The arrangement of elements in the periodic table contains lots of clues about the characteristics of those elements, their families, and their relation to each other. The table is like a multi-facetted gemstone and the more you study it, the more you discover. It’s hard to think of a more meaningfully arranged two-dimensional table (although in some ways it’s three-dimensional). For any given element in the table, its location – column and row – tells you a lot about it, as do its neighbors – above, below, left side, right side.

A good place to start when introducing students to the periodic table is with the number that appears above each?symbol. That number is three things simultaneously. I like to use the acronym A.P.E: Atomic number, number of Protons, number of Electrons. Take selenium (Se), element 34. The fact that 34 is its atomic number tells us that selenium has 34 protons in its nucleus, and 34 electrons orbiting that nucleus. This is a pretty good starting point in visualizing the structure of selenium – what we’d see if we had a really powerful microscope, one that could also snap a photo and freeze the electrons in motion.

The element to the left of selenium has one less proton and one less electron. And the element to the right of selenium has one more proton and one more electron. Simple as that.

The periodic table can get complicated, and there are many exceptions to its rules. So it’s important?to start with those facts and trends which can be trusted?to give consistently accurate and revealing information.

]]>

Translate to an equation and then solve.

Solution: “What number” translates into a variable, I’ll use “n”. The word “is” translates into an equals sign. The three percents get translated into their decimal equivalents: .8, .5, and .2. The “of’s” gets translated into multiplication signs. And 400 is an integer which needs no translation. So the equation looks like this:

n = .8 x .5 x .2 x 400

n = 32

]]>The freshness matters a lot because freshness determines the moisture content, and this impacts cooking times.

**Now to the Algebra part.** Suppose you wanted to use the data above to create a ?recipe? for microwaving chestnuts ? a formula that you or someone else could use for cooking batches of six, seven, eight chestnuts or more (theoretically).

Think of the left column as your x-values and the right column as your y-values. Notice how for each additional chestnut the number of seconds added increases each time,?it’s?not constant. So we know this would not be a linear function. Let’s assume that it is an exponential function?? that it follows the form y = ab?. How could you use this data to create an exponential function, and what would the resulting “recipe” be?

]]>It is 6:20.

a) What is the acute angle formed by the hands?

b) What will that angle be when the time is 6:54?

SOLUTION:

a) As we established (see Problem 1), the big hand moves at a rate of 6? per minute, while the small hand moves at .5? per minute. Right now the big hand is on the 4 and the small hand is somewhere between the 6 and 7. So, the angle they form is: the angle from 4 to 5, plus the angle from 5 to 6, plus the sliver beyond 6 that the small hand has moved. The wedges between the numbers on an analog clock are all 30? (360?/12 = 30?). So from 4-5 is 30?, and from 5-6 is 30?; we just need to add to this whatever ground was covered by the small hand in the 20 minutes since it was 6:00 (and the small hand was exactly on the 6). If the small hand covers .5?/minute, in 20 minutes it sweeps out an angle of .5?(20) which is 10?. So, adding up our sectors, 30 + 30 + 10 = 70. The current angle formed by the hands is 70?.

b) At 6:54, the big hand will be just below the 11, and the small hand will be just before the 7. Counting up the full sectors, we have three of them: from 7-8, from 8-9, and from 9-10, all 30 each?. So the sector from the 7 to the 10 is 90?.

We have to add to this the angle between the 10 and the big hand, and the small angle between the 7 and the small hand. If the big hand moves at 6?/minute, then at 54 minutes past the hour it has covered 6?(54) = 324?. We know the gap from the 10 to the 12 is 60?. So think of the 10 as being 300? past the hour (360?-60?). That’s a gap of 324-300 = 24?. Adding that to 90 we are now at 114?. Lastly, if the hour hand moves at .5?/minute, then at 54 minutes past the hour it has moved .5?(54) or 27? past the 6, and is only 3? away from the 7. So the little sliver between the 7 and the hour hand is 3?; adding that to 114 we get 117.

At 6:54, the obtuse angle between the hour hand and the minute hand is 117?.

]]>The time is 5:20. The acute angle formed by the clock hands is 40?. In exactly how many minutes will the big hand be right on top of the little hand? Show your work. Express your answer as a mixed number or as a decimal rounded to the nearest hundredth. Let N stand for “New angle”, let S stand for “Starting angle”, let m stand for time in minutes.

