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3 years ago

A motorboat can maintain a constant speed of 17 miles

Mathematics

1 answer:

maxonik [38]3 years ago

6 0

Answer:

10mph

Step-by-step explanation:

by adding total time taken to travel and return and simply doing math

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A chemical refinery needs a vat. They want the vat to be a rectangular prism (with square bases) that has a maximum volume of 1,

Alex17521 [72]

Answer:

For least material to be used lengths of square base and sides = 10 units.

Step-by-step explanation:

Let the lengths of the square base and the sides = x feet, x feet and y feet

Area of the square base = x² feet

Volume of the rectangular prism = Area of the square base × Height

= x²y cubic feet

1000 = x²y

y = $\frac{1000}{x^2}$ -------(1)

Material used in the prism = Surface area of the rectangular prism

= 2(lb + bh + hl)

Here, h = height of the prism

l = length of the base

w = Width of the base

Material to be used (S) = 2(xy + x² + xy) - Area of lid

S = 2(x² + 2xy) - x²

S = x² + 2xy

Now by substituting the value of y from equation (1),

S = x² + $2x(\frac{1000}{x^{2} })$

= x² + $\frac{2000}{x}$

For least amount of material used,

We will find the derivative of the given function and equate it to zero.

S' = 2x - $\frac{2000}{x^{2} }$

2x - $\frac{2000}{x^{2} }$ = 0

2x³ = 2000

x³ = 1000

x = 10 feet

From equation (1),

y = $\frac{1000}{(10)^2}$

y = 10 feet

Therefore, for least amount of the material used lengths of square base and sides will be 10 feet.

7 0

3 years ago

What is the form of the two squares identity?

zvonat [6]

Answer:

Step-by-step explanation:

Consider all options:

A. False, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ab-cd)^2+(ac+bd)^2=a^2b^2-2abcd+c^2d^2+a^2c^2+2abcd+b^2d^2=\\ =a^2b^2+c^2d^2+a^2c^2+b^2d^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2\neq a^2b^2+c^2d^2+a^2c^2+b^2d^2$

B. False, because

$(a^2-b^2)(c^2+d^2)=a^2c^2+a^2d^2-b^2c^2-b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2-b^2c^2-b^2d^2\neq a^2c^2+b^2d^2+a^2d^2+b^2c^2$

C. False, because

$(a^2+b^2)(c^2-d^2)=a^2c^2-a^2d^2+b^2c^2-b^2d^2\\ \\(ac+bd)^2-(ad+bc)^2=a^2c^2+2abcd+b^2d^2-a^2d^2-2abcd-b^2c^2=\\=a^2c^2+b^2d^2-a^2d^2-b^2c^2\\ \\a^2c^2-a^2d^2+b^2c^2-b^2d^2\neq a^2c^2+b^2d^2-a^2d^2-b^2c^2$

D. True, because

$(a^2+b^2)(c^2+d^2)=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ \\(ac-bd)^2-(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=\\=a^2c^2+b^2d^2+a^2d^2+b^2c^2\\ \\a^2c^2+a^2d^2+b^2c^2+b^2d^2= a^2c^2+b^2d^2+a^2d^2+b^2c^2$