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stiks02 [169]
2 years ago
8

If you could put wings on any species of animal, what animal would you choose?

Mathematics
2 answers:
denis-greek [22]2 years ago
5 0

Answer:

Humans

Explanation:

So I can have wings :)

Crank2 years ago
5 0

Answer:

If I could put wings on any species of animal, I would choose a Unicorn because then the Unicorn can fly and look at all the enviroment.

Step-by-step explanation:

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11:
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A graph titled Swimmers at City Pool has Temperature (degrees Fahrenheit) on the x-axis, and swimmers on the y-axis. A trend lin
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25 is the answer

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this is non linear because the rate of change is not constant.

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Is this a function ? Yes or no and why or why not
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Step-by-step explanation:

Yes it is a function, because graph is intersecting y - axis only once.

3 0
3 years ago
The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?
Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

A(-6, 2)) = (x_1, y_1)

B(2, 6) = (x_2, y_2)

AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}

AB = \sqrt{(8)^2 + (4)^2}

AB = \sqrt{64 + 16} = \sqrt{80}

AB = 8.9 (nearest tenth)

Distance between B(2, 6) and C(7, 1):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

B(2, 6) = (x_1, y_1)

C(7, 1) = (x_2, y_2)

BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}

BC = \sqrt{(5)^2 + (-5)^2}

BC = \sqrt{25 + 25} = \sqrt{50}

BC = 7.1 (nearest tenth)

Distance between C(7, 1) and D(3, -5):

CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

C(7, 1) = (x_1, y_1)

D(3, -5) = (x_2, y_2)

CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}

CD = \sqrt{(-4)^2 + (-6)^2}

CD = \sqrt{16 + 36} = \sqrt{52}

CD = 7.2 (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

D(3, -5) = (x_1, y_1)

A(-6, 2) = (x_2, y_2)

DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}

DA = \sqrt{(-9)^2 + (7)^2}

DA = \sqrt{81 + 49} = \sqrt{130}

DA = 11.4 (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0
3 years ago
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