Yo are absolutely correct. Also, the other answers cannot possibly be correct.

@@nyneeveanya8861 yep. This works because the root of a number equals the product of the roots of it's factors. So solving using root3(8) * root3(10) works as well obtaining all the prime factors (root3(2) * root3(2) * root3(2) * root3(2) * root3(5)) and is quicker. But yeah, best to read the problem a couple of times in case he wants all the factors. :-) Eric Had a track coach who also taught math. After running enough 400 meters to puke, he'd start asking us about that night's homework! Had some Voodoo Mojo working, because we learned his math and improved our running. And learned how to hurl for distance and accuracy. 😜

😊 3:57 3:58 😊

Nicely reasoned and splained😊❤

I GET IT! You seldom have to take cube root to: (-1-) change TV channel, although there's likely some equation that predicts the coverage the remote has. (-2-) To set the temperature of the thermostat, although the predictive capability of modern thermostats might need cube roots. (-3-) The automobile?? With over 30 different onboard computers, who knows whether a car need cube root. IN FACT... The volume (V) of a CUBE is side×side×side. Therefore, given that volume = X the side dimension is: X = Cuberoot(V)

Greetings. There are many applications for which the cube root would be required, engineering, mechanical and electrical, construction, and medicine.

HERE GOES -- a previous problem about --> a cone shaped glass vs height of fluid of in glass. The proof to be achieved is that glass filled to 80% height contains 50% of volume.. The volume of cone is: V = (1/3)(pi)(r^2)(h) By similar triangles the h and r of the cone are related such that: h.1/r.1 = h.2/r.2 Restated: If h is 80% then r is 80% Incorporating into equation: V.full = (1/3)(pi)(r)(h) V.50%= (1/3)(pi)(xr^2)(x.h) = (1/3)(pi)(x^3)(r)(h) Since V.full = 2•V.50% (V.50%/V.full) = 0.5 Then 0.5= (x^3•pi•r^2•h)/(pi•r^2•h) 0.5= x^3 SO..... x = cube-rt(0.5) = 0.7937 =~ 80%

For example, the equation for Kepler's Third Law is P² = a³, so the period of a planet's orbit (P) squared is equal to the length of the semi-major axis of the orbit (a) cubed.

How about our two great 400M Hurdlers, Sidney Mac and Anna Cockrell? Another WR for Sidney, and Cockrell was less than a second behind Sidney. Gabby Thomas blazed the curve and scorched the straight for gold. Rai Benjamin took his first gold it in the 400m hurdles as he raced past the Swedish WR holder missing a new WR of his own by just 0.30 seconds. And in an effort I've never seen, Quincy, ran a poor 2nd curve to trail 5 runners with 50 meters to go. Somehow, tight, with no form left, he passed five good Olympians and just leaned enough for the gold. I've never seen someone so spent, absolutely command his body to do what seemed absolutely impossible. And, as icing, his time was 3rd fastest in history: 43.29 seconds. I ran the 400m in HS and College, and know why they call this race the Beast. It doesn't matter how much stronger, faster and smarter you get, the point is to sprint faster and longer than humans we're really made to run. The training can help estimate when you're going to hit the "wall," but it's going to find everyone. But every once in awhile, it hits at the wire and you set a new PR. For once, you've tamed the Beast. Except for Sidney, she must be an Alien😊❤ As to the dimension question: We must calculate stuff in our universe with our limited senses and brains. As we experience only 3 spatial dimensions and 1 temporal one (time), we use these. Of course, we could be entirely wrong. But until we develop more senses and brain parts, and they way many folks behave, we might not be around long enough to see that‼️ BTW, theoretical physicists developed a model the dimensions of our Universe in the late 90s that uses 11 dimensions. Most are aren't detectable, so are just a mental exercise. I can see the need to stretch our minds, as that's how we stumble into new realities, but if you can't see or measure something with an experiment, it's mere speculation. This was another attempt to find a Grand Unification Theory or Gut, which unites our two greatest theories which have explained the mechanics of both large and small objects in our universe, except for some aspects of gravitation: Quantum Mechanics and General Relativity. Both explain and predict behaviors almost perfectly and have been confirmed experimentally for 100 years. But there exists a hole in each, which makes them each partial theories. This means that there is an unexplained Hidden Reality. When both theories attempt to describe the nature of black holes, which we can see to their event horizons, beyond which light photons can't escape the gravitational warpage of spacetime - often called a gravity well, both theories spit out odd but similar results. Each theory calculates the radius as an infinity or series of infinities. While useful in limits, infinite results on finite systems are not very helpful. We can see the affects of black holes on stars and such around them, and they appears to have finite diameters. The missing parts seem to involve a deeper understanding of a still undetected particle predicted by Einstein, which also should be part of the Standard Model of Physics for a Quantum Theory of Gravity. This missing criter, is a Boson, or force carrying particle, and probably has no mass. A more familiar Boson is the massless Photon, which is responsible for carrying the electromagnetic force or light. We see objects in our narrow wavelength/frequency band when photons contact the Red, Green, and Blue cones in our retinas. The colors we see are created by our occipital lobes at the back of our skulls. Red, Green and Blue colors can be mixed to allow us to see billions of colors. You've probably heard of RGB with respect to certain monitors. They use circuitry to simulate the color mixing our brains do. Too bad we don't have Raptor vision of our Eagles and other birds of prey. Not only can the see close and far (eg a hare running from half a mile above and a mile behind), but they see a wider band of colors. They can see into the near ultraviolet and into the other side of the infrared. One last bit of info. The JWST, looks in several bands, but primarily in the infrared, and deep infrared. We experience this as thermal heat. But it has a longer wavelenth than visible, ultraviolet, xRays and Gamma rays. This longer wavelength can pass through most seller gas and dust to see light within and behind it. Most of the light from the early universe has cooled over billions of years and is mostly in infrared or microwave wavelengths. To allow the 21-foot gold plated mirrors to see this light, the infrared from earth and sun are blocked by shields. This allowed the temperature of JWST to cool about 2°C, the ambient temp of interstellar space. And boy have we seen stuff we didn't expect. Early galaxies more defined, which more mature stars. Clearly, the universe cooled enough for matter like hydrogen and helium to coelesce and stars to form, ignite and spread their nuclear light into the cosmos, faster than previously thought. Eric

