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A farmer put 28 fish in a new pond. The population of fish increased by a factor of 3 every year. Which of the following expressions shows the population of the fish after 2 years?

Mathematics

ScanSolve AI Answer

Answer

B

Step-by-step solution

16 steps

Here are the solutions to your questions:

:
The initial population of fish is 28. The population increases by a factor of 3 every year. We need to find the expression for the population after 2 years.

  1. Understand the growth.

    After 1 year, the population will be 28×328 \times 3.
    After 2 years, the population will be (28×3)×3(28 \times 3) \times 3.

  2. Write the expression.

    Population after 2 years=28×32\text{Population after 2 years} = 28 \times 3^2

    The correct option is B.
    The expression is 28×32\boxed{28 \times 3^2}.

    :
    The initial temperature of the vaccine fridge is −4.5∘C-4.5^\circ \text{C}. The temperature rises by 534∘C5 \frac{3}{4}^\circ \text{C}.

  3. Convert the mixed fraction to a decimal.

    534=5+34=5+0.75=5.75∘C5 \frac{3}{4} = 5 + \frac{3}{4} = 5 + 0.75 = 5.75^\circ \text{C}

  4. Calculate the new temperature.

    New temperature=−4.5∘C+5.75∘C\text{New temperature} = -4.5^\circ \text{C} + 5.75^\circ \text{C} New temperature=1.25∘C\text{New temperature} = 1.25^\circ \text{C}

    The correct option is A.
    The new temperature in the fridge is 1.25∘C\boxed{1.25^\circ \text{C}}.

    :
    The tabletop is 1.5 m long and 0.9 m wide. A border xx cm wide is added along the edges. We need the area of the part of the table without the border.
    Given the options, it is implied that xx is in meters, not centimeters, for the coefficients to match. Let's assume xx is the border width in meters.

  5. Determine the dimensions of the table without the border.

    Original length L=1.5 mL = 1.5 \text{ m}
    Original width W=0.9 mW = 0.9 \text{ m}
    If the border is xx meters wide on all four sides, the length and width of the inner part will be reduced by 2x2x.
    New length L′=L−2x=1.5−2xL' = L - 2x = 1.5 - 2x
    New width W′=W−2x=0.9−2xW' = W - 2x = 0.9 - 2x

  6. Calculate the area of the part without the border.

    Area=L′×W′=(1.5−2x)(0.9−2x)\text{Area} = L' \times W' = (1.5 - 2x)(0.9 - 2x) Area=1.5×0.9−1.5×2x−2x×0.9+(2x)2\text{Area} = 1.5 \times 0.9 - 1.5 \times 2x - 2x \times 0.9 + (2x)^2 Area=1.35−3x−1.8x+4x2\text{Area} = 1.35 - 3x - 1.8x + 4x^2 Area=1.35−4.8x+4x2\text{Area} = 1.35 - 4.8x + 4x^2

    The correct option is D.
    The expression for the area of the part of the table without the border is 1.35−4.8x+4x2\boxed{1.35 - 4.8x + 4x^2}.

    :
    Let the width of the flower garden be WW meters.
    The length LL is 2 m less than twice the width, so L=2W−2L = 2W - 2.
    The area of the garden is 60 m260 \text{ m}^2.

  7. Set up the equation for the area.

    Area=L×W\text{Area} = L \times W 60=(2W−2)W60 = (2W - 2)W 60=2W2−2W60 = 2W^2 - 2W

  8. Rearrange the equation into a standard quadratic form.

    2W2−2W−60=02W^2 - 2W - 60 = 0

    Divide by 2:

    W2−W−30=0W^2 - W - 30 = 0

  9. Solve the quadratic equation for WW.

    Factor the quadratic:

    (W−6)(W+5)=0(W - 6)(W + 5) = 0

    This gives two possible values for WW: W=6W = 6 or W=−5W = -5.
    Since width cannot be negative, W=6 mW = 6 \text{ m}.

  10. Calculate the length of the flower garden.

    L=2W−2L = 2W - 2 L=2(6)−2L = 2(6) - 2 L=12−2L = 12 - 2 L=10 mL = 10 \text{ m}

    The correct option is D.
    The length of the flower garden is 10 m\boxed{10 \text{ m}}.

    :
    Let the age of the heifer be xx years.
    The cow is 4 years 8 months older than the heifer.

  11. Convert 8 months to years.

    8 months=812 years=23 years8 \text{ months} = \frac{8}{12} \text{ years} = \frac{2}{3} \text{ years}

  12. Express the cow's age in terms of xx.

    Cow’s age=x+4+23=x+123+23=x+143 years\text{Cow's age} = x + 4 + \frac{2}{3} = x + \frac{12}{3} + \frac{2}{3} = x + \frac{14}{3} \text{ years}

  13. Set up the equation for the product of their ages.

    The product of their ages is 8 years.

    x(x+143)=8x \left(x + \frac{14}{3}\right) = 8

  14. Expand and simplify the equation.

    x2+143x=8x^2 + \frac{14}{3}x = 8

    Multiply the entire equation by 3 to eliminate the fraction:

    3x2+14x=243x^2 + 14x = 24

    Rearrange into standard quadratic form:

    3x2+14x−24=03x^2 + 14x - 24 = 0

    The correct option is C.
    The equation that represents the information is 3x2+14x−24=0\boxed{3x^2 + 14x - 24 = 0}.

    :
    Kawira stands 1.4 m in front of a plane mirror.

  15. Understand image formation in a plane mirror.

    For a plane mirror, the image formed is virtual, erect, and located as far behind the mirror as the object is in front of it.

  16. Calculate the distance of the image from Kawira.

    Distance of Kawira from mirror = 1.4 m1.4 \text{ m}.
    Distance of image from mirror = 1.4 m1.4 \text{ m}.
    The total distance from Kawira to her image is the sum of these two distances.

    Distance from Kawira to image=1.4 m+1.4 m\text{Distance from Kawira to image} = 1.4 \text{ m} + 1.4 \text{ m} Distance from Kawira to image=2.8 m\text{Distance from Kawira to image} = 2.8 \text{ m}

    The correct option is C.
    The image is 2.8 m\boxed{2.8 \text{ m}} from Kawira.

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