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2 years ago

How can a kepler-47c happen?? Plz help :(

Mathematics

1 answer:

Natali5045456 [20]2 years ago

6 0

Answer:

I'm not SUPER sure but, it only can exist where there is a region in a planetary system where liquid water might exist on the surface of a planet. That puts it in the "habitable zone" which is where it can exist.

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2 years ago

Evaluate the expression 5x4+2x6

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The answer to this is: 132

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Here is a picture of a circle. Each square represents 1 square unit.

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2 years ago

1. A scale model of a building is 12 inches tall. The actual building is 120 feet tall. What is the scale factor of the model.

JulijaS [17]

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Step-by-step explanation:

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2 years ago

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Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams

hram777 [196]

Answer:

$X(16)=25.71grams$

Step-by-step explanation:

let $X(t)$ denote grams of $C$ formed in $t$ mins.

For $X grams$ of $C$ we have:

$\frac{2}{3}Xg$ of A and $\frac{1}{3}Xg$ of B

Amounts of A,B remaining at any given time is expressed as:

$40-\frac{2}{3}Xg$ of A and $50-\frac{1}{3}Xg$ of B

Rate at which C is formed satisfies:

$\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}$

Apply the initial condition, $X(0)=0$ ,to the expression above

$0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}$

Now at $X(8)=15$ :

$15=90-\frac{90}{90\times 8k+1} \ ->75=\frac{90}{720k+1}\\k=0.0002778$

Substitute in X(t) to get

$X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71$

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2 years ago