I went back to school (college) in my fifties. After two remedial math classes I aced College Algebra and Statistics, but I have forgotten a lot in the last fifteen plus years. I enjoy this refresher.
This is a fun review. I have gone as far as 3 semesters of calculus and differential equations over 53 years ago. Can't believe how much i have forgotten but indeed its coming back. I am enjoying this and may take advantage of the rebuilder Couse
Using the exponents method, you mention factoring numerator before dividing by denominator. But we can also divide first. For example (a-b)/c = a/c - b/c. We just have to make sure to divide each term in numerator by denominator: (2^6 - 2^5) / 2 = (2^6 / 2^1) - (2^5 / 2^1) = 2^5 - 2^4 = 2^4 (2^1 - 1) = 2^4 (1) = 2^4
@@eagle-eye29 Yes, and the final step would be 2^4 = 16. If I'm trying to figure out and write down instructions on how big to make a piece of fabric, for example, I would not write: "cut it to 2^4 inches by 3^3 inches".
I’m now 70 years, I got 16 and then I looked and there wasn’t any 16, so I thought it was wrong, I didn’t realize that I should put the 16 to powers. Thank you
I'm a little younger and just calculated the value of 2^4 in my head to see if it matched. You would think after years of working with powers of 2 I could just look at it and know.🤦 I need to give my brain a bit more exercise.
To hide the correct answer. In a form that it should not be in. But he is teaching, solving methods for problems that you can't do in your head. Which tends to be many algebra problems.
@@djohannsson8268 My math teacher refused to give multiple choice questions for exams. He believed that most students would just be lazy and luck their way into the right answer. Finally he decided to prove his theory and he told us he would do a multiple choice exam, but you had to show each and every step to prove you know how to do it and anybody who does NOT get 100% will have their grade lowered one full letter. But if you get 100% on it he would raise it a whole letter grade. Long story short, 5 questions and not a single one of them had the right answer as a choice not even like how this example technically had the right answer. Well only 3 of us, my self included, got 100% on it and he managed to prove his point. That man was the toughest cunt of a teacher I ever had, but I respected him for how tough he was and I am just as tough with my students. (Though I teach history lol)
I memorized most powers of 2 because of computer programming and bit-positions, and combined with good old "times tables" from when I was 5 the low powers of four were pretty easy as well. I came up with the right answer in about 10 seconds but was slowed down a few because you listed the correct one un-resolved to a plain number which required more thought.
Good teaching, every teacher teach differently but comes with the same correct answers. He is breaking down so you can understand it better, once it it lock in you got it.
I got the answer but I had to do it the long way: 4*4*4 = 64 then 2*2*2*2*2 = 32; then 64 - 32 = 32; then 32/2 = 16. Forget that squaring and cubing stuff… I have enough problems with the basic stuff 😢! 😂😂😂😂😂
cubes are a little tougher but you should at minimum know your squares for the first 13 numbers with out thinking. Now for cubes it is just an extra one. this means 4*4*4 is the same as 16 * 4 . 4 *6 is 24 so put down the 4 and carry a 2. 4 * 1 is 4 now add the 2 from before you get 6. So 4 * 4 * 4 is 64. I am a slow typist so the length of time it took me to type out that explanation is probably 3 times longer than it actually took to do it in my head. (it took me 2 minutes roughly to type it out) Mind you I had those squares drilled into me for 3 months straight. (I was a little slower than the other kids. After enough practice he could say for example 13² then in two seconds I would say 169.) It is tough but if you work hard it becomes second nature like breathing.
@5:24 you said "so 2 to the fourth power says take 2 and multiply it by itself 4 times" So 2 to the first power, you would take 2 and multiply it by itself ONE time which means 2 time 2 = 2? I don't think so. 2 to the fourth power equals 16 but you multiply 2 by itself three time 2 times 2 = 4 and 4 times 2 = 8 and 8 times 2 = 16. I think what you meant is you take four 2s and multiply them together, 2x2x2x2=16. I sure hope I never misspeak.
