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3 years ago

11

a horticulturalist is building a fence around a rectangular garden using the side of a building for one side of the enclosure. s

he has 81 feet of fencing. what should the dimensions of the enclosre be so that she can maxmize the gardens area?

1 answer:

Harman [31]3 years ago

5 0

The garden area is maximum when the enclosure is a square.
If a is the length of the side of the square then the length of the building is also a.
The perimeter length is 4a made up of 81+a feet, so 81+a=4a and 3a=81 making a=27 feet.

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Which purchase may people be required to save a small amount of money invested regularly that grows over time?

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A ?Or b I’m not really sure

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2 years ago

Read 2 more answers

Question 1,2, and 3 how do i factor those? Can you show the work and explain how?

DiKsa [7]

1: $3n^{2}+9n+6$

notice that each part is divisible by 3

$3n^{2}$ ÷ 3 = $n^{2}$

9n ÷ 3 = 3n

6 ÷ 3 = 2

so it becomes $3(n^{2} +3n+2)$

3n can be rewritten as 2n+n

-you want to rewrite it into two numbers that multiply to the number that's alone (in this case 2)

which would get you

$3(n^{2} +2n+n+2)$

Now that it's rewritten, you can factor out n + 2 from the equation.

<u><em>the answer is </em></u>

3(n+2)(n+1)

And you can check that by multiplying (n+2)(n+1) which is $n^{2} +2n+n+2$ and then each of those by 3, which is $3n^{2} +6n+3n+6$ or $3n^{2}+9n+6$ , our origional equation

2: $28+x^{2} -11x$

So I rewrote this as $x^{2} -11x+28$ (it's the same thing, just reordered using the commutative property)

now -11x can be rewritten as -4x-7x

(remember, the two numbers should multiply to equal 28, which is our constant.)

$x^{2} -4x-7x+28$

now we can factor out x from the first expression and -7 from the second

$x(x-4)-7(x-4)$

and lastly you factor out x-4,

<u><em>which would give you</em></u>

(x-4)(x-7)

Make sure to check your work and make sure it multiplies to $x^{2} -11x+28$

3: $9x^{2} -12x+4$

The first thing I notice when looking at this problem is that both 9 and 4 are perfect squares. Not only that, but they are the squares of 2 and 3, which are products of -12

So if you rewrite 9 as $3^{2}$ and 4 as $2^{2}$ , the equation becomes

$3^{2} x^{2} -12x+2^{2}$

now that $3^{2} x^{2}$ is ugly so it can be turned into $(3x)^{2}$

and -12x can be rewritten as $-2*3x*2$

so our equation now looks like $(3x)^2-2*3x*2+2^{2}$

There's a rule that says $a^{2} -2ab+b^{2} = (a-b)^{2}$

In our case, a=3x and b=2

<u><em>so the final answer is</em></u>

$(3x-2)^2$

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3 years ago

The formula for the future value V (in dollars) of an investment earning simple interest is V=p+prt, where p (in dollars) is the

zhannawk [14.2K]

a)

V=p+prt

now we solve for P

V=P (1+rt)

Divide both sides by (1+rt)

P=V÷(1+rt)...answer

b)

P=V÷(1+rt)

P=3,000÷(1+0.06×5)

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4 years ago

Kayla is pricing whole wheat flour.

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3 years ago

The quantity demanded x of a certain brand of DVD player is 3000/week when the unit price p is $485. For each decrease in unit p

ANTONII [103]

Answer:

Demand Equation: q = 15125 - 25p.

Supply Equation: 80p - 24000 = 9q.

Equilibrium Price: $525.

Equilibrium Quantity: 2000 units.

Step-by-step explanation:

To solve this question, first, the demand equation has to be calculated. Let price be on the y-axis and quantity by on the x-axis. It is given that p = $485 when q = 3000 units. This can also be written as (q₁, p₁) = (3000, 485). It is also given that when the price goes down by $20, the quantity increases by 500 units. Therefore, (q₂, p₂) = (3500, 465). Due to the statement "For each decrease in unit price of $20 below $485, the quantity demanded increases by 500 units", the demand function will be linear i.e. a straight line. This is because the word "for each" has been used. Therefore, finding the equation of the demand function:

(p - p₁)/(q - q₁) = (q₂ - q₁)/(p₂ - p₁).

(p - 485)/(q - 3000) = (465 - 485)/(3500 - 3000).

(p - 485)/(q - 3000) = (-20)/(500).

(p - 485)/(q - 3000) = -1/25.

25*(p - 485) = -1*(q - 3000).

25p - 12125 = -q + 3000.

25p + q = 15125.

q = 15125 - 25p. (Demand Equation).

Second, find the supply equation. It is given that the suppliers will not sell the product below or at the price of $300. Therefore, (q₁, p₁) = (0, 300). Also, the suppliers will sell 2000 units if the price is $525. Therefore, (q₂, p₂) = (2000, 525). Since it is mentioned that supply equation is linear, therefore:

(p - p₁)/(q - q₁) = (q₂ - q₁)/(p₂ - p₁).

(p - 300)/(q - 0) = (525 - 300)/(2000 - 0).

(p - 300)/(q) = (225)/(2000).

(p - 300)/(q) = 9/80.

80*(p - 300) = 9*(q).

80p - 24000 = 9q. (Supply Equation).

Equilibrium exists when demand = supply. This means that the demand and the supply equations have to be solved simultaneously. Therefore, put demand equation in supply equation:

80p - 24000 = 9(15125 - 25p).

80p - 24000 = 136125 - 225p.

305p = 160125.

p = $525.

Put p = $525 in demand equation:

q = 15125 - 25p.

q = 15125 - 25(525).

q = 15125 - 13125.

q = 2000 units.

To summarize:

Demand Equation: q = 15125 - 25p!!!

Supply Equation: 80p - 24000 = 9q!!!

Equilibrium Price: $525!!!

Equilibrium Quantity: 2000 units!!!

8 0

3 years ago