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Alex73 [517]
3 years ago
7

A square vegetable garden has side lengths of 14 feet. You plant flowers in the center portion of the garden,

Mathematics
2 answers:
Eva8 [605]3 years ago
8 0

Answer:

45ft

Step-by-step explanation:

Usimov [2.4K]3 years ago
3 0

Answer:

Area of tomato section is 45 ft^2

Step-by-step explanation:

The total area for garden is the area of a square with side 14 ft.

That means the garden's total area is the square of its side:

Area of a square of side "x" = x*x=x^2, then for our case where the square's side is 14 ft, the square's area becomes: Area=(14\,ft)^2 = 196\, ft^2

Inside this area, there is a square of side 4 ft reserved for flowers (and what is left -the rest- for veggies). The area of the flower section is then: (4\,ft)^2=16\,ft^2. Therefore the rest of the garden that is intended for veggies is (the total garden area minus the flower section area) = 196\,ft^2-16\,ft^2=180\,ft^2

This remaining veggie are is to be divided in four (4) equal parts, and one of them destined for tomato plants. The area of each of the veggie sections is one fourth of the remaining area: \frac{180\,ft^2}{4} =45\, ft^2

So in particular the tomato plant area is 45 ft^2

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A very large tank initially contains 100L of pure water. Starting at time t = 0 a solution with a salt concentration of 0.8kg/L
Scorpion4ik [409]

Answer:

1. \dfrac{dy}{dt}=4-\dfrac{3y(t)}{100+2t}

2. y(40) = 110.873 \ kg

Step-by-step explanation:

Given that:

A very large tank initially contains 100 L of pure water.

Starting at time t = 0 a solution with a salt concentration of 0.8kg/L is added at a rate of 5L/min.

. The solution is kept thoroughly mixed and is drained from the tank at a rate of 3L/min.

As 5L/min is entering and 3L/min is drained out, there is a 2L increase per minute. Therefore, the amount of water at any given time t = (100 +2t) L

t = (50 + t ) L

Since it is given that we should  consider y(t) to be the  amount of salt (in kilograms) in the tank after t minutes.

Then , the differential equation that  y satisfies can be computed as follows:

\dfrac{dy}{dt}=rate_{in} - rate_{out}

\dfrac{dy}{dt}=(0.8)(5) -\dfrac{y(t)}{100+2t} \times3

\dfrac{dy}{dt}=(0.8)(5) -\dfrac{3y}{100+2t}

\dfrac{dy}{dt}=4-\dfrac{3y(t)}{100+2t}

How much salt is in the tank after 40 minutes?

So,

suppose : e^{\int \dfrac{3}{100+2t} \ dt} = (t+50)^{3/2}

Then ,

( t + 50)^{3/2} y' + \dfrac{3}{2}(t+50)^{1/2} y = 4(t+50)^{3/2}

( t + 50)^{1.5} y' + \dfrac{3}{2}(t+50)^{0.5} y = 4(t+50)^{3/2}

[y\ (t + 50)^{1.5}]' = 4(t+ 50)^{1.5}

Taking the integral on both sides; we have:

[y(t + 50)^{1.5}] = 1.6 (t + 50)^{2.5} + C

y = 1.6 (t+50)+C(t+50)^{-1.5}

y(0) = 0 = 1.6(0+50) + C ( 0 + 50)^{-1.5}

0 = 1.6(50) + C ( 50)^{-1.5}

C= -1.6(50)^{2.5}

y(40) = 1.6 (40 + 50)^1  - 1.6 (50)^{2.5}(50+40)^{-1.5}

y(40) = 144  - 1.6 \times 17677.66953 (90)^{-1.5}

y(40) = 144  - 1.6 \times 17677.66953 \times 0.001171213948

y(40) = 144  - 33.12693299

y(40) = 110.873 \ kg

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