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Lostsunrise [7]
3 years ago
6

A farmer has 100 m of fencing to enclose a rectangular pen. Which quadratic equation gives the area (A) of the pen, given its wi

dth (w)?
Mathematics
2 answers:
Nady [450]3 years ago
7 0

Answer:

A(w)=50w-w^2

Step-by-step explanation:

Let w be the width of rectangular pen.

We have been given that a framer has 100 m of fencing to enclose a rectangular pen. This means that perimeter of pen is 100 meter.

Since the perimeter is 2 times the length and width of rectangle.

\text{Perimeter of rectangle}=2\text{ (Width+Length})

Upon substituting our given values in above formula we will get,

100=2(w\text{+Length})

\frac{100}{2}=\frac{2(w\text{+Length})}{2}

50=w\text{+Length}

50-w=w-w\text{+Length}

50-w=\text{Length}

\text{Area of rectangle}=\text{Width*Length}

Upon substituting our given values in area formula we will get,

\text{Area of rectangle}=w(50-w)

\text{Area of rectangle}=50w-w^2

Let us represent area in terms of width of rectangle as:

A(w)=50w-w^2

Therefore, the area of our given pen will be A(w)=50w-w^2.

inn [45]3 years ago
5 0

A(w)=50w-w2 if the perimeter is 100, I plugged in possible values for the width...I used w=15 and length = 35 for a perimeter of 100. then I plugged these values in the possible equations until I found the correct one giving me an area of 525 (15x35) . I could not think of an easier way.

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2 x 5

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Think of it as "the value 2 is being added together 5 times" being added together so it could be written as 2 times 5.

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In a recent year, 31.6% of all registered doctors were female. If there were 53,000 female registered doctors that year, what wa
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Answer:

The total number of registered doctors was 167,722.

Step-by-step explanation:

Total number of doctors:

The total number of doctors is given by x.

31.6% of all registered doctors were female. 53,000 female doctors.

This means that:

0.316x = 53000

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We have to solve the above equation for x. So

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3 years ago
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3 years ago
The students in Mr. Wilson's Physics class are making golf ball catapults. The
Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is 0 \leq x \leq 48.6\ ft and the range is 0 \leq y \leq 8.3\ ft

Step-by-step explanation:

we have

y=-0.014x^{2} +0.68x

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's  distance from the catapult in feet

y is the flight of the balls in feet

Part a)  How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-0.014x^{2} +0.68x=0

so

a=-0.014\\b=0.68\\c=0

substitute in the formula

x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}

x=\frac{-0.68(+/-)0.68} {(-0.028)}

x=\frac{-0.68(+)0.68} {(-0.028)}=0

x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

y=-0.014x^{2} +0.68x

Find out the derivative and equate to zero

0=-0.028x +0.68

Solve for x

0.028x=0.68

x=24.3

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint  between the x-intercepts

x=(0+48.6)/2=24.3\ ft

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

y=-0.014(24.3)^{2} +0.68(24.3)

y=8.3\ ft

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A  reasonable domain is the distance between the two x-intercepts

so

0 \leq x \leq 48.6\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A  reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

so

we have the interval -----> [0,8.3]

0 \leq y \leq 8.3\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

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