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3 years ago

Ben is 12 years older than Ishaan. Ben and Ishaan first met two years ago. Three years ago, Ben was 4 times as old as Ishaan.

2 answers:

5 0

Let x be the age of ishaan so the age of ben will be 12+x. the equation will become x=x+12
Going 3 years back, the age of ishaan was x-3 so that of ben would be x+12-3 before 3 years. at that time the age of ben was 4 times the age of ishaan. it means x+12-3(the age of ben)=4(x-3(the age of ishaan))
solving this equation, we get the answer x=7. this is the age of ishaan in the present.

butalik [34]3 years ago

5 0

Х years - Ishaan
(x + 12) years - Ben

3 years ago :
(х - 3) years - Ishaan
((x + 12) - 3) years - Ben
-------------------------------------------------
(x + 12) - 3 = (x - 3) * 4
x + 9 = 4x - 12
x - 4x = - 12 - 9
- 3x = - 21
x = 21 : 3
х = 7 years - is Ishaan now.

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posledela

Step-by-step walkthrough:

Well a standard half-life equation looks like this.

$N = N_0 * (\frac{1}{2})^{t/p$

$N_0$ is the starting amount of parent element.

$N$ is the end amount of parent element

$t$ is the time elapsed

$p$ is a half-life decay period

We know that the starting amount is 74g, and the period for a half-life is 2.8 days.

Therefore you can create a function based off of the original equation, just sub in the values you already know.

$N(t) = 74g * (\frac{1}{2})^{t/2.8days$

This is easy now that we have already made the function. Here we just reuse it, but plug in 2.8 days.

$N(t) = 74g * (\frac{1}{2})^{t/2.8days} = N(2.8days) = 74g * (\frac{1}{2})^{2.8days/2.8days}\\= 74g * \frac{1}{2} = 37g$

Now we just gotta do some algebra. Use the original function but this time, replace $N(t)$ with 10g and solve algebraically.

$10g = 74g * (\frac{1}{2})^{t/2.8days}\\\\\frac{10g}{74g} = (\frac{1}{2})^{t/2.8days}$

Take the log of both sides.

$log(\frac{5}{37}) = log((\frac{1}{2})^{t/2.8days})$

Use the exponent rule for log laws that, $log(b^x) = x*log(b)$

$log(\frac{5}{37}) = \frac{t}{2.8days} * log(\frac{1}{2})$

$\frac{log(\frac{5}{37})}{log(\frac{1}{2})} = \frac{t}{2.8days}$

$2.8 * \frac{log(\frac{5}{37})}{log(\frac{1}{2})} = t$

slap that in your calculator and you get

$t = 8.1 days$

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1 year ago

Foram prescritos 500mg de dipirona para uma criança com febre.Na unidade tem disponivel ampola de 1g/2ml.Quantos g vão ser admin

Valentin [98]

De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL

O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.

Fazendo a classica regra de 3, podemos chegar no volume desejado:

(atentar que 500mg = 0,5g)

g mL

1 --------- 2

0,5 --------- X

1 . X = 0,5 . 2

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