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3 years ago

5

If a figure is dilated by a factor of 1.5, the area of the figure will change by a factor of

2 answers:

Pachacha [2.7K]3 years ago

6 0

Answer:

The figure will change by a factor of 2.25.

Step-by-step explanation:

When the lengths (1-dimensional) change by a factor 1.5 (the definition of <em>dilating </em>a figure), the area (2-dimensional) will change by a factor 1.5 to the power of the quotient of the values of the dimensions, which is 2/1 = 2. So the area will change by a factor of 1.5^2 = 2.25.

babunello [35]3 years ago

4 0

The area of the figure will change by a factor of 2.25

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Item 16 A sphere has a radius of 8 centimeters. A second sphere has a radius of 2 centimeters. What is the difference of the vol

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Answer:

$672\pi \text{ cm}^3$ .

Step-by-step explanation:

We have been given that a sphere has a radius of 8 centimeters. A second sphere has a radius of 2 centimeters. We are asked to find the difference of the volumes of the spheres.

We will use volume formula of sphere to solve our given problem.

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$\text{Difference of volumes}=\frac{4}{3}\pi(\text{8 cm})^3-\frac{4}{3}\pi(\text{2 cm})^3$

$\text{Difference of volumes}=\frac{4}{3}\pi(512)\text{ cm}^3-\frac{4}{3}\pi(8)\text{ cm}^3$

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3 years ago

I don’t know how to answer this?

dalvyx [7]

Answer: a bigger or equal to 20

Explanation:

2a/5 - 2 _> a/4 + 1

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b. $I = 2. 6$ × $10^-3$ A

c. $I = 0. 04$ A

The LED which would glow brightest is LED C with the greatest current and voltage

The LED which would be the most dim is LED B with low voltage and consequently low current.

<h3>How to determine the current</h3>

The formula for finding current

I = V/R

Where v = voltage

R = resistance

A. V = 12V

R = 4. 7 + 15 = 19. 7 kΩ = 19700 Ω in series

$I = \frac{12}{19700}$

$I = 6. 1$ × $10^-4$ A

B. V = 9V

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$I = \frac{12}{4700}$

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It is important to note that the brightness of a bulb depends on both current and voltage depending on whether the bulb it is in parallel or series.

The LED which would glow brightest is LED C with the greatest current and voltage

The LED which would be the most dim is LED B with low voltage and consequently low current.

Learn more about Ohms law here:

brainly.com/question/14296509

#SPJ1

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