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2 years ago

12

Find the area(not to scale)

1 answer:

makvit [3.9K]2 years ago

6 0

Answer: no

Step-by-step explanation: no

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a bag contains only a red and blue marbles there are a total of 36 marbles in the back there are five red marbles for every four

Harrizon [31]

Let R and B be the number of red and blue marbles respectively.

Then,
R+B = 36
R:B = 5:4

Therefore,
Ratio of red marbles = 5/9
Ratio of blue marbles = 4/9
This means,
R = 5/9*36 = 20 marbles
B = 4/9*36 = 16 marbles

If one blue marble is removed from the bag, the new B= 16-1 = 15 blue marbles and the new total of marbles in the bag = 36-1 = 35 marbles.
The new ratios are
R = 20/35 =4/7
B = 15/35 = 3/7
That is, R:B = 4:3

The reasoning of the student is wrong. When a marble is removed, both the number of blue marbles changes as well as the total of the marbles in the bag. In other words, both the values in the ratio reduce by 1.

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3 years ago

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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.

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Olin [163]

Answer:

23.32

Step-by-step explanation:

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