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Papessa [141]
3 years ago
7

Stomach cells are moderately well adapted to the acidity and protein-digesting activities in the stomach by having: a) Stomach c

ells are moderately well adapted to the acidity and protein-digesting activities in the stomach by having a high level of secretion by chief cells. b) a thick, mucous secretion and active mitosis of epithelial cells. c) secretions that enter the stomach from the pancreas. d) a thriving colony of H. pylori. e) a high level of secretion from parietal cells.
Biology
1 answer:
Tcecarenko [31]3 years ago
5 0

Answer:

The correct answer will be option-B

Explanation:

The environment of the stomach is acidic in nature maintaining value less than seven as it kills the microbes and helps activate the pepsin to digest the proteins.

The acidic nature of the gastric juice does not affect the stomach as the mucosa layer of the stomach secretes mucous and the if the cells get affected then the cells are repaired by the process of mitosis taking in the epithelial cells.

Thus, Option-B is the correct answer.

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Three lots with parallel side boundaries extend from the avenue to the boulevard as shown answer
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Three lots with parallel side boundaries extend from the avenue to the boulevard as shown in the diagram attached.

Given;
            x + y + z = 140 meters


According to the proportion theorem,

\frac{x}{40} = \frac{y}{30}

Solving for x, using cross multiplication:

\frac{x}{1} = \frac{40y}{30}

Similarly,

x/40 = z/35

solving for z by cross multiplication,

z = 35x/40
z = 7x/8

Now, the third proportions,
y/30 = z/35

Solving for y,

y = 30z/35
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(224y + 144z + 147x)/168 = 140

224y + 144z + 147x = 23520

Substitutoing z = 7x/8,

224y + (144 × 7x/8) + 147x = 23520

224y + 126x + 147x = 23520

224y + 273x = 23520
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Substitute y = 3x/4

(224×3x/4) + 273x = 23520

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