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3 years ago

The evolutionary force that operates on the basis of variation is ___

1 answer:

5 0

Were there any answer choices? I feel that this question can have many answers..

Natural selection
Genetic Mutations
DNA (OR)
Survival of the Fittest

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Which of the following tips would you recommend to a friend who is in credit card debt?

denis-greek [22]

All of the above is your answer

4 0

4 years ago

Ang kakapusan ay umiiral dahil limitado ang pinagkukunang-yaman at walang katapusang pangangailangan at kagustuhan ng tao. Bakit

Dmitriy789 [7]

D. dahil ito ay INAASAHAN NA BABALIK SA NORMAL ANG SUPLAY NG ISANG PRODUKTO SA PAMILIGAN SA SANDALING BUMUTI ANG PANAHON.

Explanation:

YEA ITS REAL NMAN KASE MAY MGA TAO NA NINANAKAW ANG MGA DAGAT OR LIKAS NA YAMAN NATIN NA PINAGKUKUNAN NATIN NG MGA MAKAKAIN NATIN SA PANG ARAW ARAW  KUNG IKAW AY NASA TABING DAGAT. 

I THINK IM CORRECT:)

5 0

3 years ago

URGENT EXTRA POINTS

ss7ja [257]

Answer:

1.6% growth rate

43.75 years = doubling time.

Explanation:

40-24 = 1616/1000 = 0.016 x 100 = 1.6% growth rate

70/1.6 = 43.75 years = doubling time.

8 0

3 years ago

Catalina Island is 8 miles wide and 21 miles long. If one inch of rain falls on Catalina Island, how many cubic feet of rain fel

Luba_88 [7]

Answer:

Explanation:

4 0

3 years ago

Hello, I need help with a calculus FRQ. My teacher has given a hint that this last part has to do with the intermediate value th

lesya [120]

Answer:

Yes, at a time t such that (√2)/2 ≤ t ≤ 2.

Explanation:

To answer the question

Therefore, where the domain of the function is the set of all real numbers x for which f(x) is a real number we have

For Chloe's velocity

$C(t) = t\times e^{4-t^2} \ for \ 0\leq t\leq 2$

Finding the boundaries of the function gives;

$0\times e^{4-0^2} = 0$ and $2\times e^{4-2^2} = 2$

At t = 1, we have $1\times e^{4-1^2} = e^{3} = 20.086$

We find the maximum point as follows;

$\frac{\mathrm{d} \left (t\times e^{4-t^2} \right )}{\mathrm{d} x}=0$

From which we have;

$\frac{\mathrm{} e^{4-t^2} - t\times e^{4-t^2} \times2\times t }{(e^{4-t^2} )^2}=0$

$e^{4-t^2} - t\times e^{4-t^2} \times2\times t }=0$

$e^{4-t^2}(1 - t\times2\times t })=0\\e^{4-t^2}(1 - 2\times t^2 })=0\\$

$e^{4-t^2}=0$ or $(1 - 2\times t^2 })=0$

∴ 1 = 2·t² and from which t = (√2)/2

Hence the function C(x) is decreasing from t = (√2)/2 to t = 2

For Brandon

For 0 ≤ t ≤ 1, 1 ≤ B(t) ≤8 and for 1 < t ≤ 2, 8 < B(t) ≤ 1.5

1 ≤ f(x) ≤ 1.5

Given that the function B(t) is differentiable, therefore, continuous, there exists a point at which the function C(t) and B(t) intersects given that;

For 0 ≤ t ≤ (√2)/2, 0 ≤ C(t) ≤ 23.416 for (√2)/2 < t ≤ 2, 23.416 > C(t) ≥ 2

and for 0 ≤ t ≤ 0 1 ≤ B(t) ≤ 8 and for 1 < t ≤ 2, 8 > B(t) ≥ 1.5

Therefore, the curves intersect at in between (√2)/2 ≤ t ≤ 2.

8 0

3 years ago