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   We have a piece of wire.  We pass $500$ volts through this wire and measure $25$ milliamps of current.     

   If I cut the wire in half and pass $175$ volts through it, how many milliamps of current will I measure?     

   Part 1.  Use Volts and Amps to determine Ohms         E = I • R     ==>>     R  =  E / I     

                                                                                                                         R  =  500 V / 0.025 A     

                                                                                                                         R  =  20,000 Ω     

   Part 2:  Use Volts and Ohms to determine Amps         E = I • R     ==>>     I  =  E / R     

                Note:  Cutting the wire in half halves the R                                      I  =  175 V / 10,000 Ω     

                                                                                                                          I  =  0.0175 A     

                Note:  1 A = 1,000 mA                                                                       I  =  17.5 mA     

   _.     

   Jon teaches ... at an elementary school where class sizes are always at least 100 students and at most 200.     

   One day Jon decides that he wants to arrange the students in their desks in a rectangular grid with no gaps.     

   Unfortunately for Jon he discovers that doing so could only result in one straight line of desks.     

   What is the largest number of students that Jon could have in his class?     

   The number is 199 because 199 is the largest prime number less than 200.     

   If one straight line is the only possible resultant, then the number has to be     

   prime.  A prime number has no factors other than 1 and itself.  A non-prime     

   number could be factored, thus having the possibility of more than a single     

   row of desks.     

   _.     

   the graph of the line with equation 3x+6y+=8 crosses the y-axis when y=?     

   The line will cross the y-axis when x = 0.       So,    3x + 6y = 8     

                                        substitute 0 for x               3(0) + 6y = 8     

                                                                                            6y = 8     

                                                                                              y = ⁸/₆     

                                                                                              y = 1 ¹/₃     

   _.     

Levans writes a positive fraction in which the numerator and denominator are integers, and the numerator is 2 greater than the denominator. He then writes several more fractions. To make each new fraction, he increases both the numerator and the denominator of the previous fraction by 1. He then multiplies all his fractions together. He has 3 fractions, and their product equals 10. What is the value of the first fraction he wrote? 

      x+2           x+3            x+4            10     

   ––––––  •  ––––––  •  ––––––  =  ––––––     

        x             x+1            x+2              1            cancel the x+2's     

       1             x+3            x+4             10     

   ––––––  •  ––––––  •  ––––––  =  ––––––     

        x             x+1             1                 1            cross multiply     

                               x² + 7x + 12  =  10x² + 10x     

                            –9x² – 3x + 12  =  0                multiply by –1     

                              9x² + 3x – 12  =  0                divide by 3     

                              3x² + x – 4  =  0     

                         (3x + 4)(x – 1)  =  0     

                                              x  =  –4/3     discard – the problem specifies an integer     

                                              x  =  1     

   Therefore, the first fraction is (1+2) / 1  =  3 / 1     

   _.     

   Two circles intersect at two points, $P$ and $Q$.     

   The equations of the two circles are $x^2 + (y - 1)^2 = 1$ and $(x - 1)^2 + y^2 = 1$.     

   Find the length PQ.     

                x² + (y – 1)²  = 1 

       and   (x – 1)² + y²  = 1     

   Use Desmos to plot both circles and note     

   that they intersect at the origin and at (1,1)     

   That makes PQ = sqrt(2)     

   _.     

   Jon teaches a fourth grade class at an elementary school where class sizes are always at least $20$ students and at most $60$.     

   One day Jon decides that he wants to arrange the students in their desks in a rectangular grid with no gaps.     

   Unfortunately for Jon he discovers that doing so could only result in one straight line of desks.     

   What is the largest number of students that Jon could have in his class?     

   This one is a breeze.  The number is 59 because 59 is the largest (20 < x < 60) that's a prime number.     

   A prime number has no factors other than 1 and itself.  So, there can be only a single row of 59 desks.     

   _.     

   What is the square root of 64     

   The sqrt(64) = + 8 . . . everybody knows it's 8, but     

   sometimes they forget that negative 8 is also a root.     

   _.     

   How do I solve a y= system using the combination method? For example, y = 6x +4 and y = 4x - 2?     

   The combination method, more commonly called the elimination method, is a way to solve a      

   system of equations by adding or subtracting the equations to eliminate one of the variables.     

   Using the provided example:              y  =  6x + 4     

                                                               y  =  4x – 2     

                                  subtract                0  =  2x + 6     

                                                          – 2x  =  6     

                                                             2x  =  – 6     

                                                               x  =  – 3     

                OR set the y's equal      6x + 4  =  4x – 2     

                                                      2x         =      – 6     

                                                               x  =  – 3     

   _.     

   All members of our painting team paint at the same rate.     

   If 20 members can paint a 2000 square foot wall in 40 minutes, then how long would it take     

   the 20 members to paint a 6000 square foot wall, in minutes?     

   20 members   ==>>    paint 2000 ft² / 40 min  =  50 ft² per min     

   20 members   ==>>    paint 6000 ft² /  x          =  50 ft² per min     

                                               6000 ft²                =  50 ft² per min • x     

                                          6000 ft² / 50 ft²/min  =  x     

                                                                        x  =  120 min     

   Another way to do it is to consider the 6000 ft² wall as equivalent to three 2000 ft² walls     

   So, because one wall takes 40 minutes, three walls would take (3 • 40 min) = 120 min     

   _.     

