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2 years ago

12

A neighbor needs to put up a square fence around her garden to keep the deer out. If the area is 500 square feet, how much fence

does she need to buy, rounded to the nearest tenth of a foot? Show your work.

1 answer:

kolezko [41]2 years ago

8 0

Area of a square = length * width

And since it’s a square the length and width is the same amount

So width=length=L
500 ft^2=L*L=L^2

Square root both sides
Sqrt(500 ft^2) =L= 22.36067977

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Perpendicular to y = 2x – 6<br> and passing through (4,3)

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y = 2x – 6 slope = 2

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3 years ago

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What is the simplified form of 30 times x to the sixth power over 14 times y to the fifth power times the fraction 7 times y-squ

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If we convert the given in its mathematical form, we have,

(30x⁶/14y⁵)(7y²/6x⁴)
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((6x⁴)(5x²)/(7y²)(2y³))(7y²/6x⁴)

Cancellation of the common terms will give us an answer of,

<em>5x²/2y³

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3 years ago

A bridge in the shape of an arch connects two cities separated by a river. The two ends of the bridge are located at (–7, –13) a

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Answer:

$y=-\dfrac{13}{49}x^2$

Step-by-step explanation:

The shape of an arch corresponds to a parabola.

the general equation for a parabola is:

$y=ax^2+bx+c$

we're given three coordinates: (-7,-13),(7,-13) and (0,0)

so we can plug these values in the general equation to make 3 separate equations:

(x,y) = (-7,-13)

$-13=a(-7)^2+b(-7)+c$

$49a-7b+c=-13$

(x,y) = (7,-13)

$-13=a(7)^2+b(7)+c$

$49a+7b+c=-13$

(x,y) = (0,0)

$0=a(0)^2+b(7)+c$

$c=0$

so we have three equations. and we can solve them simultaneously to find the values of a,b, and c.

we've already found c = 0, let's use substitute it to other equations.

$49a-7b+c=-13\quad\Rightarrow\quad49a-7b=-13$

$49a+7b+c=-13\quad\Rightarrow\quad49a+7b=-13$

we can solve these two equation using the elimination method, by simply adding the two equations

$\quad\quad49a-7b=-13\\+\quad49a+7b=-13$

------------------------------

$\quad\quad 98a=-26$

$\quad\quad a=-\dfrac{13}{49}$

Now we can plug this value of a in any of the two equations.

$49a-7b=-13$

$49\left(-\dfrac{13}{49}\right)-7b=-13$

$-13-7b=-13$

$-7b=0$

$b=0$

We have the values of a,b, and c. We can plug them in the general equation to find the equation of the arch.

$y=\left(-\dfrac{13}{49}\right)x^2+0x+0$

$y=-\dfrac{13}{49}x^2$

$49y=-13x^2$

This our equation of the arch!

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3 years ago