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Anna35 [415]
2 years ago
11

A 3 cm x 3 cm rectangle sits inside a circle with radius of 4 cm.

Mathematics
1 answer:
Ivanshal [37]2 years ago
8 0

Step-by-step explanation:

you did not show us the picture or described what the shaded area is.

I assume it is the area of the circle without the inner square (as the "rectangke" is actually a square, because its sides are equally long).

so, we need to calculate the area of the circle and subtract the area of the square.

the area of the square is 3×3 = 9 cm².

the area of the circle is

pi×r²

with r being the radius.

so,

pi×4² = pi×16 = 16pi

the shaded area is then

16pi - 9 = 41.26548246... cm² ≈ 41.27 cm²

nothing of the answer options seems to be in that area, so maybe my assumption about the shaded area is wrong. or the specified answer options are mistyped. or both.

anyway, without additional information that is the best I can do.

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A function is shown: f(x) = 36x² - 1.
givi [52]

Based on the given function, the equivalent function that best shows the x-intercepts on the graph is f(x) = (6x - 1)(6x + 1)

<h3>What are equivalent functions?</h3>

Equivalent functions are different functions that have equal values when evaluated and compared

<h3>How to determin the equivalent function that best shows the x-intercepts on the graph?</h3>

The function is given as:

f(x) = 36x^2 - 1

Express 1 as 1^2

f(x) = 36x^2 - 1^2

Express 36x^2 as (6x)^2

f(x) = (6x)^2 - 1^2

Apply the difference of two squares.

This is represented as:

(a + b)(a - b) = a^2 - b^2

So, we have the following equation

f(x) = (6x - 1)(6x + 1)

Based on the given function, the equivalent function that best shows the x-intercepts on the graph is f(x) = (6x - 1)(6x + 1)

Read more about equivalent function at:

brainly.com/question/2972832

#SPJ1

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2 years ago
The range is the set of
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Hope this helps  :)

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Suppose this relation holds for n=k, i.e. 3\vert k^3-k. We then hope to show it must also hold for n=k+1.

You have

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We assumed that 3\vert k^3-k, and it's clear that 3\vert 3(k^2+k) because 3(k^2+k) is a multiple of 3. This means the remainder upon divides (k+1)^3-(k+1) must be 0, and therefore the relation holds for n=k+1. This proves the statement.
4 0
3 years ago
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