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2 years ago

Rry YUVIVU W

Mathematics

2 answers:

KIM [24]2 years ago

4 0

Rip juice wrld man what an inspirational person

Lapatulllka [165]2 years ago

3 0

Answer:

Subtract 4 from both sides

Step-by-step explanation:

Ap ex

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Find the approximate perimeter of ABC plotted below.

maksim [4K]

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

8 0

3 years ago

Ask your teacher find the limit. use l'hospital's rule where appropriate. if there is a more elementary method, consider using i

liberstina [14]

Answer:

ln(5/3)

Step-by-step explanation:

The desired limit represents the logarithm of an indeterminate form, so L'Hopital's rule could be applied. However, the logarithm can be simplified to a form that is not indeterminate.

<h3>Limit</h3>

We can cancel factors of (x-1), which are what make the expression indeterminate at x=1. Then the limit can be evaluated directly by substituting x=1.

$\diplaystyle \lim\limits_{x\to1}{(\ln(x^5-1)-\ln(x^3-1))}=\lim\limits_{x\to1}\ln{\left(\dfrac{x^5-1}{x^3-1}\right)}\\\\=\lim\limits_{x\to1}\ln\left(\dfrac{x^4+x^3+x^2+x+1}{x^2+x+1}\right)=\ln{\dfrac{5}{3}}$