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3 years ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar.

Mathematics

2 answers:

UkoKoshka [18]3 years ago

5 0

Answer:

treyeryrthgfhfgh

Step-by-step explanation:

hgfhgfhfhggh

OverLord2011 [107]3 years ago

4 0

Answer:

aaaaaaaaaaaaaaaaaaaaaAaaaaaaaaaaa

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Read 2 more answers

Jasmin, Quinn, and Maddy are rock climbing. Jasmin is 42 ½ feet below Quinn. Maddy is 67 ¾ feet above Jasmin. What is Maddy's po

ladessa [460]

Answer:

$25\frac{1}{4}$ feet

Step-by-step explanation:

Given

$Maddy = 67\frac{3}{4}$ above Jasmin

$Jasmin = 42\frac{1}{2}$ below Quinn

Required

Determine Maddy's position relative to Quinn's

From the given parameter, one can easily deduce that Quinn is in between Maddy and Jasmin (See attachment)

Using the attachment as a point of reference;

$z = 67\frac{3}{4}$

$x = 42\frac{1}{2}$

The relationship between x, y and z is

$z = y + x$

Make y the subject of formula

$y = z - x$

Substitute the values of z and x

$y = 67\frac{3}{4} - 42\frac{1}{2}$

Convert to improper fractions

$y = \frac{271}{4} - \frac{85}{2}$

Take LCM

$y = \frac{271 - 170}{4}$

$y = \frac{101}{4}$

$y = 25\frac{1}{4}$

Hence; Maddy's position relative to Quinn's is  $25\frac{1}{4}$  feet

8 0

3 years ago

The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much

Y_Kistochka [10]

Answer:

x = ∛ 2*V/5

y = ∛ 2*V/5

h = V/ ∛ 4*V²/25

Step-by-step explanation:

Dimensions of the aquarium base is x*y

We call c₁ cost per unit area of the sides, then cost per unit area of slate is equal 5c₁.

let call h the height of the aquarium then volume of the aquarium is:

V = x*y*h where h = V / x*y

As the base is a rectangular one there are 2 sides x*h . and 2 sides y*h

According to this:

Ct (cost of aquarium ) = cost of the base + cost of the sides

cₐ ( cost of the base) = 5*c₁*x*y

c₆ (cost of the sides ) = c₁*2*x*h + c₁*2*y*h

C(t) = 5*c₁*x*y +2* c₁*x* V/x*y + 2* c₁*y* V/x*y or

C(t) = 5*c₁*x*y + 2*c₁*V/y *2*c₁* V/x

Taking partial derivatives en x and y we have:

C´(x) = 5*c₁*y - 2*c₁*V/x²

C´(y) = 5*c₁*x - 2*c₁*V/y²

C´(x) = C´(y) ⇒ 5*c₁*y - 2*c₁*V/x² = 5*c₁*x - 2*c₁*V/y²

or 5*y - 2*V/x² = 5*x - 2*V/y²

(5*y*x² - 2*V)/x² = ( 5*y²x - 2*V) /y²

(5*y*x² - 2*V)*y² = ( 5*y²x - 2*V)*x²

5*y³*x² - 2*V*y² = 5*y²x³ - 2*V*x²

5*y³*x² - 5*y²x³ = 2*V * ( y² - x²)

by symmetry x = y

Then using x = y and plugging that value on the derivatives

C´(x) = 5*c₁*y - 2*c₁*V/x²

C´(x) = 5*c₁*x - 2*c₁*V/x²

C´(x) = 0 ⇒ 5*c₁*x - 2*c₁*V/x² = 0

5*x - 2*V/x² = 0 ⇒ 5*x³ - 2*V = 0 ⇒ 5*x³ = 2*V ⇒ x³ = 2*V/5

x = ∛ 2*V/5 and y = ∛ 2*V/5 and h = V/ x*y h = V/ ∛ 4*V²/25

7 0

3 years ago