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2 years ago

6

The back of Tom's property is a creek. Tom would like to enclose a rectangular area, using the creek as one side and fencing for

the other three sides, to create a pasture. If there is 920 feet of fencing available, what is the maximum possible area of the pasture

1 answer:

pickupchik [31]2 years ago

4 0

Answer:

Max Area = 105,800 sq.ft

Step-by-step explanation:

A square will always give us the maximum area.

Thus, one side would be;

920/4 = 230 feet

So, we want a square 230 ft by 230 ft

however, from the question, we are to use the creek as one side. So, we'll take the 230 ft that we don't need because of the creek and then add it to the opposite side to get 230 + 230 = 460 ft.

Thus,we now have a rectangle with dimensions: 230 ft by 460 ft

Area is given by;

area = length × width

Maximum Area = 230 × 460

Max Area = 105,800 sq.ft

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What is the domain of f(x) = (x + 2)2 – 1?

larisa [96]

Simplify

$\\ \tt\Rrightarrow f(x)=(x+2)^2-1$

$\\ \tt\Rrightarrow f(x)=x^2+4x+2-1$

$\\ \tt\Rrightarrow f(x)=x^2+4x+1$

The Domain is set of all real numbers

$\\ \tt\Rrightarrow Domain\in R$

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1 year ago

The sum of twelve and a number is no greater than sum of twice the number and -8

Leona [35]

1. Write as an algebraic equation:

12 + n ≤ 2n - 8

2. Solve for n.

12 + n - 12 ≤ 2n - 8 - 12

n ≤ 2n - 20

n - 2n ≤ 2n - 20 - 2n

-n ≤ -20

Multiply both sides by -1 (this means you must flip the inequality sign).

n ≥ 20

FINAL ANSWER: n ≥ 20

Hope this helps! Feel free to ask for clarification.

7 0

2 years ago

Need help please now

ZanzabumX [31]

It’s A letter A hope this helped

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2 years ago

In the 1st generation, there are 6 squirrels in a forest. Every generation after

igomit [66]

Answer:

You are looking for a geometric series equation. The first term of the sequence is 6 and the common ratio is 4 hence the equation is 6(4)^n-1 where n is the number of generations.

Step-by-step explanation:

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3 years ago

The general form of an parabola is 2x2−12x−3y+12=0 .

Murljashka [212]

Answer:

The standard form of the parabola is $(x-3)^2=4*\frac{3}{8}(y+2)$

Step-by-step explanation:

The standard form of a parabola is

$(x-h)^2=4p(y-k)$ .

In order to convert $2x^2-12x-3y+12=0$ into the standard form, we first separate the variables:

$2x^2-12x+3y+12=0\\\\2x^2-12x+12=3y$

we now divided both sides by 2 to remove the coefficient from $2x^2$ and get:

$x^2-6x+6=\frac{3}{2}y$ .

We complete the square on the left side by adding 3 to both sides:

$x^2-6x+6+3=\frac{3}{2}y+3$

$x^2-6x+9=\frac{3}{2}y+3$

$(x-3)^2=\frac{3}{2}y+3$

now we bring the right side into the form $4p(y-k)$ by first multiplying the equation by $\frac{2}{3}$ :

$\frac{2}{3} *(x-3)^2=\frac{2}{3} *(\frac{3}{2}y+3)\\\\\frac{2}{3} *(x-3)^2=y+2$

and then we multiplying both sides by $\frac{3}{2}$ to get

$(x-3)^2=\frac{3}{2} (y+2)$ .

Here we see that

$4p=\frac{3}{2}$

$\therefore p=\frac{3}{8}$

Thus, finally we have the equation of the parabola in the standard form:

$\boxed{(x-3)^2=4*\frac{3}{8}(y+2)}$

5 0

2 years ago