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3 years ago

8

En un videojuego, Marta ha conseguido 36.450 puntos capturando 11 monedas de oro. ¿ cuántos puntos vale cada moneda de oro?

1 answer:

sukhopar [10]3 years ago

3 0

Answer:

3313.64 puntos

Step-by-step explanation:

Podemos interpretar la pregunta anterior Matemáticamente como:

11 monedas de oro = 36.450 puntos

1 moneda de oro = x

Multiplicar cruzada

11 monedas de oro × x = 36.450 puntos × 1 monedas de oro

x = 36.450 puntos × 1 monedas de oro / 11

x = 3313.6363636 puntos

Aproximadamente

1 moneda de oro = 3313.64 puntos

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2 years ago

How could you change the equation from Part B so it has infinitely many solutions? What would infinitely many solutions mean in

Andrews [41]

Answer:

We do not have the equation for the part B, so we can not aswer it correctly.

But i will give a general answer.

We could have infinite answers always when we have more variables than linear independent equations:

This is, if we have one variable, x, we can have infinite solutions if we have no equations (or equations with no restrictions for our variable)

So if we have an equation like:

x*4 = √16*x

you can see that both sides of the equation are exactly the same, so this equation actually does not have any value, and x could take infinite different values and the equation will remain true.

If we have two variables, x and y, we will have infinite solutions if we have only one equation:

y = a*x + b

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3 years ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago

Simplify three over five divided by negative four over nine..

gladu [14]

3/5 divided by -4/9
3/5 times the reciprical -9/4
-27/20=-1 7/20

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3 years ago

Ivan has $8.75 in nickels and quarters in his desk drawer. The number of nickels is twice the number of quarters. How many coins

LuckyWell [14K]

When Ivan divides his coins into groups of 2 nickels and 1 quarter, worth $0.35, he find that he has $8.75/$0.35 = 25 such groups.

Ivan has 25 quarters and 50 nickels.

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3 years ago

Read 2 more answers