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3 years ago

Conducted in Paris, France, equal amounts of pigeon feed

Mathematics

1 answer:

Vedmedyk [2.9K]3 years ago

7 0

Answer:

6.122 Be Nice to Pigeons, As They Remember Your Face In a study30 conducted in Paris, France, equal amounts of pigeon feed were spread on the ground in two adjacent locations.

Step-by-step explanation:

https://www.chegg.com/homework-help/questions-and-answers/6122-nice-pigeons-remember-face-study30-conducted-paris-france-equal-amounts-pigeon-feed-s-q25636011

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Molly runs each morning once around a circular path in her neighborhood. if the diameter of the path is 1.4 miles, how far does

Lemur [1.5K]

To work out the distance of the path, you need to find out the circumference.
Formula for circumference = pi multiplied by the diameter

So you simply do:
pi multiplied by 1.4 miles

And you should get:
4.39822971503... miles

Simply round the result to an appropriate value (e.g. 3 decimal places).

Answer: 4.398 miles (to 3dp)
Hope the answer helps :)

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2 years ago

For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perim

ICE Princess25 [194]

Answer:

Part a) The area of the figure is $\frac{9}{2}(4+\pi )\ cm^{2}$

Part b) The perimeter of the figure is $3(2+2\sqrt{2}+ \pi)\ cm$

Step-by-step explanation:

Step 1

Find the area of the figure

In this problem we have that

The figure ABC is the half of a square and the other figure is a semicircle

<u>Find the area of the figure ABC</u>

we have

$AB=6\ cm, BC=6\ cm$

The area of the half square ABC is equal to find the area of triangle ABC

$A1=\frac{1}{2}*6*6=18\ cm^{2}$

<u>Find the area of the semicircle</u>

The area of the semicircle is equal to

$A2=\pi r^{2}/2$

we have that

$r=6/2=3\ cm$

substitute

$A2=\pi (3)^{2}/2$

$A2=(9/2) \pi\ cm^{2}$

The area of the figure is equal to

$18\ cm^{2}+(9/2) \pi\ cm^{2}= \frac{9}{2}(4+\pi )\ cm^{2}$

Step 2

Find the perimeter of the figure

The perimeter of the figure is equal to

$P=AB+AC+length\ CB$

we have

$AB=6\ cm$

Applying Pythagoras theorem

$AC=\sqrt{6^{2}+6^{2}}\\AC=6\sqrt{2}\ cm$

Remember that

the circumference of a semicircle is equal to

$C=\frac{1}{2}2\pi r=\pi r$

$r=6/2=3\ cm$

$C=\pi(3)$

$C=3 \pi\ cm$

The perimeter of the figure is equal to

$P=6\ cm+6\sqrt{2}\ cm+3 \pi\ cm$

Simplify

$P=3(2+2\sqrt{2}+ \pi)\ cm$

5 0

3 years ago

Please Help!!!

IgorC [24]

A) 1 - 7p - 2p^2 + 5p^3

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3 years ago

Pythagorean triples are given by the formulas x2 - y2, 2xy, and x2 + y2. Use the formulas for the Pythagorean triples to prove w

Arte-miy333 [17]

Let $k$ be an integer. Suppose there is a triangle with legs of length 16 and $2k+1$ . Then by the Pythagorean theorem, the length of the hypotenuse should be

$\sqrt{16^2+(2k+1)^2}=\sqrt{4k^2+4k+257}$

The formulas for Pythagorean triples say that if the legs are integers, then so must be the hypotenuse, because if $x=16$ and $y=2k+1$ are integers, then so are $x^2-y^2$ , $2xy$ , and $x^2+y^2$ .

However, $4k^2+4k+257$ is not a perfect square trinomial, which means for any integer $k$ , the length of the hypotenuse is not an integer, so such a triangle doesn't exist.

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2 years ago

Answers please.....

prisoha [69]

Step-by-step explanation:

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3 years ago