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3 years ago

8

A^2+2ab+b^2. solve for a=10 and b=50

1 answer:

qaws [65]3 years ago

3 0

A² + 2ab + b² ; Solve when a=10 & b=50

Simply plug in 10 for a and 50 for b in our given equation :)

(10)² + 2(10)(50) + (50)²

Simplify.

100 + 20(50) + 2500

Simplify.

(100 + 2500) + 1000

Simplify.

2600 + 1000

Simplify.

3600

Hope I helped! :)

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3 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How

DedPeter [7]

Answer:

a) $Q(t) = 180e^{-0.023t}$

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

$Q(t) = Q(0)e^{-rt}$

$0.5Q(0) = Q(0)e^{-30r}$

$e^{-30r} = 0.5$

Applying ln to both sides of the equality.

$\ln{e^{-30r}} = \ln{0.5}$

$-30r = \ln{0.5}$

$r = \frac{\ln{0.5}}{-30}$

$r = 0.023$

So

$Q(t) = Q(0)e^{-0.023t}$

180-mg sample, so Q(0) = 180

$Q(t) = 180e^{-0.023t}$

(b) How much of the sample remains after 120 years?

This is Q(120).

$Q(t) = 180e^{-0.023t}$

$Q(120) = 180e^{-0.023*120}$

$Q(120) = 11.4$

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

$Q(t) = 180e^{-0.023t}$

$1 = 180e^{-0.023t}$

$e^{-0.023t} = \frac{1}{180}$

$e^{-0.023t} = 0.00556$

Applying ln to both sides

$\ln{e^{-0.023t}} = \ln{0.00556}$

$-0.023t = \ln{0.00556}$

$t = \frac{\ln{0.00556}}{-0.023}$

$t = 225.8$

225.8 years.

8 0

3 years ago

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