Start by decomposing the number inside the root into primes
Then group the terms into cubes if possible
rewrite the root
![\sqrt[3]{80}=\sqrt[3]{10\cdot2^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B80%7D%3D%5Csqrt%5B3%5D%7B10%5Ccdot2%5E3%7D)
then cancel the terms that are cubes and bring them out of the root
![\sqrt[3]{80}=2\sqrt[3]{10}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B80%7D%3D2%5Csqrt%5B3%5D%7B10%7D)
Answer:
The point will move to the left 4 place values.
Step-by-step explanation:
When you add or subtract from the x value, you are moving the point parallel to the x-axis, the amount you need to go. If subtracting, you move left, if adding, you move right.
When you add or subtract from the y value, you are moving the point parallel to the y-axis, the amount you need to go. If subtracting, you move down, if adding, you move up.
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Answer:
100
Step-by-step explanation:
Mixed candy question... Skittles jar... to be filled with Jelly beans.
Let's first calculate the volume of the jar. We'll assume it's a regular cylindrical prism jar, unlike the one on the photo which is narrower on top.
V = π * r² * h = π * (3.5)² * 11.5 = 140.875 π = 442.6 cubic cm
Now, we don't have the precise measurement of a jelly bean, but we know it's roughly 2-3 cubic cm. The precision isn't needed to answer this question, just to have a rough idea... it's no 300 cu cm per jelly bean.
So, let's assume a 3 cu cm per jelly bean (2 cu cm wouldn't the final answer)....
442.6 / 3 = 147.5 jelly beans, approximately.
So, can they fit 100,000? No
Can we fit 10,000 in there? No
Can we fit 100? Yes.
Can we fit 1? Certainly
The most reasonable lower-limit would then be 100.
Answer:
19 4/5
Step-by-step explanation:
