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kindricko

kindricko

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kindricko
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  • First year income: .04(U) = .01(400000 - U) .04U = 4000 - .01U .05U = 4000 U = 80000 S = 400000 - 80000 = 320000 Y = .04(80000) + .01(320000) = 3200 + 3200 = 6400 This will remain constant, year to year, unless the interest is added back …
    in algebra problem? Comment by kindricko June 2021
  • First year income: .04(U) = .01(400000 - U) .04U = 4000 - .01U .05U = 4000 U = 80000 S = 400000 - 80000 = 320000 Y = .04(80000) + .01(320000) = 3200 + 3200 = 6400 This will remain constant, year to year, unless the interest is added back …
    in algebra problem? Comment by kindricko June 2021
  • First year income: .04(U) = .01(400000 - U) .04U = 4000 - .01U .05U = 4000 U = 80000 S = 400000 - 80000 = 320000 Y = .04(80000) + .01(320000) = 3200 + 3200 = 6400 This will remain constant, year to year, unless the interest is added back …
    in algebra problem? Comment by kindricko May 2021
  • First year income: .04(U) = .01(400000 - U) .04U = 4000 - .01U .05U = 4000 U = 80000 S = 400000 - 80000 = 320000 Y = .04(80000) + .01(320000) = 3200 + 3200 = 6400 This will remain constant, year to year, unless the interest is added back …
    in algebra problem? Comment by kindricko May 2021
  • First year income: .04(U) = .01(400000 - U) .04U = 4000 - .01U .05U = 4000 U = 80000 S = 400000 - 80000 = 320000 Y = .04(80000) + .01(320000) = 3200 + 3200 = 6400 This will remain constant, year to year, unless the interest is added back …
    in algebra problem? Comment by kindricko May 2021
  • First year income: .04(U) = .01(400000 - U) .04U = 4000 - .01U .05U = 4000 U = 80000 S = 400000 - 80000 = 320000 Y = .04(80000) + .01(320000) = 3200 + 3200 = 6400 This will remain constant, year to year, unless the interest is added back …
    in algebra problem? Comment by kindricko May 2021
  • First year income: .04(U) = .01(400000 - U) .04U = 4000 - .01U .05U = 4000 U = 80000 S = 400000 - 80000 = 320000 Y = .04(80000) + .01(320000) = 3200 + 3200 = 6400 This will remain constant, year to year, unless the interest is added back …
    in algebra problem? Comment by kindricko May 2021
  • First year income: .04(U) = .01(400000 - U) .04U = 4000 - .01U .05U = 4000 U = 80000 S = 400000 - 80000 = 320000 Y = .04(80000) + .01(320000) = 3200 + 3200 = 6400 This will remain constant, year to year, unless the interest is added back …
    in algebra problem? Comment by kindricko May 2021
  • First year income: .04(U) = .01(400000 - U) .04U = 4000 - .01U .05U = 4000 U = 80000 S = 400000 - 80000 = 320000 Y = .04(80000) + .01(320000) = 3200 + 3200 = 6400 This will remain constant, year to year, unless the interest is added back …
    in algebra problem? Comment by kindricko May 2021