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3 years ago

Look at this graph:

Mathematics

2 answers:

IRINA_888 [86]3 years ago

5 0

I can’t see the picture of the graph

sergejj [24]3 years ago

4 0

The answer would be 4/3 because you have to find the change in y over the change in x.

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A yardstick placed perpendicular to the ground casts a 4 foot shadow. At the same time, a grain silo casts a 34 foot shadow. Wri

kobusy [5.1K]

There is a 30 ft difference how can you not tell

4 0

3 years ago

13 + what = 23.6? I need help i will give u brainlest if you help plz

Bas_tet [7]

Answer:

10.6

Step-by-step explanation:

13+10.6= is 23.6

Hope this helps

3 0

3 years ago

Read 2 more answers

Please help me with theses 4 questions

MA_775_DIABLO [31]

1. 864 mm^3
2.1640 cm^3
3. 4.0 cm
4. 150pi ft^3

3 0

4 years ago

What is 100.023424234 divided by 10.32342 then added by 1000.293920 and then subtracted by 20.2323?

Nata [24]

989.75060139
100.023424234 divided by 10.32342 = 9.68898138737
9.68898138737 + 1000.293920 = 1009.98290139
1009.98290139 - 20.2323 = 989.75060139

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

4 years ago