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1 month ago

Zoe draws ΔABC on the coordinate plane. What is the approximate perimeter of ΔABC

Mathematics

2 answers:

zzz [12.3K]1 month ago

6 0

Answer:

Correct answer is Option C.

Step-by-step explanation:

Provided: Triangle ABC plotted on the coordinate plane.

The coordinates given are A( 1, 1 ), B( 3, 5 ), and C( 5, 1 ).

We need to determine the approximate perimeter of the triangle.

We will utilize the distance formula for two points $(x_1,y_1)\:and\:(x_2,y_2)$

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Thus,

$AB=\sqrt{(3-1)^2+(5-1)^2}=\sqrt{(2)^2+(4)^2}=\sqrt{4+16}=\sqrt{20}=4.47\:(approx.)$

$CB=\sqrt{(5-3)^2+(1-5)^2}=\sqrt{(2)^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}=4.47\:(approx.)$

$AC=\sqrt{(5-1)^2+(1-1)^2}=\sqrt{(4)^2+(0)^2}=\sqrt{16}=4$

Therefore, the Perimeter = AB + AC + CB = 4.47 + 4.47 + 4 = 12.94 units

Consequently, Option C is correct.

lawyer [12.5K]1 month ago

5 0

B. 12 units because the total length calculated together results in 12.

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An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount o

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During an archaeological excavation, an ancient campfire is uncovered. The charcoal is determined to have significantly less than 1/1000 of the standard amount of $^{14}\text{C}$ . Calculate the minimal age of the charcoal, taking into account that $2^{10} = 1024$

Response:

57300 years

Step-by-step breakdown:

Using the relationship of half-life time against fraction, which can be expressed as:

$\dfrac{N}{N_o} = (\dfrac{1}{2})^{\frac{t}{t_{1/2}}$

In this context,

N indicates the current atom

represents the initial atom $N_o$

t signifies the time

denotes the half-life $t_{1/2}$

Since the charcoal was found to contain less than 1/1000 of the typical amount of $^{14}\text{C}$ .

Thus;

$\dfrac{N}{N_o} = \dfrac{1}{1000}$

However; the objective is to estimate the minimum age of the charcoal while noting $2^{10} = 1024$

this means $2^{10} = 1024$ , then:

$\dfrac{1}{1000}> \dfrac{1}{1024}$

$\dfrac{1}{1000}> \dfrac{1}{2^{10}}$

$\dfrac{1}{1000}> (\dfrac{1}{2})^{10}$

$\dfrac{N}{N_o} = \dfrac{1}{1000}$

Then

$\dfrac{N}{N_o} > (\dfrac{1}{2})^{10}$

Consequently, it can be estimated that the minimum time elapsed is 10 half-lives.

For $^{14}\text{C}$ , the standard half-life time is 5730 years

Thus, the estimation of the minimum age of the charcoal is 5730 years × 10

= 57300 years

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