The perimeter of a rectangle is twice the

Mathematics

1 answer:

masya89 [10]3 years ago

5 0

Answer:

The length is 14 m.

Step-by-step explanation:

length = L

width = W

perimeter = P

"The perimeter of a rectangle is twice the sum of its length and its width."

P = 2(L + W)

"The perimeter is 40 meters"

40 = 2(L + W)

"and its length is 2 meters more than twice its width."

L = 2W + 2

Replace L with 2W + 2 in equation 40 = 2(L + W):

40 = 2(2W + 2 + W)

40 = 2(3W + 2)

20 = 3W + 2

3W = 18

W = 6

The width is 6 meters.

L = 2W + 2 = 2(6 m) + 2 m = 12 m + 2 m = 14 m

Answer: The length is 14 m.

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Use the definition of continuity and the properties of limit to show that the function f(x)=x sqrtx/ (x-6)^2 is continuous at x=

jasenka [17]

Answer:

The function $\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$ is continuous at x = 36.

Step-by-step explanation:

We need to follow the following steps:

The function is:

$\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$

The function is continuous at point x=36 if:

The function $\\ f(x)$ exists at x=36.
The limit on both sides of 36 exists.
The value of the function at x=36 is the same as the value of the limit of the function at x = 36.

Therefore:

The value of the function at x = 36 is:

$\\ f(36) = \frac{36*\sqrt{36}}{(36-6)^{2}}$

$\\ f(36) = \frac{36*6}{900} = \frac{6}{25}$

The limit of the $\\ f(x)$ is the same at both sides of x=36, that is, the evaluation of the limit for values coming below x = 36, or 33, 34, 35.5, 35.9, 35.99999 is the same that the limit for values coming above x = 36, or 38, 37, 36.5, 36.1, 36.01, 36.001, 36.0001, etc.

For this case:

$\\ lim_{x \to 36} f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$