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2 years ago

What kind of relationship is it when a tree needs fungus to be present to grow normally

1 answer:

galina1969 [7]2 years ago

5 0

I think the question is describing the situation where the tree
needs the fungus AND the fungus needs the tree. Each one
needs the other one in order to survive and grow normally.

This is called "symbiosis", or a "symbiotic" relationship.

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Correct shape will reduce fluid friction. True False

AURORKA [14]

I believe that is true.

hope this helps!

8 0

2 years ago

Read 2 more answers

A grindstone increases in angular speed from 6.00 rad/s to 12.20 rad/s in 16.00 s. Through what angle does it turn during that t

Akimi4 [234]

Answer:

Explanation:

Given that the grand stone has initial angular velocity of

w(ini)= 6rad/

And it has a final angular velocity of

w(fin)=12.20rad/sec

Time taken is t=16s

Using equation of angular motion

To get angular acceleration (α)

w(fin)=w(ini)+αt

12.20=6+16α

16α=12.20-6

16α=6.2

α=6.2/16

α=0.3875rad/sec²

The angular acceleration is 0.39rad/s²

Angle that he turn using

w(fin)²=w(ini)²+2αθ

12.2²=6²+2×0.3875θ

12.2²-6²=0.775θ

0.775θ=112.84

Then, θ=112.84/0.775

θ=145.6radian

The angular displacement is 145.6rad

6 0

2 years ago

Why is the sky blue?

Naya [18.7K]

Molecules in the air scatter blue light from the sun more than they scatter red light.

7 0

2 years ago

Read 2 more answers

Leoni is participating in four drawing competitions. If the probability of her losing any

Alex17521 [72]

Answer:

(a) $Probability = 0.7599$

(b) $Probability = 0.2646$

Explanation:

Represent losing with L and winning with W.

So:

$L = 0.7$ --- Given

$n = 4$

Probability of winning would be:

$W = 1 - L$

$W = 1 - 0.7$

$W = 0.3$

The question illustrates binomial probability and will be solved using the following binomial expansion;

$(L + W)^4 = L^4 + 4L^3W + 6L^2W^2 + 4LW^3 + W^4$

So:

Solving (a): Winning at least 1

We look at the above and we list out the terms where the powers of W is at least 1; i.e., 1,2,3 and 4

So, we have:

$Probability = 4L^3W + 6L^2W^2 + 4LW^3 + W^4$

Substitute value for W and L

$Probability = 4 * 0.7^3*0.3 + 6*0.7^2*0.3^2 + 4*0.7*0.3^3 + 0.3^4$

$Probability = 0.7599$

Hence, the probability of her winning at least one is 0.7599

Solving (a): Wining exactly 2

We look at the above and we list out the terms where the powers of W is exactly 2

So, we have:

$Probability = 6L^2W^2$

Substitute value for W and L

$Probability = 6*0.7^2*0.3^2$

$Probability = 0.2646$

Hence, the probability of her winning exactly two is 0.2646

6 0

2 years ago

HELPPPPPPPPPPPP PPPPPPPPPPPPPLLLLLLLLLLLEEEEEEEEEAAAAAAAASSSSSSSSSSEEEEEEEEEE... THNAKS

zavuch27 [327]

Yes I pretty sure it's c

6 0

3 years ago