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3 years ago

A square field has an area of 479 ft what is the approximate length of a size of the field

Mathematics

1 answer:

MissTica3 years ago

4 0

Answer:

21.9 ft

Step-by-step explanation:

The area of a square is the square of the side length, so the side length is the square root of the area:

... s = √(479 ft²) ≈ 21.886 ft

This is approximately 21.9 ft.

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Pls help 25h²- 16t² factorise it

noname [10]

Answer:

(5h-4t)(5h+4t)

Step-by-step explanation:

25h² - 16t²

adding and subtracting 20ht from it (√25 and √16 = 5x4 = 20):

25h² + 20ht - 20ht -16t²

factor:

(5h-4t)(5h+4t)

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Answer:

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Find the length of the following curve. If you have a grapher, you may want to graph the curve to see what it looks like.

stepladder [879]

The length of the curve $y = \frac{1}{27}(9x^2 + 6)^\frac 32$ from x = 3 to x = 6 is 192 units

<h3>How to determine the length of the curve?</h3>

The curve is given as:

$y = \frac{1}{27}(9x^2 + 6)^\frac 32$ from x = 3 to x = 6

Start by differentiating the curve function

$y' = \frac 32 * \frac{1}{27}(9x^2 + 6)^\frac 12 * 18x$

Evaluate

$y' = x(9x^2 + 6)^\frac 12$

The length of the curve is calculated using:

$L =\int\limits^a_b {\sqrt{1 + y'^2}} \, dx$

This gives

$L =\int\limits^6_3 {\sqrt{1 + [x(9x^2 + 6)^\frac 12]^2}\ dx$

Expand

$L =\int\limits^6_3 {\sqrt{1 + x^2(9x^2 + 6)}\ dx$

This gives

$L =\int\limits^6_3 {\sqrt{9x^4 + 6x^2 + 1}\ dx$

Express as a perfect square

$L =\int\limits^6_3 {\sqrt{(3x^2 + 1)^2}\ dx$

Evaluate the exponent

$L =\int\limits^6_3 {3x^2 + 1} \ dx$

Differentiate

$L = x^3 + x|\limits^6_3$

Expand

L = (6³ + 6) - (3³ + 3)

Evaluate

L = 192

Hence, the length of the curve is 192 units