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2 years ago

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An open box is to be made from a square piece of cardboard, 18 inches by 18 inches, by removing a small square from each corner

and folding up the flaps to form the sides. What are the dimensions of the box of greatest volume that can be constructed in this way

1 answer:

seraphim [82]2 years ago

6 0

The dimension of the box of the greatest volume that can be constructed in this way is 12x12x3 and the volume is 432.

<h3>How to solve the dimension?</h3>

Let x be the side of the square to remove. Then the volume of the box is:

V(x) = (18 - 2x)² * x = 324x - 72x² + 4x³

To find the maximum volume, differentiate and set it to 0:

V'(x) = 324 - 144x + 12x²

0 = x² - 12x + 27

0 = (x - 9)(x - 3)

x = 3 or 9

When x = 3,

V"(x) =-144+24x

V"(3) =-144+72=-72<0

so volume is maximum at x=3

Therefore the box is 12x12x3 and the volume is 432.

Learn more about dimension on:

brainly.com/question/26740257

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Please helppp 3/4+(1/3÷1/6)-(-1/2)=

geniusboy [140]

Answer:

your answer should be 3 1/4

Step-by-step explanation:

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2 years ago

What is the product of (X4/3)(X2/3) ?

katrin [286]

Hi, the answer to this would be x6/9. I'm assuming the x4/3 and x2/3 are fractions and the x's aren't exponents. Now how I got x6/9 is shown here.

1st Step: Started off by regrouping the terms
1/3x3 x^4x^2

2nd Step: we can easily simplify 3x3 to just 9. And now we're left with 1/9x^4x^2

3rd Step: Now we can simplify the 1/9 to just x^4x^2/9

4th Step: Now we can use the product rule which is simple. So We add the exponents and simplify it to just one exponent. So x4+2=6 that simplifies to just x^6.

Final Answer: x^6/9.
Hope this helped you :)

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3 years ago

(1 point) Suppose F⃗ (x,y)=⟨2y,−sin(y)⟩ and C is the circle of radius 3 centered at the origin oriented counterclockwise. (a) Fi

g100num [7]

Answer:

The required vector parametric equation is given as:

r(t) = <3cost, 3sint>

For 0 ≤ t ≤ 2π

Step-by-step explanation:

Given that

f(x, y) = <2y, -sin(y)>

Since C is a cirlce centered at the origin (0, 0), with radius r = 3, it takes the form

(x - 0)² + (y - 0)² = r²

Which is

x² + y² = 9

Because

cos²β + sin²β = 1

and we want to find a vector parametric equations r(t) for the circle C that starts at the point (3, 0), we can write

x = 3cosβ

y = 3sinβ

So that

x² + y² = 3²cos²β + 3²sin²β

= 9(cos²β + sin²β) = 9

That is

x² + y² = 9

The vector parametric equation r(t) is therefore given as

r(t) = <x(t), y(t)>

= <3cost, 3sint>

For 0 ≤ t ≤ 2π

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3 years ago

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3 years ago

The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?

Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

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

Let,

$A(-6, 2)) = (x_1, y_1)$

$B(2, 6) = (x_2, y_2)$

$AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}$

$AB = \sqrt{(8)^2 + (4)^2}$

$AB = \sqrt{64 + 16} = \sqrt{80}$

$AB = 8.9$ (nearest tenth)

Distance between B(2, 6) and C(7, 1):

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

Let,

$B(2, 6) = (x_1, y_1)$

$C(7, 1) = (x_2, y_2)$

$BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}$

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

$BC = \sqrt{25 + 25} = \sqrt{50}$

$BC = 7.1$ (nearest tenth)

Distance between C(7, 1) and D(3, -5):

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

Let,

$C(7, 1) = (x_1, y_1)$

$D(3, -5) = (x_2, y_2)$

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

$CD = \sqrt{(-4)^2 + (-6)^2}$

$CD = \sqrt{16 + 36} = \sqrt{52}$

$CD = 7.2$ (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

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

Let,

$D(3, -5) = (x_1, y_1)$

$A(-6, 2) = (x_2, y_2)$

$DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}$

$DA = \sqrt{(-9)^2 + (7)^2}$

$DA = \sqrt{81 + 49} = \sqrt{130}$

$DA = 11.4$ (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0

3 years ago