feet   =   12.4 / 2   =   6.2

area of inner circle in sq feet  =   pi * 6.2²   =   38.44pi

radius of outer circle in feet   =   6.2 + 2.1   =   8.3

area of outer circle in sq feet   =   pi * 8.3²   =   68.89pi

area of walk in sq feet   =   68.89pi - 38.44pi   =   30.45pi   ≈   95.7

Find the fourth arithmetic means between -21 and -36

\(\text{We use $a_n = a_1 + (n-1)d$ to find the common difference $d$.} \)

Solution

\(\text{The first term is $a_1 = - 21$ and the fourth term is $a_4 = - 36 $.}\\ \text{We must find the common difference so that the terms }\)

\(\begin{array}{cccc} -21, & -21+d, & -21+2d, & -36 \\ \uparrow & \uparrow & \uparrow & \uparrow \\ a_1 & a_2 & a_3 & a_4 \end{array}\)

\(\text{form an arithmetic sequence.} \)

\(\text{To find the common difference $d$, we substitue $-21$ for $a_1$; 4 for $n$, and $-36$ for $a_n \\$in the formula for the 4th term:}\)

\(\begin{array}{rcll} a_4 &=& a_1 + (n-1)d \qquad & \text{This gives the 4th term of any arithmetic sequence.} \\ -36 &=& -21 + (4-1)d \qquad & \text{Substitute.} \\ -36 &=& -21 + 3d \qquad & \text{Subtract within the parentheses.}\\ -15 &=& 3d \qquad & \text{Subtract -21 from both sides.} \\ -5 &=& d \qquad & \text{To isolate d, divide both sides by 3.} \\ \end{array}\)

\(\text{To find the two arithmetic means between $-21$ and $-36$, $\\$we add the common difference $-5$, as shown:}\)

\(\begin{array}{rcll} -21+d &=& -21 +(-5) \\ &=& -26 \qquad & \text{This is $a_2$.} \\ \end{array}\)

\(\begin{array}{rcll} -21+2d &=& -21 +2(-5) \\ &=& -21 -10 \\ &=& -31 \qquad & \text{This is $a_3$.} \\ \end{array}\)

\(\text{Two arithmetic means between $-21$ and $-36$ are $-26$ and $-31$.} \)

If x, y, and z are positive with xy = 20(cube root 2), xz = 35 (cube root 2), and yz = 14(cube root 2),

then what is xyz?

\(\begin{array}{|rcll|} \hline xy &=& 20\sqrt[3]{2} \\ xz &=& 35\sqrt[3]{2} \\ yz &=& 14\sqrt[3]{2} \\\\ xy\cdot xz\cdot yz &=& 20\sqrt[3]{2}\cdot 35\sqrt[3]{2} \cdot 14 \sqrt[3]{2} \\ x^2y^2z^2 &=& 20\cdot 35 \cdot 14 \cdot (\sqrt[3]{2} )^3 \\ (xyz)^2 &=& 20\cdot 35 \cdot 14 \cdot 2 \\ xyz &=& \sqrt{20\cdot 35 \cdot 14 \cdot 2} \\ xyz &=& \sqrt{19600} \\ \mathbf{xyz} &\mathbf{=} & \mathbf{140} \\ \hline \end{array}\)

The solution to this question requires analysis of mutually exclusive (conditional) sets. Each set has an individual probability. The sum of these probabilities times the probability of choosing one of the sets determines the overall probability of a “factorific” coloring.

From the question: The condition requires all divisors of a blue number to be blue. Note a number can be blue without it being a divisor of all greater numbers as long as its divisors are blue.

*Grogg chooses to color none (0) of the numbers.

Probability of factorific: Zero (0)

*Grogg chooses to color one (1) of the six (6) numbers.

Probability of colorific: (1/6).

Explanation: Number one (1) is the only positive integer that has one divisor.

*Grogg chooses to color two (2) of the six (6) numbers.

