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Elena L [17]
3 years ago
9

Which scenario could be represented by the algebraic expression 365/y?

Mathematics
2 answers:
Valentin [98]3 years ago
6 0
Barb wants to divide the days in a year by a number, option 3.
snow_tiger [21]3 years ago
4 0

Answer: Barb wants to divide the days in a year by a number.

Step-by-step explanation:

The given expression is \frac{365}{y}  [365 divided by y], where y be any arbitrary number.

The given expression has numerator as 365 and denominator as y which can be any number.

We know that , the number of days in one year = 365

Hence, from the given option the right scenario which could be represented by the algebraic expression is "Barb wants to divide the days in a year by a number.

"

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<h3>What is the surface area of a composite figure formed by two right prisms?</h3>

According to the image, we have a <em>composite</em> figure formed by two <em>right</em> prisms. The <em>surface</em> area of this figure is the sum of the areas of its faces, represented by squares and rectangles:

A = 2 · (4 cm) · (5 cm) + 2 · (2 cm) · (4 cm) + (2 cm) · (5 cm) + (3 cm) · (5 cm) + (5 cm)² + 4 · (3 cm) · (5 cm)

A = 166 cm²

By definition of <em>surface</em> area and the <em>area</em> formulae for squares and rectangles, the <em>surface</em> area of the <em>composite</em> figure is equal to 166 square centimeters.

To learn more on surface areas: brainly.com/question/2835293

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Preview Activity 3.5.1. A spherical balloon is being inflated at a constant rate of 20 cubic inches per second. How fast is the
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The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

Because \frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi } the radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Let r be the radius, d the diameter, and V the volume of the spherical balloon.

We know \frac{dV}{dt}=20 \:{in^3/s} and we want to find \frac{dr}{dt}

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\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=\frac{20}{4\pi(6)^2 } =\frac{5}{36\pi }\approx 0.044

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

When d = 16, r = 8 and \frac{dr}{dt} is:

\frac{dr}{dt}=\frac{20}{4\pi(8)^2}=\frac{5}{64\pi }\approx 0.025

The radius is increasing at a rate of approximately 0.025 in/s when the diameter is 16 inches.

Because \frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi } the radius is changing more rapidly when the diameter is 12 inches.

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