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3 years ago

Please help! What is the least amount of material needed to make the mailing tube?

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

3 0

Material needed = πD x height

π x 3 x (3 x 12) //multiply by 12 to change the 3 feet to inches

= 339.29 in²

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Answer: 339.29 in²
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Solve for x. −4≤16−4x

Nitella [24]

Answer:

the answer for x<5

Step-by-step explanation:

Step 1: Simplify both sides of the inequality.

−4<−4x+16

Step 2: Flip the equation.

−4x+16>−4

Step 3: Subtract 16 from both sides.

−4x+16−16>−4−16

−4x>−20

Step 4: Divide both sides by -4.

−4x

−4

−20

−4

x<5

4 0

3 years ago

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5. Explain how you would classify if a square root is rational or irrational.

Alex73 [517]

Answer:

if its not a perfect square(like u can't square it) then its irrational

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Use the method of Lagrange multipliers to find the dimensions of the rectangle of greatest area that can be inscribed in the ell

Tanzania [10]

Answer:

Length (parallel to the x-axis): $2 \sqrt{2}$ ;

Height (parallel to the y-axis): $4\sqrt{2}$ .

Step-by-step explanation:

Let the top-right vertice of this rectangle $(x,y)$ . $x, y >0$ . The opposite vertice will be at $(-x, -y)$ . The length the rectangle will be $2x$ while its height will be $2y$ .

Function that needs to be maximized: $f(x, y) = (2x)(2y) = 4xy$ .

The rectangle is inscribed in the ellipse. As a result, all its vertices shall be on the ellipse. In other words, they should satisfy the equation for the ellipse. Hence that equation will be the equation for the constraint on x and y.

For Lagrange's Multipliers to work, the constraint shall be in the form: $g(x, y) =k$ . In this case

$\displaystyle g(x, y) = \frac{x^{2}}{4} + \frac{y^{2}}{16}$ .

Start by finding the first derivatives of $f(x, y)$ and $g(x, y)$ with respect to x and y, respectively:

$f_x = y$ ,
$f_y = x$ .

$\displaystyle g_x = \frac{x}{2}$ ,
$\displaystyle g_y = \frac{y}{8}$ .

This method asks for a non-zero constant, $\lambda$ , to satisfy the equations:

$f_x = \lambda g_x$ , and

$f_y = \lambda g_y$ .

(Note that this method still applies even if there are more than two variables.)

That's two equations for three variables. Don't panic. The constraint itself acts as the third equation of this system:

$g(x, y) = k$ .

$\displaystyle \left\{ \begin{aligned} &y = \frac{\lambda x}{2} && (a)\\ &x = \frac{\lambda y}{8} && (b)\\ & \frac{x^{2}}{4} + \frac{y^{2}}{16} = 1 && (c)\end{aligned}\right.$ .

Replace the $y$ in equation $(b)$ with the right-hand side of equation $(b)$ .

$\displaystyle x = \lambda \frac{\lambda \cdot \dfrac{x}{2}}{8} = \frac{\lambda^{2} x}{16}$ .

Before dividing both sides by $x$ , make sure whether $x = 0$ .

If $x = 0$ , the area of the rectangle will equal to zero. That's likely not a solution.

If $x \neq 0$ , divide both sides by $x$ , $\lambda = \pm 4$ . Hence by equation $(b)$ , $y = 2x$ . Replace the $y$ in equation $(c)$ with this expression to obtain (given that $x, y >0$ ) $x = \sqrt{2}$ . Hence $y = 2x = 2\sqrt{2}$ . The length of the rectangle will be $2x = 2\sqrt{2}$ while the height will be $2y = 4\sqrt{2}$ . If there's more than one possible solutions, evaluate the function that needs to be maximized at each point. Choose the point that gives the maximum value.

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3 years ago

Help me out question #9

Alex777 [14]

The original height of the plant is 20 inches, and each week for 4 weeks its height increases by 2h inches, so at the end of 4 weeks its height would be:
20 + 4(2h) = 20 + 8h
Now for the next two weeks the plant grows in height by h inches each week, so we need to add a further 2h inches to the equation. Thus the expression that shows the height of the plant after 6 weeks is:
20 + 8h + 2h = 20 + 10h, therefor A is the correct answer

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Is it grabbed or grab in this sentence

ki77a [65]

Answer:

grabbed

Step-by-step explanation:

just lose the is

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