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andrew11 [14]
4 years ago
7

Zeus Industries bought a computer for $2593. It is expected to depreciate at a rate of 30% per year. What will the value of the

computer be in 3 years?
Mathematics
1 answer:
Natalija [7]4 years ago
8 0
First year:
.3 x 2593 = 777.9

$2593 - $777.90 = $1815.10

Second year:
.3 x 1815.10 = 544.53

$1815.10 - $544.53 = $1270.57

Third year:
.3 x 1270.57 = <span>381.171

$1270.57 - $381.171 = $</span>889.39900

$889.39900 rounded to the nearest hundredth is $889.40
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Answer:

a is x>3

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3 years ago
On a coordinate plane, a circle has a center at (0, 0). Point (3, 0) lies on the circle.
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Answer:

The correct option is;

No, the distance from (0, 0) to (2, √6) is not 3 units

Step-by-step explanation:

The given parameters of the question are as follows;

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Point on the circle (x, y) = (3, 0)

We are required to verify whether point (2, √6) lie on the circle

We note that the radius of the circle is given by the equation of the circle as follows;

Distance \, formula = \sqrt{\left (x_{2}-x_{1}  \right )^{2} + \left (y_{2}-y_{1}  \right )^{2}}

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(3 - 0)² + (0 - 0)² = 3²

Hence r² = 3² and r = 3 units

We check the distance of the point (2, √6) from the center of the circle (0, 0) as follows;

\sqrt{\left (x_{2}-x_{1}  \right )^{2} + \left (y_{2}-y_{1}  \right )^{2}} = Distance

Therefore;

(2 - 0)² + (√6 - 0)² = 2² + √6² = 4 + 6 = 10 = √10²

\sqrt{\left (2-0 \right )^{2} + \left (\sqrt{6} -0 \right )^{2}} = 10

Which gives the distance of the point (2, √6) from the center of the circle (0, 0) = √10

Hence the distance from the circle center (0, 0) to (2, √6) is not √10 which s more than 3 units hence the point  (2, √6), does not lie on the circle.

4 0
3 years ago
Read 2 more answers
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Answer:

Where 0 < x < 3

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The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

f'(x) = \dfrac{(-x + 2)^4}{dx}  = -4 \cdot (-x + 2)^3 = 0

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f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx}  = -12 \cdot (-x + 2)^2

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We have, f(2) = (-2 + 2)⁴ = 0

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Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

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The location of the local maximum is at (0, 16).

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Mekhanik [1.2K]

Answer:

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