Probability of factorific: = (1/5)

Explanation: Number of ways to choose two numbers from 6 numbers = (nCr(6, 2)) = 15.

Number one (1) must be in the set because it’s a divisor of all integers. The other number has to be a prime number (2, 3, or 5)

{1, 2}

{1, 3}

{1, 5}

Number of sets that have a one (1) and a prime: Three (3). 

Probability (3/15) = (1/5)

*Grogg chooses to color three (3) of the six (6) numbers.

Probability of factorific: (5/20) = (1/5)

Explanation: Number of ways to choose 3 numbers from 6 numbers = (nCr(6, 3)) = 20

Number one (1) must be in the set because it’s a divisor of all integers.

These 4 sets meet the conditions:

{1, 2, 3} Two (2) is blue here, it’s not a divisor of three, but one (1) is its divisor, so this is valid.

{1, 2, 4} all divisors are blue for all blue numbers.

{1, 2, 5}                       ‘’

{1, 3, 5}                       ‘’

{1, 5, 6}                       ‘’

Five of 20 sets meet the conditions: (1/5)

*Grogg chooses to color four (4) of the 6 numbers.

Probability of factorific: (4/15)

Explanation: Number of ways to choose 4 numbers from 6 numbers = (nCr(6, 4)) = 15

Number one (1) must be in the set because it’s a divisor of all integers.

These 3 sets meet the conditions

{1, 2, 3, 4} all divisors are blue for all blue numbers.

{1, 2, 3, 5}                  ‘’

{1, 2, 3, 6}                  ‘’

{1, 2, 4, 5}                  ‘’

Four of 15 sets meet the conditions: (4/15)

*Grogg chooses to color five (5) of the six (6) numbers.

Probability of factorific: (1/2)

Explanation: Number of ways to choose 5 numbers from 6 numbers = (nCr(6, 5)) = 6

Number one (1) must be in the set because it’s a divisor of all integers. Number of sets of five (5) that have one (1) as an element = (nCr(5, 4)) = 5.

Three sets meet the conditions

{1,2,3,4,5} all divisors are blue for all blue numbers.

{1,2,3,4,6}                  ‘’

{1,2,3,5,6}                  ‘’

Three sets of 6 sets meet the conditions (3/6) = (1/2)

*Grogg chooses to color six of the six (6) numbers.

Probability of factorific: (1) or 100%

Explanation:All numbers are blue and all numbers have all their divisors colored blue.

---------

Sum of individual (mutually exclusive) probabilities: ((0)+(1/6)+(1/5)+(1/4)+(4/15)+(1/2)+(1))

Grogg has a (1/7) probability of picking one of the seven sets (including the empty set).

(1/7)*((0)+(1/6)+(1/5)+(1/4)+(4/15)+(1/2)+(1)) = (143/420) ≈ 34.05%

(1/7)*((0)+(1/6)+(1/5)+(1/5)+(4/15)+(1/2)+(1)) = (143/420) ≈ 33.33%

The overall probability that Grogg’s coloring is factorific is ≈ 33.33%

Sources: A genetically enhanced chimp brain and comprehensive programming of basic set theory from Lancelot Link.

GA

Edits: Error corrections

                Algebra      Not in Algebra     Total

Chemistry             5               6                        11

Not in Chemistry  14            15                        29

Total                     29            21                        40

A)  N(Algebra  or Chemistry)  =  N (Algebra)  + N (Chemistry) - N (In Both) =

                                                      29          +         11             -           5            =     35   

B)   Ratio of those in just Chemistry to total  =   6 / 40   =  3/20

 the ratio we would expext is 2000  =  3/20 * 2000  =  (2000/20) * 3 =  600    

This is a 30-60-90 right triangle....the side across from the 30° angle  =12

The side across from the 60° angle   (what we are interested in ) is √3  times  this

