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What is the quotient when the square of 4x2y2 is divided by the square of -2xy2?

Mathematics

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Step-by-step solution

23 steps
  1. Apply the zero exponent rule.

    Any non-zero base raised to the power of zero is 1.

    (2x3y2)0=1(2x^3y^2)^0 = 1

    The expression 2x3y22x^3y^2 is assumed to be non-zero.

    The final answer is 1\boxed{\text{1}}.
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  2. Apply the negative exponent rule.

    A term with a negative exponent can be rewritten as its reciprocal with a positive exponent.

    2k−1=2×1k12k^{-1} = 2 \times \frac{1}{k^1}
  3. Simplify the expression.

    2×1k=2k2 \times \frac{1}{k} = \frac{2}{k}

    The final answer is 2k\boxed{\frac{2}{k}}.
    What's next?


  4. Multiply the numerators and denominators.

    [3x12y2][4y28x2]=3x1⋅4y22y2⋅8x2=12x1y216x2y2\left[\frac{3x^1}{2y^2}\right] \left[\frac{4y^2}{8x^2}\right] = \frac{3x^1 \cdot 4y^2}{2y^2 \cdot 8x^2} = \frac{12x^1y^2}{16x^2y^2}
  5. Simplify the numerical coefficients and variables using exponent rules.

    Divide 12 by 16: 1216=34\frac{12}{16} = \frac{3}{4}.
    For xx: x1x2=x1−2=x−1\frac{x^1}{x^2} = x^{1-2} = x^{-1}.
    For yy: y2y2=y2−2=y0=1\frac{y^2}{y^2} = y^{2-2} = y^0 = 1.

    34x−1y0\frac{3}{4} x^{-1} y^0
  6. Remove zero and negative exponents.

    y0=1y^0 = 1 and x−1=1xx^{-1} = \frac{1}{x}.

    34⋅1x⋅1=34x\frac{3}{4} \cdot \frac{1}{x} \cdot 1 = \frac{3}{4x}

    The final answer is 34x\boxed{\frac{3}{4x}}.
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  7. Move the decimal point to the right until there is one non-zero digit to its left.

    The number is 0.000000427. Moving the decimal point past 4 gives 4.27.

  8. Count the number of places the decimal point was moved.

    The decimal point moved 7 places to the right.

  9. Write the number in scientific notation.

    Since the decimal point was moved to the right, the exponent of 10 will be negative.

    4.27×10−74.27 \times 10^{-7}

    The final answer is 4.27×10−7\boxed{4.27 \times 10^{-7}}.
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  10. Identify the total amount of bacteria and the number of days.

    Total bacteria =2.5×108= 2.5 \times 10^8.
    Number of days =5= 5.

  11. Calculate the average number of bacteria per day.

    Average per day =Total bacteriaNumber of days= \frac{\text{Total bacteria}}{\text{Number of days}}

    2.5×1085\frac{2.5 \times 10^8}{5}
  12. Perform the division.

    Divide 2.5 by 5: 2.5÷5=0.52.5 \div 5 = 0.5.

    0.5×1080.5 \times 10^8
  13. Express the answer in scientific notation.

    The coefficient 0.5 is not between 1 and 10. Move the decimal point one place to the right to get 5.0. Since the decimal point was moved one place to the right, decrease the exponent of 10 by 1.

    5.0×108−1=5.0×1075.0 \times 10^{8-1} = 5.0 \times 10^7

    The final answer is 5.0×107\boxed{5.0 \times 10^7}.
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  14. Simplify the expression inside the brackets using the quotient rule for exponents (am/an=am−na^m/a^n = a^{m-n}).

    2x1y3x2y1=2x1−2y3−1=2x−1y2\frac{2x^1y^3}{x^2y^1} = 2x^{1-2}y^{3-1} = 2x^{-1}y^2
  15. Apply the outer exponent to each term inside the brackets using the power rule for exponents ((ab)m=ambm(ab)^m = a^m b^m and (am)n=amn(a^m)^n = a^{mn}).

    (2x−1y2)2=22(x−1)2(y2)2=4x−2y4(2x^{-1}y^2)^2 = 2^2 (x^{-1})^2 (y^2)^2 = 4x^{-2}y^4
  16. Remove negative exponents.

    x−2=1x2x^{-2} = \frac{1}{x^2}.

    4⋅1x2⋅y4=4y4x24 \cdot \frac{1}{x^2} \cdot y^4 = \frac{4y^4}{x^2}

    The final answer is 4y4x2\boxed{\frac{4y^4}{x^2}}.
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  17. Simplify the expression inside the brackets using the quotient rule for exponents (am/an=am−na^m/a^n = a^{m-n}).

    x1y2x3y1=x1−3y2−1=x−2y1\frac{x^1y^2}{x^3y^1} = x^{1-3}y^{2-1} = x^{-2}y^1
  18. Apply the outer exponent to each term inside the brackets using the power rule for exponents ((ab)m=ambm(ab)^m = a^m b^m and (am)n=amn(a^m)^n = a^{mn}).

    (x−2y1)2=(x−2)2(y1)2=x−4y2(x^{-2}y^1)^2 = (x^{-2})^2 (y^1)^2 = x^{-4}y^2
  19. Remove negative exponents.

    x−4=1x4x^{-4} = \frac{1}{x^4}.

    1x4⋅y2=y2x4\frac{1}{x^4} \cdot y^2 = \frac{y^2}{x^4}

    The final answer is y2x4\boxed{\frac{y^2}{x^4}}.
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  20. Find the square of 4x3y24x^3y^2.

    (4x3y2)2=42(x3)2(y2)2=16x3×2y2×2=16x6y4(4x^3y^2)^2 = 4^2 (x^3)^2 (y^2)^2 = 16x^{3 \times 2}y^{2 \times 2} = 16x^6y^4
  21. Find the square of −2xy2-2xy^2.

    (−2xy2)2=(−2)2(x1)2(y2)2=4x1×2y2×2=4x2y4(-2xy^2)^2 = (-2)^2 (x^1)^2 (y^2)^2 = 4x^{1 \times 2}y^{2 \times 2} = 4x^2y^4
  22. Divide the result from Step 1 by the result from Step 2.

    16x6y44x2y4\frac{16x^6y^4}{4x^2y^4}
  23. Simplify the expression.

    Divide the coefficients: 16÷4=416 \div 4 = 4.
    For xx: x6x2=x6−2=x4\frac{x^6}{x^2} = x^{6-2} = x^4.
    For yy: y4y4=y4−4=y0=1\frac{y^4}{y^4} = y^{4-4} = y^0 = 1.

    4x4y0=4x4⋅1=4x44x^4y^0 = 4x^4 \cdot 1 = 4x^4

    The final answer is 4x4\boxed{4x^4}.

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