1^5+3-(2×3^2)÷20÷4+2

Asha and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is 512 the full and Samuel’s shelf is 1

Mnenie [13.5K]

Answer:

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3 0

2 years ago

What is the difference between a constant and a variable

denis-greek [22]

Answer:

A constant is a number that stays the same. A variable is number that doesn't remain the same.

7 0

3 years ago

Read 2 more answers

Find the length and width of 4 different rectangles such that each rectangle has an area of 24 square units. Write the length an

Ksenya-84 [330]

Answer:

77

Step-by-step explanation:

because im smart btw no treats involved

5 0

3 years ago

MARSVECTORCALC6 3.4.020. My Notes A rectangular box with no top is to have a surface area of 64 m2. Find the dimensions (in m) t

ioda

Answer:

We would have

$l =w =\frac{8\sqrt{3}}{3} \\h = \frac{8\sqrt{3}}{6}$

where " l " is length, " w" is width and "h" is height.

Step-by-step explanation:

Step 1

Remember that

Surface area for a box with no top = $lw+2lh+2wh = 64$

where " l " is length, " w" is width and "h" is height.

Step 2.

Remember as well that

Volume of the box = $l*w*h$

Step 3

We can now use lagrange multipliers. Lets say,

$F(l,w,h) = lwh$

and

$g(l,w,h) = lw+2lh+2wh = 64$

By the lagrange multipliers method we know that

$\nabla F = \lambda \nabla g$

Step 4

Remember that

$\nabla F = (wh,lh,lw)$

and

$\nabla g = (w+2h,l+2h , 2w+2l)$

So basically you will have the system of equations

$wh = \lambda (w+2h)\\lh = \lambda (l+2h)\\lw = \lambda (2w+2l)$

Now, remember that you can multiply the first eqation, by "l" the second equation by "w" and the third one by "h" and you would get

$lwh = l\lambda (w+2h)\\\\lwh = w\lambda (l+2h)\\\\lwh = h\lambda (2w+2l)$

Then you would get

$l\lambda (w+2h) = w\lambda (l+2h) = h\lambda (2w+2l)$

You can get rid of $\lambda$ from these equations and you would get

$lw+2lh = lw+2wh = 2wh+2lh$

And from those equations you would get

$l = w =2h$

Now remember the original equation

$lw+2lh+2wh = 64$

If we plug in what we just got, we would have

$l^{2} + l^{2} + l^2 = 64 \\3l^{2} = 64 \\l = w = \frac{8\sqrt{3} }{3} \\h = \frac{8\sqrt{3} }{6}$

7 0

3 years ago

2. In circle 0, points A and B lie on the circle such that OA = 2x + 9 and OB = 27 – x. Which of the following

malfutka [58]

6 because you set them equal and solve

4 0

2 years ago