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Thread: Algebra ... Exponents HW

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    Algebra ... Exponents HW

    Radio signals trvel at a rate of 3 x 10^8 meters per secoond. How many seconds would it take for a radio signal to travel from a satellite to the surface of Earth if the satellite is obiting at a height of 9.6 x 10^6 meters?
    3.2 x 10^2 sceonds
    3.2 x 10^2 seconds
    3.13 x 10^1 seconds
    2.88 x 10^15 seconds

    (y^-5)^-10y^10
    y^-60
    y^60
    y^-150
    y^150

    (2/5n^9)^2
    10n^18
    4/5n
    4/25n^18
    4/10n^18

    (3/5y^4)^-2
    6/10y^16
    10y^16/6
    9/25y^8
    25y^8/9

    A $6,300.00 principle earns 6% interest , compounded anually. After 3 years, what is the balance in the account
    1,360,800
    7,503.40
    7,434
    25,804.80

    A $3,500 principle earns 3% interest , compounded anually. After 20 years, what is the balance in the account
    7,700
    4,713.99
    5,600
    6349.06
    A boat costs $92,000 and depreciates in value by 15% per year. How muuch will the boat be worth after 10 years?
    18,112.45
    78,200
    18,941.98
    69,000



    If you can help Ill rep

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    rachel's Avatar
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    I can edit as I do the other ones.

    Radio signals trvel at a rate of 3 x 10^8 meters per secoond. How many seconds would it take for a radio signal to travel from a satellite to the surface of Earth if the satellite is obiting at a height of 9.6 x 10^6 meters?
    3.2 x 10^2 sceonds
    3.2 x 10^2 seconds
    3.13 x 10^1 seconds
    2.88 x 10^15 seconds

    I'm assuming one of the first two answers is supposed to be 3.2 x 10^(-2), because that would be the correct answer. If it can travel 3 x 10^8 and has to travel a distance of 9.6x10^6, you'd have to take 9.6x10^6 divided by 3 x 10^8. When you divide exponents, you actually subtract their values, so in this case...

    You'd take 9.6 divided by 3, which would give you the first value (3.2). Then, to take 10^6 divided by 10^8, you'd take 6-8, which would give you -2. Thus, your answer is 3.2x10^(-2).

    If it's still confusing to figure out which is divided by which, think of it on simpler terms. If you have to travel 10 feet, and you can travel 2 feet per second, how many seconds does it take you to travel 10 feet? It would take you 5 seconds.

    10 feet = 9.6 x 10^6 meters (the distance it has to travel)
    2 feet per second = 3 x 10^8 meters per second (how quickly it can travel)
    Answer = 5 seconds = 3.2 x 10^(-2) seconds (how long it would take)

    ------------

    (y^-5)^-10y^10
    y^-60
    y^60
    y^-150
    y^150

    Honestly, this one isn't making much sense to me. When you raise a power to a power, you multiply the exponents. Therefore, this should break down:

    (y^-5)^-10y = y^(-5 x -10y) = y^(50y)
    y^(50y)^10 = y^(50y x 10) = y^(500y)

    But that obviously isn't any of your answers, LOL. Am I reading the problem right? If so, someone else maybe have to step in, but yeah.

    -----------------

    (2/5n^9)^2
    10n^18
    4/5n
    4/25n^18
    4/10n^18

    So, now that there are numbers involved, you treat them the same way that you normally would, so for the first part, you'd do (2/5)^2 which would become (4/25)

    But, when you take a power to a power, you multiply their values, so (n^9)^2 = n^(9 x 2) = n^18

    Put the two parts together for (4/25)n^18

    ------------------

    (3/5y^4)^-2
    6/10y^16
    10y^16/6
    9/25y^8
    25y^8/9

    For this one, if you have a negative exponent, you flip the numbers in front. So, instead of trying to solve (3/5)^-2, you'd be solving (5/3)^2, which would give you (25/9).

    For the exponents, you'd take (y^4)-2, which would become y^(4 x -2), or y^-8. But, when you have a negative exponent in the top, you can push it to the bottom of a fraction to make it positive. That would leave you, when you combined the two, with:

    25 in the top over 9y^8 in the bottom, or 25/(9y^8).

    Again, either I'm reading the problem incorrectly or something because, using the rules of exponents, that's what it should be, but that's not one of your options. xD

    --------------

    A $6,300.00 principle earns 6% interest , compounded anually. After 3 years, what is the balance in the account
    1,360,800
    7,503.40
    7,434
    25,804.80

    For these problems, you'll be using the equation:

    A = P (1 + r/n) ^ (nt)

    A = ending amount (what we're looking for). P = starting amount (6300). r = interest rate (6% = 0.06). n = number of "compoundings" per year (so compounded annually = once per year = 1). t = number of years.

    A = 6300 (1+(0.06/1)) ^ (1 x 3) = 6300 (1+0.06) ^ 3 = 6300 (1.06)^3 = 6300(1.191016) = $7503.40

    ----------------

    A $3,500 principle earns 3% interest , compounded anually. After 20 years, what is the balance in the account
    7,700
    4,713.99
    5,600
    6349.06

    Same equation as the last problem:

    A = 3500 (1+(0.03)/1)^(1x20) = 3500 (1+0.03)^(20) = 3500 (1.03)^20 = 3500 (1.806111234669) = $6321.39

    Although this isn't any of the answers you listed, sometimes where you round can make a big difference. I'd guess that the 6349.06 is what your teacher is looking for, depending on where you round.

    --------

    A boat costs $92,000 and depreciates in value by 15% per year. How muuch will the boat be worth after 10 years?
    18,112.45
    78,200
    18,941.98
    69,000

    For this one, you're using the same equation, but instead of adding 1+r/n, you'll be doing 1-r/n, since the value is depreciating.

    A = 92,000 (1-0.15/1)^(1 x 10) = 92,000 (1-0.15)^10 = 92,000(0.85)^10 = 92000 x (0.1968744043) = $18,112.45

    --------------

    Hopefully someone else can come here in and clear up the ones that are confusing, but there ya go. ^_^
    Last edited by rachel; 02-04-2014 at 10:50 PM.

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    Thank You C: and yes it was supposed to be (-2) exponent

    ---------- Post added at 12:18 AM ---------- Previous post was at 12:08 AM ----------

    Really all I need help on now is the second one and the last one @rachel

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    Quote Originally Posted by Sk8ter View Post
    Thank You C: and yes it was supposed to be (-2) exponent

    ---------- Post added at 12:18 AM ---------- Previous post was at 12:08 AM ----------

    Really all I need help on now is the second one and the last one @rachel
    Ooooooops
    My cK didn't update at all, so that's awkward.
    Otherwise I would have skipped all the other mumbo-jumbo, my bad.

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