Answer:
- The diagram bar is attached.
- Addition equation: 
- Multiplication equation: 
- How are the equation related? Each equation shows 3 groups of 7.
Step-by-step explanation:
We know that Jan buys 3 bags of beads and each bag contains 7 beads, then, you can draw the bar diagram shown attached.
Observe that the diagram has 3 blocks (each block represents a bag of bead) and there is a number 7 inside of each block (which is the number of beads contained in a bag).
Therefore:
- Add the numbers inside the blocks in order to get the addition equation that shows the number of beads Jan buys. This is:
- Multiply 3 blocks by 7 in order to get multiplication equation that show the number of beads Jan buys:
The equations are related. Each one shows 3 groups of 7.
7^5. Found by trial and error.
Okay so this is going to be confusing but work from right to left. since you can’t do 0-5, borrow from the one next to the zero on the top, cross out the one and put zero and add a one on top of the zero which makes it 10. now you can do 10-5 to get 5. put the five down below the line. next solve the second column of numbers ( second to the right). since you cannot do 0- 3, like the last one, you have to borrow from the 7 over to the left. take one from there and cross the seven out and put a six. now you have 10-3 which is 7. put the 7 down. now do move to the next column and do 6-2 to get 4. put the four down below the line. move to the next column and do 2-2 to get 0. put the zero down below the line. lastly, we know the answer has to be negative so we are going to do 2-7 to get -5. put -5 down below the line and together to get the answer -502,275. i know that was very confusing but i hope it helped a little bit. i added a picture to clarify a little
Answer:
John is 29, Myles is 17.
Step-by-step explanation:
Make a substitution:
Then the system becomes
![\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bu%7D%7D%7Bu-v%7D%2B%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bu%7D%7D%7Bu%2Bv%7D%3D%5Cdfrac%7B81%7D%7B182%7D%5C%5C%5C%5C%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bv%7D%7D%7Bu-v%7D-%5Cdfrac%7B2%5Csqrt%5B3%5D%7Bv%7D%7D%7Bu%2Bv%7D%3D%5Cdfrac1%7B182%7D%5Cend%7Bcases%7D)
Simplifying the equations gives
![\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bu%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac%7B81%7D%7B182%7D%5C%5C%5C%5C%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac1%7B182%7D%5Cend%7Bcases%7D)
which is to say,
![\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bu%5E4%7D%7D%7Bu%5E2-v%5E2%7D%3D%5Cdfrac%7B81%5Ctimes4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7Bu%5E2-v%5E2%7D)
![\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81](https://tex.z-dn.net/?f=%5Cimplies%5Csqrt%5B3%5D%7B%5Cleft%28%5Cdfrac%20uv%5Cright%29%5E4%7D%3D81)
Substituting this into the new system gives
![\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1](https://tex.z-dn.net/?f=%5Cdfrac%7B4%5Csqrt%5B3%5D%7Bv%5E4%7D%7D%7B%28%5Cpm27v%29%5E2-v%5E2%7D%3D%5Cdfrac1%7B182%7D%5Cimplies%5Cdfrac1%7Bv%5E2%7D%3D1%5Cimplies%20v%3D%5Cpm1)
Then
(meaning two solutions are (7, 13) and (-7, -13))
