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nadya68 [22]
3 years ago
11

Plants could not live if they were unable to get water and nutrients from soil. In some places, especially deserts, these resour

ce are in short supply so plants must compete with each other for them. Which of the following is an actual strategy used by plants to keep competitors from being a problem?
A) Having very spiky parts above ground

B) Releasing ‘plant poisons’ (herbicides) from their roots into the soil
Biology
2 answers:
galben [10]3 years ago
7 0
This would be A, herbicides is a chemical used by humans not plants!
Sati [7]3 years ago
3 0

Answer:

A is your answer.

Explanation:

Hope this helps!

Please mark me brainliest!!!

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Please help will give brainless !!!!!:)
Anestetic [448]

Answer:

they start to run out of places to stay cool and then will eventually run out of food to eat then they will die. hope this helps

7 0
3 years ago
The Magnetosphere Is Invisible. The Aurora Borealis (Northern Lights) Is The Evidence Of The Magnetophere.
IrinaK [193]
The answer is true can you tell me answer my question
5 0
3 years ago
What force is needed to accelerate a 200 kg mass across a surface at 75.5 m/sec2?
stich3 [128]
By using the equation F=ma (Force = Mass*acceleration), the answer is 15,100 N for the first one and 1.25 m/s^2 for the second.
3 0
3 years ago
Tres ejemplos en los que se puedan emplear de polea movil
galben [10]

Answer:

rueda, maciza y acanalada en su borde

6 0
3 years ago
Suppose that we are dealing with four genes, each gene consisting of a dominant allele (capital letter) and a recessive allele (
Sauron [17]

Answer: The probability of obtaining an individual who is CcmmLLPp 1/32.

Explanation: This can be achieved by crossing similar genes to obtain individual probability as shown in the attached image.

When you cross Cc in the first genotype with CC in the second genotype, the following probabilities will be obtained;

P (CC) = ¾, P(Cc) = ¼.

Similarly, crossing Mm with mm, we get;

P (Mm) = ½, P (mm) = ½

Crossing Ll with Ll, we get

P (LL) = ¼, P (Ll) = ½, P (ll) = ¼

Crossing PP with pp, we get

P (Pp) = 1

Therefore, the probability of individual with genotype CcmmLLPp will be;

P (Cc) x P (mm) x P (LL) x P (Pp)

= ¼ x ½ x ¼ x 1

= 1/32

3 0
3 years ago
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