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3 years ago

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A cardboard box without a top is to have volume 62500 cubic cm. find the dimensions which minimize the amount of material used.

1 answer:

bezimeni [28]3 years ago

8 0

Answer: A cardboard box without a lid is to have a volume of 32,000 cubic cm. Find the dimensions that minimize the amount of cardboard used

. ans: base 40 x 40, height = 20

Solution:

The typical box might look like the one below

where . In addition we have xyz = 32000 ,so we need to minimize .

We have,

From the geometry of the problem so y = x. So or x = 40.

Finally, y = x = 40 and z = 32000/(xy)=20.

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The multipliers of this system of equations have been determined to create opposite terms of the y-variable.

sergiy2304 [10]

Answer:

-48

Step-by-step explanation:

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5 0

3 years ago

The surface area of a given cone is 1,885.7143 square inches. What is the slang height?

kap26 [50]

Answer:

If $r >> h$ , the slang height of the cone is approximately 23.521 inches.

Step-by-step explanation:

The surface area of a cone (A) is given by this formula:

$A = \pi \cdot r^{2} + 2\pi\cdot s$

Where:

$r$ - Base radius of the cone, measured in inches.

$s$ - Slant height, measured in inches.

In addition, the slant height is calculated by means of the Pythagorean Theorem:

$s = \sqrt{r^{2}+h^{2}}$

Where $h$ is the altitude of the cone, measured in inches. If $r >> h$ , then:

$s \approx r$

And:

$A = \pi\cdot r^{2} +2\pi\cdot r$

Given that $A = 1885.7143\,in^{2}$ , the following second-order polynomial is obtained:

$\pi \cdot r^{2} + 2\pi \cdot r -1885.7143\,in^{2} = 0$

Roots can be found by the Quadratic Formula:

$r_{1,2} = \frac{-2\pi \pm \sqrt{4\pi^{2}-4\pi\cdot (-1885.7143)}}{2\pi}$

$r_{1,2} \approx -1\,in \pm 24.521\,in$

$r_{1} \approx 23.521\,in \,\wedge\,r_{2}\approx -25.521\,in$

As radius is a positive unit, the first root is the only solution that is physically reasonable. Hence, the slang height of the cone is approximately 23.521 inches.

3 0

2 years ago

What equations have z=4

natita [175]

<h2>Just some examples...</h2>

2z+4=12

2(4)+4=12

<em>8+4=12</em>

____________

5z*2=40

5(4)*2=40

<em>20*2=40</em>

<h2 />

6 0

3 years ago

In triangle PQR, PR = 23mm, QR = 39mm, and m<R = 163 degrees. Find the area of the triangle to the nearest tenth.

Schach [20]

Answer: 131.1287 square mm (approx)

Step-by-step explanation:

The area of a triangle,

$A=\frac{1}{2} \times s_1\times s_2\times sin \theta$

Where $s_1$ and $s_2$ are adjacent sides and $\theta$ is the include angle of these sides,

Here PR and QR are adjacent sides and ∠R is the included angle of these sides,

Thus, we can write,

$s_1 = PR= 23\text{ mm}$ , $s_2=QR=39\text{ mm}$ and $\theta = 163^{\circ}$ ,

Thus, the area of the triangle PQR,

$A=\frac{1}{2} \times 23\times 39\times sin163^{\circ}$

$A=\frac{262.257419136}{2} = 131.128709568\approx 131.1287\text{ square mm}$

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3 years ago

What is the equation for the line of reflection <br>a. x= 3 <br>b. y = 3 <br>c. y = x <br>d. x = 6

Stells [14]

Answer:

The answer to that question is c. y = x

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3 years ago