So

12√3    =   20.78   ft

Note  cos B  = adjacent side  / hypotenuse  = a / c

So

cos B   = 6 / 11

Take the arccos  (cosine  inverse) to find  B

arccos   (6/11)  = 56.944268849146°

Degrees  =  56

Minutes  = 60 * .944268849146   =  56.65613094876  =   57 minutes

So

B  = 56° 57'

P (D l M)  = P(D and M) / P(M)        rearrange as

P ( D and M )  =  P (D l M)  * P (M)

P (D and M)  =   0.7  *  0.6  =      0.42

If A  =  60°.....B  =  30°

This is a  30-60-90  right triangle.....the side across from the 30°  is   1/√3  as long as the side opposite the  60° angle

So

b  = 15 * 1/√3  ≈   8.66

a  = 10

b = 4

Just a simple  Pythagorean Triangle problem  ;

c  = √ (10^2  + 4^2 )   =  √ (100 + 16)  = √ 116  ≈  10.77

B = 38°

c  =  18

So...C   = 90  - 38  =   52°

Using the Law of Sines

b/ sin B  = c / sin C

b /sin (38)  = 18 / sin (52 )  

b  =   18*sin(38)  / sin( 52 )   = 14.06 

sin^-1  = .2945   ≈  17.127557612801°

minutes  = .127557612801* 60  = 7.65345676806

seconds = .65345676806  * 60 =  39.21" 

So

17 ° 7'  39.21"

tan (50°)  ≈  1.1918

cos (134°) ≈ -0.6947

If x + 1/x = 7, then what is the value of x^2 + 1/(x^2) + 1?

Note that

(x + 1/x)^2   = 7^2

x^2  + 2 (x * 1/x)  + (1/x)^2  = 49

x^2  + 2(1)  + 1/x^2  =  49

x^2  +  2  +  1/x^2   =  49      subtract 2 from both sides

x^2  + 1/x^2   =  47   ......so.....

x^2 + 1/x^2  + 1   =  47  + 1

x^2 + 1/x^2 +  1   = 48

thanks cool kid

Well...don't know about the "cool"  part...but....here goes nothing......

                       Algebra      Not in Algebra     Total

Chemistry             5               6                        11

Not in Chemistry  14            15                        29

Total                     29            21                        40

A)  N(Algebra  or Chemistry)  =  N (Algebra)  + N (Chemistry) - N (In Both) =

                                                      29          +         11             -           5            =     35   

B)   Ratio of those in just Chemistry to total  =   6 / 40   =  3/20

So.....we would expect the same ratio  in  2000  =  3/20 * 2000  =  (2000/20) * 3 =  600    

I'll let you write the "complete sentences" part based on  my answers...... 

Thank You!

If j, k, and l are positive with jk = 24, jl = 48, and kl = 18, find j + k + l.

jk  = 24 ⇒  24/j  = k

jl = 48  ⇒  48/j  = l

kl = 18   

So

kl  = 18

(24/j) (48/j)  = 18

1152 / j^2  = 18

j^2  = 1152/18

j^2  = 64   take the positive root

j =  8

So.....24 / j = k   ⇒    24 / 8  = k ......  3  = k

And  48 / j  = l ⇒   48 / 8  = l    ......  6  = l

So

j + k + l    = 8 + 3  +  6  =   17

Sorry, Manuel...I got distracted...I meant to write:

"Keep  the degrees....divide the minutes by 60   and the seconds  by 3600"

See my corrected answer!!

Here's the second one

39  +  50/60  + 35/ 3600   =  39.8431°

Thanks so much, CPhill!

                       O     A       B     AB         Total
Male (M)    210    174      74   42          500
Female (F) 315    261    111   63          750
Total           525    435   185   105       1250

P (A / M)   =  "probability that the student has blood type A  given a male"  =   174 / 500   =  87 / 250

P(A ∩ M)  = P(A)  * P(M)  = 

174/1250     =      (435/1250) *  ( 500/ 1250 ) 

87/625   =  87/625

"C"  must be correct.......