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4 years ago

A single-celled organism is represented below. Structure X carries out a function most similar to which structure in a human?

Mathematics

1 answer:

tester [92]4 years ago

4 0

Answer: lung

Step-by-step explanation:

Attachedfile/picture shows the structure.

Respiration is a process of degradation of complex organic compound with the production of carbon dioxide, water and energy.

Respiration involves two phases, which are;

(1). External Respiration or Breathing: this is a process in which

animals take oxygen in and release carbon dioxide.

(2). Internal Respiration or Cellular Respiration: this process involve the release of energy from food substance with the release of carbondioxide and and water.

Single celled animals or unicellular animals such as amoeba exchange gases through cell surface. The STRUCTURE X IS THE PLASMA MEMBRANE. There is absorption of of Oxygen from the surrounding air or water,hence, giving out carbondioxide through plasma membrane by Diffusion.

PS: Lung is used in respiration process in Human

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Liono4ka [1.6K]

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3 years ago

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3 years ago

Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ

atroni [7]

Answer: $y=Ce^(^3^t^{^9}^)$

Step-by-step explanation:

Beginning with the first differential equation:

$\frac{dy}{dt} =27t^8y$

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

$\frac{1}{y} \frac{dy}{dt} =27t^8$

Multiply both sides by 'dt' to get:

$\frac{1}{y}dy =27t^8dt$

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

$\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt$

$ln(y)=27(\frac{1}{9} t^9)+C$

$ln(y)=3t^9+C$

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

$e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)$

$y=e^(^3^t^{^9} ^+^C^)$

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

$y=e^(^3^t^{^9}^)e^C$

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

$y=Ce^(^3^t^{^9}^)$

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

$\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)$

$\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8$

Now check if the derivative equals the right side of the original differential equation:

$(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)$

$Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)$

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

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2 years ago

PLEASE HELP!! Algebra 2 Review !!

Kamila [148]

Answer:

$\frac{28-3i}{26}$

Step-by-step explanation:

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