The correct answer is D. You turn 4^3 into s^6. When you subtract 2^5 from 2^6 you get 2^5 because 2^7 = 2(2^5) or 2^5 + 2^5. That leaves 2^5/2. Since 2^5 = 2(2^4) the answer is 2^4.
Stating the answer as 2^4 is inappropriate. Most teachers insist on resolution (simplification) of answers, unless there is a specific reason for doing otherwise. Most teachers assigning this problem would consider the correct answer to be 16 and most students would try to simplify the result to its most basic form, i.e. 16. Many teachers would give full credit for 16 and would take off points for 2^4.
Experiment and try different solutions. It's how you get comfortable. I preferred the following after several runs and fun: We know 4^3 = 2^6 Splitting the numerator and writing 2 as 2^1: 2^6 / 2^1 - 2^5 /2^1 Wanting to keep them as exponentials and find a common factor in both LHs and RHS 2^(6-1) - 2^(5-1) ==> 2^5 - 2^4 ==> 2^4(2^1) - 2^4. Factoring out 2^4 2^4(2^1 - 1) (easy to see (2^1 - 1) is 1) 2^4(1)
At the 5:30 mark he incorrectly says that 2 to the 4th is multiplying 2 by itself 4 times. In fact, it is multiplying 2 by itself 3 times. Easy mistake to make verbally bc you line up 4 2’s in a row, but the act of multiplication only takes place 3 times.
The problem is easiest to solve if the 4 ^ 3 is first transformed to power of 2 Ex; ((2 ^ 2) ^ 3 = 2 ^ 6. Then divide both terms of the numerator by 2 (in the denominator). This will have the effect of greatly simplifying the problem by eliminating the denominator from (2 ^ 6) - (2 ^ 5) / 2 to (2 ^ 5) - (2 ^ 4). Then factor out the common multiple (2 ^ 4) from each term to give (2 ^ 4) * ( 2 -1 ). The perform the subtraction that will yield the final answer of (2 ^ 4) * 1 = (2 ^ 4) Few steps reduces the chance of error and cuts the solution time during timed exams.
I know how to do this just wondering how John is going to teach it .There are a few ways to do this .I have been doing my own rebuilding I have an old High School Algebra 2 book from 1976 and my old College Algebra book which I found Yup I am past the Algebra 2 level umm Trig pre- calculus and Calculus …well that is a different story lol…I had to retrain myself to do these problems I just had to do problems I could do them but not fast enough as I did more my speed pick up I am 66
Une mendoj qe ushtrimet matematikore duhet te ndshen. Te cilesohen ushtrimet qe jane per shkollat fillore dhe 9 vjeçare dhe ato per shkollat e mesme dhe ato ne universitet qe vazhdojne degen e mesuesise per matematike. Ky eshte si shume te tjera , ushtrim i thjeshte mstematike. Ndjese per kete vrejtje. Falemnderit per mirekuptimin
(4 to the third minus 2 to the fifth) = (2 to the sixth minus 2 to the fifth) = (2(2 to the fifth)minus 2 to the fifth) = 2 to the fifth and then that is divided by 2 resulting in 2 to the fourth.
This "complicated" method can make solving a problem such as (4^103 - 2^205) / (2^200) much simpler than the "simple" brute force approach. [Answer: 2^5]
On your initial "blackboard". you have written four to the third minus two to the fifth, divided by two. Four to the third is 74. 2 to the fifth is 32. 74 minus 32 is 42. 42 divided by 2 is 21. Then you start showing us how to solve it using exactly the same method I used, simplifying the numerator, dividing by the denominator. Then you mysteriously change the two to the fifth to two to the fourth, and solve a completely different problem. No wonder you claim "Many will get it wrong". Well, of course we will, when you don't solve the same problem you pose.
Do you mean this actual calculation? As in, subtracting 2⁵ from 4³ and then dividing the result by 2? It would be pretty bizarre to say to mathematics teachers that before they could use anything as an example like this, they had to be able to cite a case of exactly that calculation being used by someone in a real world problem. There's one immediately obvious practical problem with that. Those of us who do calculations in the real world do not copy them all into some sort of global encyclopedia for teachers to refer to, so what you're asking isn't possible anyway. But quite apart from that, what a silly idea it is. Perhaps you meant to say, show me a real world problem that requires the ability to understand mathematical notation and the ability to manipulate expressions and the ability to have insight into problem solving. But of course that would be a very silly question too. Who doesn't benefit from a bit of mathematical ability and problem solving ability?
