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2 years ago

Round log 58 to the nearest thousandth

Mathematics

2 answers:

Vesnalui [34]2 years ago

6 0

Log 58 = 1.7634

The digit in the ten thousandth place is 4 ⇒ less than 5 , discard

Log 58 = 1.7634 ≈ 1.763

Answer: 1.763

Luda [366]2 years ago

5 0

You have $\log(58) = 1.763427...$ . Rounding this number to the nearest thousandth means to write it as the three-decimal-digits number which is nearest to this one.
The two numbers with three decimal digits are 1.763 and 1.764. Since the fourth decimal digit is 4, your number is closer to 1.763 than 1.764, so this is its approximation.

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Find the approximate perimeter of ABC plotted below.

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Answer:

B. 21.2

Step-by-step explanation:

Perimeter of ∆ABC = AB + BC + AC

A(-4, 1)

B(-2, 3)

C(3, -4)

✔️Distance between A(-4, 1) and B(-2, 3):

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

$AB = \sqrt{(-2 - (-4))^2 + (3 - 1)^2} = \sqrt{(2)^2 + (2)^2)}$

$AB = \sqrt{4 + 4}$

$AB = \sqrt{16}$

AB = 4 units

✔️Distance between B(-2, 3) and C(3, -4):

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

$BC = \sqrt{(3 - (-2))^2 + (-4 - 3)^2} = \sqrt{(5)^2 + (-7)^2)}$

$BC = \sqrt{25 + 49}$

$BC = \sqrt{74}$

BC = 8.6 units (nearest tenth)

✔️Distance between A(-4, 1) and C(3, -4):

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

$AC = \sqrt{(3 - (-4))^2 + (-4 - 1)^2} = \sqrt{(7)^2 + (-5)^2)}$

$AC = \sqrt{47 + 25}$

$AC = \sqrt{74}$

AC = 8.6 units (nearest tenth)

Perimeter of ∆ABC = 4 + 8.6 + 8.6 = 21.2 units

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Answer:

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