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3 years ago

6-4n what are the first 5 terms

Mathematics

1 answer:

NikAS [45]3 years ago

5 0

Answer:

2, - 2, - 6, - 10, - 14

Step-by-step explanation:

To find the first 5 terms, substitute n = 1, 2, 3, 4, 5 into the expression.

n = 1 : 6 - 4(1) = 6 - 4 = 2

n = 2 : 6 - 4(2) = 6 - 8 = - 2

n = 3 : 6 - 4(3) = 6 - 12 = - 6

n = 4 : 6 - 4(4) = 6 - 16 = - 10

n = 5 : 6 - 4(5) = 6 - 20 = - 14

Thus the first 5 terms are 2, - 2, - 6, - 10, - 14

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We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the c

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We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

$\frac{dT}{dt}=k(T-a)$

We now separate the time and the temperature variables as follows,

$\frac{dT}{T-a}=kdt$

Integrating both sides of the differential equation, we obtain;

$ln(T-a)=kt +c$

This natural logarithmic equation can be rewritten as;

$T-a=e^{kt +c}$

Applying the laws of exponents, we obtain,

$T-a=e^{kt}\times e^{c}$

$T-a=e^{c}e^{kt}$

We were given the initial conditions,

$T(0)=4$

Let us apply this condition to obtain;

$4-20=e^{c}e^{k(0)}$

$-16=e^{c}$

Now our equation, becomes

$T-a=-16e^{kt}$

or

$T=a-16e^{kt}$

When we substitute a=20,
we obtain,

$T=20-16e^{kt}$

b) We were also given that,

$T(5)=8$

Let us apply this condition again to find k.

$8=20-16e^{5k}$

This implied

$-12=-16e^{5k}$

$\frac{-12}{-16}=e^{5k}$

$\frac{3}{4}=e^{5k}$

We take logarithm to base e of both sides,

$ln(\frac{3}{4})=5k$

This implies that,

$\frac{ln(\frac{3}{4})}{5}=k$

$k=-0.2877$

After 15 minutes, the temperature will be,

$T=20-16e^{-0.2877\times 15}$

$T=20-0.21376$

$T=19.786$

After 15 minutes, the temperature is approximately 20°C

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