Very helpful content. A request. Is it possible to do an annual "wrong" answer Olympics. Same problems but you must correctly arrive at the wrong answer. 😂
but the original equation did not have any parentheses. So although you stumbled on the right answer you voided PEMDAS. To maintain the integrity of PEMDAS, I believe it is move correct to simplify the equation to 4^3/2 - 2^5/2.
@@reach60532there are assumed parenthesis because it's all over 2. To express that in serialized text format the parenthesis simply need to be made explicit, because there is no way to convey "all this over" anything.
@@Spudz76 My point is that this is just one in a series of videos attempting to demystify the order of operations. To suddenly introduce magical parentheses without a word of explanation is a huge disservice to the students watching these videos. Either explicitly explain where these parenthesis came from or simplify the equation to 4^3/2 - 2^5/2 and maintain the integrity of PEMDAS.
@@reach60532 You can put as many parenthesis around the part that's already first without explanation or confusion, IMO. Your suggestion is actually more confusing.
Well I got 16 and came to find out what I did wrong. DUH 2 to the 4 th = 16 so no, I didn’t think about it still being a power answer since the first thing I did was get rid of the powers. Never said I was a genius.
Misleading. If your goal is to teach order of operations you are adding confession by saying you can willy-nilly add parentheses. I believe it would be clearer to evaluate 4^3/2 - 2^5/2 and evaluate thusly. It preserves the integrity of PEMDAS and doesn't confuse the student by suggesting that you can just make stuff up and magically apply it to every situation.
I totally agree. This guy is super long winded and turns every problem in a pretty much useless review of PEMDAS. Not to mention he always starts out problems with multiple choice answers explaining the reasons for the wrong choices instead of explaining the process to calculating the correct answer. If I had to sit through his classes I would probably off myself.
I simply referred to my memorized powers of two (computer programming) and immediately eliminated the exponent parts to normal numbers and then the rest was easy. Except the bit where I had to also convert the answer to a normal number to compare with the normal number I had as the answer, which was double annoying. At least when I see exponents of two I almost just see the normal number instead like X-ray glasses.
I went back to school (college) in my fifties. After two remedial math classes I aced College Algebra and Statistics, but I have forgotten a lot in the last fifteen plus years. I enjoy this refresher.
I'm going back to college now and I'm 61, I give you a lot of credit
4^3 = 2^6
Next we have (2^6 - 2^5)/2.
We can split this fraction into two parts like this: (2^6/2) - (2^5/2).
2^5 - 2^4 = 2^4.
This is a fun review. I have gone as far as 3 semesters of calculus and differential equations over 53 years ago. Can't believe how much i have forgotten but indeed its coming back. I am enjoying this and may take advantage of the rebuilder Couse
Using the exponents method, you mention factoring numerator before dividing by denominator. But we can also divide first. For example (a-b)/c = a/c - b/c. We just have to make sure to divide each term in numerator by denominator:
(2^6 - 2^5) / 2 = (2^6 / 2^1) - (2^5 / 2^1) = 2^5 - 2^4 = 2^4 (2^1 - 1) = 2^4 (1) = 2^4
What would be the point of looking for an answer to a simple equation and stopping with 2^4?
Exactly ... I got 16 and then had to go back to 2^4
Exactly.
I think the ‘point” is to show people like me how to solve the equation 🎉
@@eagle-eye29 Yes, and the final step would be 2^4 = 16.
If I'm trying to figure out and write down instructions on how big to make a piece of fabric, for example, I would not write: "cut it to 2^4 inches by 3^3 inches".
@@eagle-eye29 You cannot say that you've solved the problem if you stop short of SIXTEEN!!!
I've been out of high school > 55 years, and never again thought about these stuff. So I am intrigued to resurrect my memory.
i'm with you. grad in 69. love this guy.
