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Fynjy0 [20]
3 years ago
10

A snake slithers 2/9 mile in 4/5 hour.

Mathematics
2 answers:
kykrilka [37]3 years ago
8 0

Answer:

Your answer is A. 8/45 mph

Step-by-step explanation:

I can provide a more detailed answer in a few but for starters, you can rule out answers C & D because that would be the fastest snake alive to be able to travel that far in such a short time.

That leaves a choice between A or B. Using the common denominator method to switch 2/9 and 4/5 to x/45, this makes the most sense. When comparing the 8/45 mph to the actual question, it's very reasonable.

I'll come back in a few and provide the actual math. I'm late for the doctor.

Good luck! Hope this helps!

9966 [12]3 years ago
6 0

Answer:

B: 5/18 miles per hour

Step-by-step explanation:

1. multiply the fractions by the reciprocal

2/9 × 4/5 = 10/36

2. Divide answer by common multiple which is 2

10/36 ÷ 2 = 5/18

5/18 miles per hour is the snakes speed

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Firdavs [7]
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Read 2 more answers
Find the area of each shaded region each outer polygon is regular <br><br><br> See the picture
PSYCHO15rus [73]

Answer:

Area of the shaded region = 3.46 square units

Step-by-step explanation:

Measure of the interior angle of a regular polygon = \frac{(n - 2)\times180}{n}

From the picture attached,

Number of sides of the given polygon 'n' = 6

Interior angle (∠BAF) of the given polygon = \frac{(6 - 2)\times 180}{6}

                                                                       = 120°

Measure of ∠BAC = \frac{120}{4}

                              = 30°

Now we apply sine rule in ΔAGB,

sin(30°) = \frac{\text{Opposite side}}{\text{Hypotenuse}}

            = \frac{BG}{AB}

BG = AB[sin(30°)]

     = 2\times \frac{1}{2}

     = 1

By applying cosine rule in ΔABG,

cos(30°) = \frac{\text{Adjacent side}}{\text{Hypotenuse}}

              = \frac{AG}{AB}

AG = AB[cos(30°)]

     = 2\times \frac{\sqrt{3} }{2}

     = \sqrt{3}

AC = 2(AG)

     = 2√3

Area of ΔABC = \frac{1}{2}(\text{Base})(\text{Height})

                       = \frac{1}{2}(AC)(BG)

                       = \frac{1}{2}(2\sqrt{3} )(1)

                       = \sqrt{3}

Area of the shaded region = Area of ΔABC + Area of ΔFED

                                             = 2(Area of ΔABC)

                                             = 2√3

                                             = 3.46 square units

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Answer:

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