I like this way of thinking also.

That's how I solved it.

You tempt me to get out my slide rule and see if I can remember how to do it on that. But the slide is stuck. First I'd have to figure out how to restore it.

@@rickhole That sounds like a good project. Interestingly, I see the cube root of 80 can also be expressed as 20*cube root of 0.01. 2*cuberoot(10) = 4*cuberoot(1.25) = 20*cuberoot(0.01). 0.01 is 1 * 1/100, so you can move the "1" outside the radical as its cube root, which is one - so you're stuck at 20*cuberoot(0.01). You could also say that equals 20*cuberoot(64*.01/64). That means the cube root of 80 =80 * cuberoot(.01/64). I think at this point I'll leave the rabbit hole to someone else! 🙂

Dang, this is interesting. 80 * cuberoot(.01/64) is food for thought. .01/64 is 4^-3 * .01. 4^-3 can move outside the radical by dividing instead of multiplying. 80*cuberoot(4^-3 * .01) becomes 20 * cuberoot(.01), which was what I was at a few steps ago. But what about solving cuberoot(16*5)? When you move 2^3 outside of a cube root radical, you're multiplying the radical by 2^(3/3). The numerator was from what it was inside the radical, the denominator is the power you're reducing it by. So, cuberoot (16*5) becomes cuberoot(4^2*5) which becomes 4^(2/3) times cuberoot of 5. In that term, 4^(2/3), the two is from the "2" power of 4 inside the radical and the denominator is from the "3" on the radical itself. Intriguing that 4^(2/3) times 5^(1/3) is the cube root of 80. It seems like there should be something to be done with that 2/3 and 1/3.

Hi guys, Need to check your log rules. For example, given the Exponential equation below, Log 3 is the exponent we raise Base 10 to equal 3. 10^x = 3 ✅ Exp Form log 10^x = log 3 x log 10 = log 3 x = log 3 log 10 x = log 3 = 0.4771 ✅ log form Also, the power rule of logs requires you to move the power, as it is (integer, variable, rational number), in front of the log. And you must take the logs of *both* sides when you use it. This question asked us to simplify an expression. It really wasn't an equation, although you could assume John's question mark was a variable. 😅 Simplification: 80^(1/3)❓ ((2^4) * (5))^(1/3) ((2^3))*(2)*(5))^(1/3) 2*(2*5)^(1/3) 2*(10)^(1/3) == 4.3089✅

@@BluesChoker01 Did I get anything wrong?

When dealing with timed/multiply choice question eliminate incorrect answers and move on. A,B and C are all to small so the answer must be D. What is the next question!

@@walter71342 That's what I taught my grandson facing a multiple choice college entry exam. I had taken so many of those multiple choice, timed tests that doing as you say is the first choice, not the last.

a) 8 is too big. Has to be between 4 and 5. b) and c) are too small.

Now, that's how Mr. KZfaq MethMan would do it😂😂❤️

B and C are too small. A is too big.

@@gavindeane3670 you are right, my mistake!

I got it too, but at least when I was in the 4th grade long long ago, we had no concept of cube roots, let alone how to simplify them

@@jald910 I was just cappin.