I am 62 and i think the answer is D.
Yes D right
I did it using your last shown method. Explained nicely by the way.
Great that you bring up math on UA-cam!
I’m now 70 years, I got 16 and then I looked and there wasn’t any 16, so I thought it was wrong, I didn’t realize that I should put the 16 to powers. Thank you
I did the same thing. I just stopped at that point and decided to let him work it out for us.
I'm a little younger and just calculated the value of 2^4 in my head to see if it matched. You would think after years of working with powers of 2 I could just look at it and know.🤦
I need to give my brain a bit more exercise.
D.) 4^3=64, and 2^5=32; subtract, get 32. Divide by 2 and get 16, but it's written in exponential form.
To hide the correct answer. In a form that it should not be in.
But he is teaching, solving methods for problems that you can't do in your head. Which tends to be many algebra problems.
@@djohannsson8268 My math teacher refused to give multiple choice questions for exams. He believed that most students would just be lazy and luck their way into the right answer. Finally he decided to prove his theory and he told us he would do a multiple choice exam, but you had to show each and every step to prove you know how to do it and anybody who does NOT get 100% will have their grade lowered one full letter. But if you get 100% on it he would raise it a whole letter grade. Long story short, 5 questions and not a single one of them had the right answer as a choice not even like how this example technically had the right answer. Well only 3 of us, my self included, got 100% on it and he managed to prove his point.
That man was the toughest cunt of a teacher I ever had, but I respected him for how tough he was and I am just as tough with my students. (Though I teach history lol)
I memorized most powers of 2 because of computer programming and bit-positions, and combined with good old "times tables" from when I was 5 the low powers of four were pretty easy as well. I came up with the right answer in about 10 seconds but was slowed down a few because you listed the correct one un-resolved to a plain number which required more thought.
It’s a pleasure to follow these kind of questions because they regard them as difficult for many when in practice are very basic.
got it 16. easy math exponents. thanks for the fun. 64 - 32 = 32 then /2 = 16 that's 2^4 simple.
16
Good teaching, every teacher teach differently but comes with the same correct answers. He is breaking down so you can understand it better, once it it lock in you got it.
I love these examples. I learnt a few years ago at 80. And now I do understand your teaching...
I got the answer but I had to do it the long way: 4*4*4 = 64 then 2*2*2*2*2 = 32; then 64 - 32 = 32; then 32/2 = 16.
Forget that squaring and cubing stuff… I have enough problems with the basic stuff 😢! 😂😂😂😂😂
Wtf I did it in 3 mn
cubes are a little tougher but you should at minimum know your squares for the first 13 numbers with out thinking. Now for cubes it is just an extra one. this means 4*4*4 is the same as 16 * 4 .
4 *6 is 24 so put down the 4 and carry a 2. 4 * 1 is 4 now add the 2 from before you get 6. So 4 * 4 * 4 is 64.
I am a slow typist so the length of time it took me to type out that explanation is probably 3 times longer than it actually took to do it in my head. (it took me 2 minutes roughly to type it out)
Mind you I had those squares drilled into me for 3 months straight. (I was a little slower than the other kids. After enough practice he could say for example 13² then in two seconds I would say 169.) It is tough but if you work hard it becomes second nature like breathing.
D =16
@5:24 you said "so 2 to the fourth power says take 2 and multiply it by itself 4 times" So 2 to the first power, you would take 2 and multiply it by itself ONE time which means 2 time 2 = 2? I don't think so. 2 to the fourth power equals 16 but you multiply 2 by itself three time 2 times 2 = 4 and 4 times 2 = 8 and 8 times 2 = 16. I think what you meant is you take four 2s and multiply them together, 2x2x2x2=16. I sure hope I never misspeak.
I was waiting for the "other thing" at the end (substracting exponents when you divide identical bases) and you came up with the goods. 👍
The correct answer is D. You turn 4^3 into s^6. When you subtract 2^5 from 2^6 you get 2^5 because 2^7 = 2(2^5) or 2^5 + 2^5. That leaves 2^5/2. Since 2^5 = 2(2^4) the answer is 2^4.
