2 years ago

find the length and width of a rectangle that has an area of 968 square feet and whose perimeter is a minimum

Mathematics

1 answer:

andrezito [222]2 years ago

3 0

The question isn't finished...

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Find the measure of angle b (image below)

prohojiy [21]

Answer:

123+b=180 (being sum of straight angle)

b=180-123

b=57

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3 years ago

3 is less than the sum of a number s and 4

monitta

3 is less than the sum of any positive number added with 4

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Zadanie optymalizacyjne z matematyki. Proszę o rozwiązanie i wyjaśnienie

Yakvenalex [24]

Volume of the pyramid:

$V=\dfrac{s^2h}3$

Perimeter of the cross-section:

$40=\sqrt2\,s+2\sqrt{\dfrac{s^2}2+h^2}=\sqrt2\left(s+\sqrt{s^2+2h^2}\right)$

$\implies h=\sqrt{\dfrac{(20\sqrt2-s)^2-s^2}2}=\sqrt{400-20\sqrt2\,s}$

Area of the cross-section:

$P=\dfrac12(\sqrt2\,s)h=\dfrac{sh}{\sqrt2}$

$\implies P=\dfrac{s\sqrt{400-20\sqrt2\,s}}{\sqrt2}=s\sqrt{200-10\sqrt2\,s}$

First derivative test:

$\dfrac{\mathrm dP}{\mathrm ds}=\dfrac{20\sqrt2-3s}{\sqrt{4-\dfrac{\sqrt2}5s}}=0\implies s=\dfrac{20\sqrt2}3$

Then the height of the cross-section/pyramid is

$h=\sqrt{400-20\sqrt2\,s}=\dfrac{20}{\sqrt3}$

The volume of the pyramid that maximizes the cross-sectional area $P$ is

$V=\dfrac{\left(\frac{20\sqrt2}3\right)^2\frac{20}{\sqrt3}}3=\dfrac{16000}{27\sqrt3}$

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3 years ago

Question about dilation attached in images.

statuscvo [17]

This is an enlarged dilation

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3 years ago

108+621 is the same as 100+

notsponge [240]

Answer:

108+621= 100+629 hhh grade 2?

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