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Nata [24]
3 years ago
15

Fill in the missing amounts

Mathematics
1 answer:
Shalnov [3]3 years ago
5 0

Answers:

                                     Mar        Abr         May

Receipts                     $1,155     $1,250    $945

Expenses                    $1,100    $990      $900

Net Cash Flow            $55        $260     $45

Cumulative Balance   $55        $315      $360


Solution:

1) Mar

Cumulative Balance Mar =Net Cash Flow Mar

Cumulative Balance Mar =$55

2) Abr

Net Cash Flow Abr=Receipts Abr-Expenses Abr

Net Cash Flow Abr=$1,250-$990

Net Cash Flow Abr=$260

3) May

Net Cash Flow May=Receipts May-Expenses May

$45=Receipts May-$900

Solving for Receipts May: Adding $900 both sides of the equation:

$45+$900=Receipts May-$900+$900

$945=Receipts May

Receipts May =$945


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2 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

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\mathrm{Multiply\:by\:the\:conjugate}\:\frac{\sqrt{2}}{\sqrt{2}}

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=\frac{\sqrt{14}}{4}

=\frac{\sqrt{14}}{4}.

3 0
3 years ago
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