I absolutely love these math games. I am 72 and you make me feel young. I understand this stuff! Thanks
Congratulaciones 😊
Stating the answer as 2^4 is inappropriate. Most teachers insist on resolution (simplification) of answers, unless there is a specific reason for doing otherwise. Most teachers assigning this problem would consider the correct answer to be 16 and most students would try to simplify the result to its most basic form, i.e. 16. Many teachers would give full credit for 16 and would take off points for 2^4.
32/2 = 16 ( d ) 2 to the power of 4
Experiment and try different solutions. It's how you get comfortable. I preferred the following after several runs and fun:
We know 4^3 = 2^6
Splitting the numerator and writing 2 as 2^1: 2^6 / 2^1 - 2^5 /2^1
Wanting to keep them as exponentials and find a common factor in both LHs and RHS
2^(6-1) - 2^(5-1) ==> 2^5 - 2^4 ==> 2^4(2^1) - 2^4.
Factoring out 2^4
2^4(2^1 - 1) (easy to see (2^1 - 1) is 1)
2^4(1)
JOhn,
Shalom
Is this a factorizing problem ?
{2to the third times 2 to the third} - {2 to the third times 2square} over 2. { 8 times 8} minus{8 times 4} over 2. {64-32} divided by 2
Solved in my head in 10 seconds.
At the 5:30 mark he incorrectly says that 2 to the 4th is multiplying 2 by itself 4 times. In fact, it is multiplying 2 by itself 3 times. Easy mistake to make verbally bc you line up 4 2’s in a row, but the act of multiplication only takes place 3 times.
d = 16
Loving it! If you could only see me now Mrs Feilding( special math teacher)!!!!
These are the small things that I don't consider. Probably got loads of problems wrong because of technicalities like this. Thanks 😊
God is in the details.
Saw it immediately in my head. Then checked it the long way
The problem is easiest to solve if the 4 ^ 3 is first transformed to power of 2 Ex; ((2 ^ 2) ^ 3 = 2 ^ 6.
Then divide both terms of the numerator by 2 (in the denominator). This will have the effect of greatly simplifying the problem by eliminating the denominator from (2 ^ 6) - (2 ^ 5) / 2 to (2 ^ 5) - (2 ^ 4).
Then factor out the common multiple (2 ^ 4) from each term to give (2 ^ 4) * ( 2 -1 ).
The perform the subtraction that will yield the final answer of (2 ^ 4) * 1 = (2 ^ 4)
Few steps reduces the chance of error and cuts the solution time during timed exams.
Its even easier (64-32)/2=16
@@Ed19601lol 😊 agreed
I know how to do this just wondering how John is going to teach it .There are a few ways to do this .I have been doing my own rebuilding I have an old High School Algebra 2 book from 1976 and my old College Algebra book which I found Yup I am past the Algebra 2 level umm Trig pre- calculus and Calculus …well that is a different story lol…I had to retrain myself to do these problems I just had to do problems I could do them but not fast enough as I did more my speed pick up I am 66
D 16
Thanks
Simple things made difficult
I remember the PEMDAS and it works well
this was a very easy math problem - no paper needed - you should be able to do this in your head - it's hardly rocket science :p
Exactly
Well designed multiple choice answer. ;)
Thanks for an easy one to start my daily math lessons. :)
I love those base 2 challenges 😊
(some assembly language skills help)
16 is right answer
Une mendoj qe ushtrimet matematikore duhet te ndshen. Te cilesohen ushtrimet qe jane per shkollat fillore dhe 9 vjeçare dhe ato per shkollat e mesme dhe ato ne universitet qe vazhdojne degen e mesuesise per matematike.
Ky eshte si shume te tjera , ushtrim i thjeshte mstematike.
Ndjese per kete vrejtje.
Falemnderit per mirekuptimin
(4 to the third minus 2 to the fifth) = (2 to the sixth minus 2 to the fifth) = (2(2 to the fifth)minus 2 to the fifth) = 2 to the fifth and then that is divided by 2 resulting in 2 to the fourth.
