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3 years ago

What is the difference between Partial Product and Regrouping? Describe.

Mathematics

1 answer:

Nonamiya [84]3 years ago

6 0

Partial Product is for multiplication and regrouping is for addition.

Hope that helped:)

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Help on this to please

d1i1m1o1n [39]

Answer:

7). up 4 over 4

8). up 1 over 2

Step-by-step explanation:

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3 years ago

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2x -2 = -200 whats the value of x

Elden [556K]

Answer:

Step-by-step explanation:

2x - 2 = - 200

Bringing like terms on one side

2x = - 200 + 2 = - 198

2x = - 198

x = - 198/2 = - 99

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3 years ago

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Max has sticks of lengths 6,7,8 and 10 inches. He wants to make a right triangle for an art project. Which 3 sticks should he us

MrMuchimi

C since that's the only triangle where A^2+B^2=C^2 (36+64=100).

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3 years ago

. 9 triangles per 6 circles Triangles Circles 9 | 24 8 30 | 50 atio.

professor190 [17]

Answer: 9840

Step-by-step explanation:

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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