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Daniel [21]
3 years ago
11

The order Cetacea includes modern-day whales, dolphins, and porpoises. These marine mammals share common characteristics that ha

ve led scientists to make a conclusion about their evolutionary history. Cetaceans:
lack gills and must breathe air
give birth to live young rather than lay eggs
have fin bones that are similarly shaped to limb bones
have spines that flex up and down rather than side to side
The characteristics of cetaceans provide evidence that

A. they evolved from land mammals.

B. the theory of evolution is incorrect.

C. they evolved from air-breathing fish.

D. the bodies of their ancestors never changed.
Biology
1 answer:
allsm [11]3 years ago
6 0

Answer:

The correct option is <u><em>A. they evolved from land mammals.</em></u>

Explanation:

The process of evolution explains how closely one organism is to another based on their common ancestors. Scientists believe that those organisms which show many similarities in their structures or functions might have evolved from a common ancestor in the past.

As in the above-mentioned question, scientists have observed that cetaceans lack gills and breath air, this is a property of land mammals. The cetaceans have fin bones which are similar to the limb bones. As limb bones are present in land mammals, hence they cetaceans and land mammals might have common ancestors in the past.

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The statement 2 + 2 = 4 would be considered a _____ in the science world.
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What is the impact of life skill to you? A. To be able to create problems B. To be self assertive. C. To be influenced by friend
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3 years ago
We are studying 3 strains of bacteria, with populations p1, p2, p3, in an environment with three food sources, A, B, C. In a day
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Answer:

The population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

Explanation:

From the given information:

For  food source A; we have:

3P₁ + P₂ + 2P₃ = 58    units of food A ---- (1)

For food source B; we have:

2P₁ + 4P₂ + 2P₃ = 70   units of food B  ---- (2)

For food source C; we have:

P₁ + P₂  = 20   units of food C    ----- (3)

From equation (1) and (2); we have:

3P₁ + P₂ + 2P₃ = 58

2P₁ + 4P₂ + 2P₃ = 70

By elimination method

 3P₁ + P₂ + 2P₃ = 58

-

 2P₁ + 4P₂ + 2P₃ = 70

<u>                                       </u>

<u> P₁  -   3P₂   + 0    = - 12    </u>

P₁ = -12 + 3P₂   ---- (4)

Replace, the value of P₁  in (4) into equation (3)

P₁ + P₂  = 20

-12 + 3P₂ + P₂  = 20

4P₂ = 20 + 12

4P₂ = 32

P₂ = 32/4

P₂ = 8

From equation (3) again;

P₁ + P₂  = 20

P₁ + 8 = 20

P₁  = 20 - 8

P₁  = 12

To find P₃;  replace the value of P₁ and P₂ into (1)

3P₁ + P₂ + 2P₃ = 58

3(12) + 8 + 2P₃ = 58

36 + 8 + 2P₃ = 58

2P₃ = 58 - 36 -8

2P₃ = 14

P₃ = 14/2

P₃ =  7

Thus, the population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

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3 years ago
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