4^3 = 4 * 4 * 4 = 64. 2^5 = 32. So, 64 - 32 = 32 and 32 divided by 2 gives 16 or 2^4. The answer is d.
Is 16 without writing unless is required to show a method.
64 - 32 whole divided by 2 gives 16 ( 2^4)
D?
64-32=32/2=16=2 to the power4
Why didn't you finish? What is 2 to the 4th power?
@@quabledistocficklepo35972 to the 4th power is the answer, he was done.
@@quabledistocficklepo3597 2 to tje 4 power is =16
@@areeratasudhasirikul952 Right!
I don't agree. Once you get to 16, there's no reason for going further.
16 ans
I'm coming up with the answer of 16 which would be 2 to the fourth power, so I have to go with d as the answer.
lol, I had it up to 2 to the 4th . . . its been a long time. THANKS! :)
Thank you
d). 2 to the 4th power
Is what?
@@quabledistocficklepo3597
The multiple choice answer to the equation in the video.
@@michaeljozwiak25 To me, the answer is 16. To me, "answer" mean the end, not the steps on the way there..
@@quabledistocficklepo3597
The answer has to be one of those stated, 16 isn't a valid answer option but 2⁴ is.
Making something very simple very complicated. I always thought that the idea was to make difficult things simple. Perhaps I'm too simple.
This "complicated" method can make solving a problem such as (4^103 - 2^205) / (2^200) much simpler than the "simple" brute force approach. [Answer: 2^5]
Snob!
@@ralphmccawley1554 why does that make me a snob?
4^3 = 2^6
2^6 = 2*2^5
2*2^5 - 2^5 = 2^5
2^5 / 2 = 2^4
2^4 = 16
d) is the correct answer;
4 cubed is 64. 2 to the power of 5 is 32. 64 Minus 32 is 32. Then 32 divided by 2 is 16. So the correct answer is (d ) which is also 16.
I got it from watching your videos
I never would have been able to do that before
I would have seen that and said heck no 😂
Thanks math man 👍❤️😁💪😎👋
4^3=64, 2^5=32.
64-32=32.
32/2=16.
16=2^4.
No calculator and
certainly not 22 minutes.
D=16
On your initial "blackboard". you have written four to the third minus two to the fifth, divided by two. Four to the third is 74. 2 to the fifth is 32. 74 minus 32 is 42. 42 divided by 2 is 21.
Then you start showing us how to solve it using exactly the same method I used, simplifying the numerator, dividing by the denominator. Then you mysteriously change the two to the fifth to two to the fourth, and solve a completely different problem. No wonder you claim "Many will get it wrong". Well, of course we will, when you don't solve the same problem you pose.
But 4to the third power is 64 not 74
Now... imagine a real world problem that would somehow require this calculation.
Do you mean this actual calculation? As in, subtracting 2⁵ from 4³ and then dividing the result by 2?
It would be pretty bizarre to say to mathematics teachers that before they could use anything as an example like this, they had to be able to cite a case of exactly that calculation being used by someone in a real world problem.
There's one immediately obvious practical problem with that. Those of us who do calculations in the real world do not copy them all into some sort of global encyclopedia for teachers to refer to, so what you're asking isn't possible anyway. But quite apart from that, what a silly idea it is.
Perhaps you meant to say, show me a real world problem that requires the ability to understand mathematical notation and the ability to manipulate expressions and the ability to have insight into problem solving. But of course that would be a very silly question too. Who doesn't benefit from a bit of mathematical ability and problem solving ability?
Why make it easy if you can make it complicated?
(4^3 - 2^5) / 2 =
(2^6) / 2 - (2^5) / 2 =
(2^4 x 2 x 2) / 2 - (2^4 x 2) / 2 =
2^4 x 2 - 2^4 x 1 =
2^4
16i.e2^4
Saying "four to the third" sounds like there is a fractional exponent, AKA the cube root of 4.
(2 power 6- 2 power 5)/2=2 power 5(2-1)/2=2 power 5/2= 16 ,So answer is (d)
Which is d.
Very helpful content.
