Answer:
The answer is to organize them and speak about them accurately.
Explanation:
A) Matter that is made up of cells and tissues is called ORGANS
B) GUARD cells are involved in the closing and opening of stomata.
C) Excess sugar produced during photosynthesis is stored in from of STARCH
D) VITAMIN D is required for absorption of calcium.
E) Absorption of water takes place in the SMALL INTESTINES
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Answer:
1. During the day, the sun heats up both the ocean surface and the land. Water is a good absorber of the energy from the sun. The land absorbs much of the sun’s energy as well. However, water heats up much more slowly than land and so the air above the land will be warmer compared to the air over the ocean.
2.The air over the ocean is now warmer than the air over the land. The land loses heat quickly after the sun goes down and the air above it cools too. This can be compared to a blacktop road. During the day, the blacktop road heats up and becomes very hot to walk on. At night, however, the blacktop has given up the added heat and is cool to the touch. The ocean, however, is able to hold onto this heat after the sun sets and not lose it as easily. This causes the low surface pressure to shift to over the ocean during the night and the high surface pressure to move over the land. This causes a small temperature gradient between the ocean surface and the nearby land at night and the wind will blow from the land to the ocean creating the land breeze.
3. The sand should both heat and cool faster than the water. This is because water has a higher specific heat ca- pacity than sand – meaning that it takes a lot of heat, or energy, to raise the temperature of water one degree, whereas it takes comparatively little energy to change the temperature of sand by one degree.
4.
Answer:
The correct answer is E. none of the above. The population will drops below 100 when t ≥ 38.
Explanation:
Given A= A0 e^kt. The population 10 years ago is A0, the population today is A(10), and we have to find the value of "k" and then the time when population drops below 100.
So, A(t) = 1700 e^kt ⇒ A(10) = 1700 e^k(10) ⇒ 800 = 1700 e^k(10) ⇒
800/1700 = e^k(10) ⇒ln (800/1700) = k(10) ln e ⇒ -0.754/10 = k ⇒
k = -0.0754.
Now you have all the parameters, so you can find the time at which the population drops below 100.
A(t) = 1700 e^kt ⇒ 100 = 1700 e^(-0.0754)t ⇒100/1700 = e^(-0.0754)t ⇒
ln(100/1700) = (-0.0754)t ln e ⇒ [ln(100/1700)]/(-0.0754) = t ⇒
t = 38.
So, the population will drops below 100 when t ≥ 38.