SOLUTION: When the big hand is right on top of the small hand with no space between them, they create an angle of 0?. So N, the new angle (i.e. the final angle) will be 0.?The time that will elapse in minutes is unknown. The starting angle is 40. The speeds of the two hands (see Problem 1) are: big hand, 6?/minute; small hand, .5?/minute. The big hand in this case is shrinking the angle, so we’ll subtract its movement from 40. The small hand will slightly grow the angle, so we’ll add its movement to 40. The general layout of the equation will be:

New angle = Starting angle – (big-hand-movement) + (small-hand-movement)

Plugging in our values:

0 = 40 – 6m + .5m . . . From here it’s just equation solving.

0 = 40 – 5.5m

5.5m = 40

m = 7 3/11 or 7.27 minutes

]]>

It’s 4:30. The acute angle formed by the big hand and the little hand is 45?. To the nearest tenth, what will that angle be when the time is 4:44? Solve by writing a formula which uses the following variables:

N = new angle

S = starting angle

SOLUTION: In the 14 minutes that transpire between 4:30 and 4:44, the big hand will greatly enlarge the angle as it moves away from the small hand; while the small hand will slightly shrink the angle as it “chases after” the big hand. So the general layout of our equation will look like this:

New angle = Starting angle *plus* Big-hand-movement *minus* Small-hand-movement

As established in Problem 1 (https://classsizeofone.com/clock-hands-math-problem-part-1/), the big hand moves at a rate of 6? per minute, while the small hand moves at .5? per minute. The time is known, and is the same for both hands: 14 minutes. So,

N = S + 6(14) – .5(14)

N = 45 + 84 – 7 = 122. . . . At 4:44, the acute angle formed by the hands will be 122?.

]]>SOLUTION: When the big hand moves from its current position, it approaches the small hand, shrinking the angle. When the small hand moves, it moves away from the big hand, enlarging the angle. So the general layout of the equation would go like this:

New angle = Starting angle *Minus* Movement-of-big-hand *Plus* Movement-of-small-hand

The hands move at different rates, so let’s figure out each of those rates. Imagine the big hand sitting on the “12” moves 5 minutes so it’s sitting on the “1”. That’s obviously a change of 5 minutes but it’s also a movement of 30?, since the original angle was 90? and moving to the “1” represents 1/3 of the distance between the “12” and the “3”. So, if the ratio is 30? per 5 minutes, that’s a unit rate of 6? per minute. (Another way to get there is this: There are 360? in a circle and 60 minutes on a clock face. 360/60 reduces to a ratio of 6:1.) So if the big hand moves 6? in 1 minute, then in 10 minutes it moves is 6?(10) or 60?. Meanwhile the small hand would take a full hour to go from its current position, “3”, to the next number, “4”, which also represents a movement of 30?. So if something moves 30? in 60 minutes, that means its unit rate would be 30/60 or .5 degrees per minute. If something moves .5 degrees in 1 minute, then in 10 minutes it moves .5?(10), which is 5?. So getting back to the general equation above,

N = S – (big-hand-movement) + (small-hand-movement)

let’s plug in the values for the degrees covered by the two hands, plus 90 for the starting angle which was given:

N = 90 – 60 + 5 … So at 3:10, the angle between the two hands is 35?.

]]>

?Keep Change Flip? is one of the more common tips used in middle school math. It guides students through the process of dividing fractions. It’s a shorthand way of reminding them to take the following three steps: 1) **Keep** the first number as is, 2) **Change** the division sign to multiplication, 3) **Flip** the second fraction upside down (…then multiply the fractions). Multiplying fractions is always taught before dividing fractions, so these steps revert division problems back to multiplication, which presumably the student has mastered.

Keep Change Flip sounds very similar to another math mnemonic…

?Keep Change Change? is a?memory aid?used to teach students how to subtract signed numbers. This one, too, lets students convert a trickier operation back to something they already know. Specifically, it converts subtraction problems to addition problems. (This is around the time when many students are taught that subtracting is ?adding the opposite.?) As an example, take a problem like -7 – (-5). The **Keep** would have you keep the -7 as is. The first **Change**?refers to the operation symbol, a subtraction sign which becomes an addition sign. The second **Change**?refers to the sign of the second number: if it’s negative make it positive; if positive make it negative. Following these steps would remake the problem into: -7 + 5.