A request. Is it possible to do an annual "wrong" answer Olympics. Same problems but you must correctly arrive at the wrong answer. 😂
Got this . . . I got 16 and figured the fraction. I thought I was seeing a trick.
There may be a ''quick way'' but let's just do it ..
(4^3 - 2^5) / 2
(64 - 32) / 2
32 / 2 which is 16 which is 2^4
but the original equation did not have any parentheses. So although you stumbled on the right answer you voided PEMDAS. To maintain the integrity of PEMDAS, I believe it is move correct to simplify the equation to 4^3/2 - 2^5/2.
@@reach60532 .. and if I'd written 4^3 - 2^5 / 2
someone like you would have said ''PEDMAS .. divide first''.
@@reach60532there are assumed parenthesis because it's all over 2. To express that in serialized text format the parenthesis simply need to be made explicit, because there is no way to convey "all this over" anything.
@@Spudz76 My point is that this is just one in a series of videos attempting to demystify the order of operations. To suddenly introduce magical parentheses without a word of explanation is a huge disservice to the students watching these videos. Either explicitly explain where these parenthesis came from or simplify the equation to 4^3/2 - 2^5/2 and maintain the integrity of PEMDAS.
@@reach60532 You can put as many parenthesis around the part that's already first without explanation or confusion, IMO. Your suggestion is actually more confusing.
D I.e. 16 or 2 raised to the power of 4
d) 2 to the fourth
(D) 2^4
This should be dead easy for computer science people. If somebody can't do this don't hire them.
d) 2^4
I got 16 then looked at the answer and my smiled turned to a frown but now I have my smile back 😂😂
Blew 2toexponent 5 =32 not 64oops!
C
Ans.D)16
Why not just 16
I did in my head answer 16.
Well I got 16 and came to find out what I did wrong. DUH 2 to the 4 th = 16 so no, I didn’t think about it still being a power answer since the first thing I did was get rid of the powers. Never said I was a genius.
No, listing the solution as a power was the wrong thing.
I ❤ cross canceling
(64-32):2=16
Very easy, in my head only - # d).
Misleading. If your goal is to teach order of operations you are adding confession by saying you can willy-nilly add parentheses. I believe it would be clearer to evaluate 4^3/2 - 2^5/2 and evaluate thusly. It preserves the integrity of PEMDAS and doesn't confuse the student by suggesting that you can just make stuff up and magically apply it to every situation.
I totally agree. This guy is super long winded and turns every problem in a pretty much useless review of PEMDAS. Not to mention he always starts out problems with multiple choice answers explaining the reasons for the wrong choices instead of explaining the process to calculating the correct answer. If I had to sit through his classes I would probably off myself.
I simply referred to my memorized powers of two (computer programming) and immediately eliminated the exponent parts to normal numbers and then the rest was easy. Except the bit where I had to also convert the answer to a normal number to compare with the normal number I had as the answer, which was double annoying. At least when I see exponents of two I almost just see the normal number instead like X-ray glasses.
Anyone who works with binary instantly knows that 2^6 - 2^5 =
.
.
wait for it
.
.
2^5, which when divided by 2 =
.
.
.
Wait for it.
.
.
.
.
2^4
16 =2^4
2^4
16 same answer for d
Simple: (64-32)/2=16
Don't need a calculator for that
Two power four.
Got this right with the second method, but it sure seems counterintuitive to me that 2^6 - 2^5 = 2^5. My lizard brain wanted to say its 2.
2^(5-1)=2^4
Many will get wrong - most will not care. Most people today would not recognize a slide rule. Most people today have zero basic math skills.
At least he changed his cover screen to not say "98% will get this wrong".
The answer is 2 to the power 4.
Why are you confusing by elaborating so long a very simple problem. Does it require calculator to find out 4^3 and 2^5. 64-32/2 = 32 which is 2^4.
64-32/2 = 48.
D, 2 to the power of 4
Answer: 2 power 4=16
Easy 4×4×4‐ 2×2×2×2×2/2= 64 - 32 /2 = 32/2 = 16= 2×2×2 ×2 = 4×4 my machine will not do the powers