   Find the maximum area of a right triangle with hypotenuse 1m.     

   The area of a quadrilater for a given perimeter is maximized when the quadrilateral is a square.     

   So let's assume the same holds true for a triangle, by considering the triangle as half a square.     

   If the triangle is half a square, that means both legs are the same length.     

   by Pythagoras' Theorem                                          a² + b²  =  c²     

   the legs are equal, which means a = b therefore           2a²  =  1²     

                                                                                            a    =  sqrt(1/2)     and b is the same     

   Area of a triangle   =  (1/2) • base • height     

                                 =  (1/2) • sqrt(1/2) • sqrt(1/2)     

                                 =  (1/2) • (1/2)     

                                 =  1/4 meter²     

   _.     

   All members of our painting team paint at the same rate.     

   If 20 members can paint a 2000 square foot wall in 40 minutes, then how long     

   would it take the 20 members to paint a 6000 square foot wall, in minutes?     

   20 mem   ==>>    paint 2000 ft² / 40 min  =  50 ft²/min     

   20 mem   ==>>    paint 6000 ft² /    x         =  50 ft²/min     

                                        6000 ft²                 =  50 ft²/min • x     

                                        6000 ft² / 50 ft²/min  =  x             

                                                                             x  =  120 min     

   Another way to do it is to consider the 6000 ft² wall as equivalent to three 2000 ft² walls     

   So, because one wall takes 40 minutes, three walls would take 3 • 40 min = 120 min     

   .     

   I have to paint one side of a wall.  The wall is 12 meters tall and 120 meters long.     

   Each gallon of paint covers 150 square feet. If a foot is approximately 0.3048 meters, then     

   what is the smallest whole number of gallons I can buy and have enough paint to cover the whole wall?     

   1 ft = 0.3048 m     ==>>     1 m = (1 / 0.3048) ft = 3.2808 ft     

   Height   =  (12 m)   • (3.2808 ft/m)  =    39.37 ft     

   Length  =  (120 m) • (3.2808 ft/m)  =  393.70 ft     ==>>     Area  =  (39.37 ft) • (393.70 ft)     

                                                                                                         =  15,499.97 ft²     

   Paint req'd     ==>>     (15,499.97 ft²) / (150 ft²/gallon)  =  103.33 gal     

   You can't buy just part of a gallon, so you will have to buy 104 gallons     

   _.     

   The following cards are split into three piles at random,     

   so that every pile contains the same number of cards.     

   What is the probability that every pile contains an Ace?     

   Ace of spades     

   Ace of hearts     

   Ace of diamonds     

   Ace of clubs     

   Two of spades     

   Two of hearts     

   Six cards divided evenly into three piles will create two cards per pile.     

   This makes 6C2 possible different piles, i.e., 15 possible combinations.     
 

   AS AH      AH AD      AD AC      AC 2S      2S 2H     

   AS AD      AH AC      AD 2S      AC 2H     

   AS AC      AH 2S      AD 2H     

   AS 2S      AH 2H     

   AS 2H     

   As illustrated above, every combination will have an Ace, except one.     

   The only pile that doesn't contain an Ace is the pile that has both 2's.     

   So the probability that a pile will contain an Ace is 14 / 15, i.e., 93.3%.     

   _.     

   how do you work out the rate of interest in the second year?     

   The interest rate is the same the second year as it was the first.     

   You have to add the first year's interest to the principal amount,     

   then figure the second year's interest using that new principal.     

   _.     

   a triangular prism has been created with two equilateral triangles and three congruent rectangles.     

   the height of the triangle is 3.5 inches. find the surface area of the triangular prism.     

   You could calculate the height of the rectangles by using     

   trigonometry and the fact that the triangles are equilateral.     

   But there’s no information given from which to deduce the     

   length of the rectangles, which is also needed to calculate     

   the area of the rectangles.     

   _.     

   how to find radius and height of cylinder given only a volume. The formula is V=1/3PIr^2h     

   No, it isn't.  The volume of a cone is         V = ¹/₃ • π • r² • h     

                      The volume of a cylinder is    V = π • r² • h     

   Either way, you can't do it.  You can't find two unknowns with only a single equation.     

   _.     

You can re-arrange this to 

e * e * e  * 2 * 5/2          then it easier to see tha this equals 

e^3 * 5 

= 5 e^3 

   When the same constant is added to the numbers 60, 120, and 160, a three-term geometric sequence arises.     

   What is the common ratio of the resulting sequence?     

      60 + x         120 + x     

   –––––––  =  –––––––     

    120 + x         160 + x     

   (120 + x) (120 + x)  =  (60 + x) (160 + x)     

   x² + 240x + 14400  =  x² + 220x + 9600     

          240x  – 220x  =  9600 – 14400     

                         20x  =  – 4800     

                             x  =  – 240     

      60 – 240          120 – 240     

   ––––––––––  =  ––––––––––     

    120 – 240          160 – 240     

       – 180                 – 120     

   ––––––––––  =  ––––––––––     =   1.5     

       – 120                   – 80     

   _.     

   A terminal zero is a $0$ that appears at the end of a number.     

   For example, the number $3,800$ has two terminal zeros.     
   How many terminal zeroes does $40 \cdot 6 \cdot 75 \cdot 12$ have?     

   I just did the multiplication, then counted the terminal zeros.     

   40 • 6 • 75 • 12  =  216,000     I count three of them.     

   _.