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bija089 [108]
3 years ago
9

A rancher has a rectangular 40 acre piece of property fronting a highway. He wants to put a barbed-wire fence along his road fro

ntage. How much fence does he need if his property depth is 1,200’?
Mathematics
1 answer:
Lyrx [107]3 years ago
8 0
An acre is a surface measure, used in agriculture in several countries.
 According to the country and the time, it is equivalent to several surfaces.
 We will use the following conversion:
 1 acre = 43560 feet ^ 2
 Thus:
 40 acre = 1,742,400 feet ^ 2
 The area of the rectangle is:
 1,742,400 = (1,200) * (x)
 Clearing the other dimension we have:
 x = (1742400) / (1200)
 x = 1452 feet
 Answer:
 He needs 1452 feet of fence.
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Suppose the initial height of a Pumpkin is 12 feet and the pumpkin is being launched with a velocity of 61 feet per second. Use
iogann1982 [59]
<h2>Hello!</h2>

The answer is:

The maximum height before landing will be 69.7804 feet.

<h2>Why?</h2>

Since there is no information about the angle of the launch, we can safely assume that it's launched vertically.

So, we can calculate the maximum height of the pumpkin using the following formulas:

y=y_o+v_{o}*t-\frac{1}{2}gt^{2}

v=vo-gt

Where,

y, is the final height

y_o, is the initial height

g, is the acceleration of gravity , and it's equal to:

g=32.2\frac{ft}{s^{2} }

t, is the time.

Now, we are given the following information:

y_{o}=12ft\\\\v=61\frac{ft}{s}

Then, to calculate the maximum height, we must remember that at the maximum height, the speed tends to 0, so, calculating we have:

Time calculation,

We need to use the following equation,

v=vo-gt

So, substituting we have:

v=61\frac{ft}{s}-32.2\frac{ft}{s^{2}}*t\\\\-61\frac{ft}{s}=-32.2\frac{ft}{s^{2}}*t\\\\t=\frac{-61{ft}{s}}{-32.2\frac{ft}{s^{2}}}=1.8944s

We know that it will take 1.8944 seconds to the pumpkin to reach its maximum height.

Maximum height calculation,

Now, calculating the maximum height, we need to use the following equation:

y=y_o+v_{o}*t-\frac{1}{2}gt^{2}

Substituting and calculating, we have:

y=y_o+v_{o}*t-\frac{1}{2}gt^{2}

y=12ft+61\frac{ft}{s}*1.8944s-\frac{1}{2}32.2\frac{ft}{s^{2}}*(1.8944s)^{2}

y=12ft+61\frac{ft}{s}*1.8944s-\frac{1}{2}32.2\frac{ft}{s^{2}}*(1.8944s)^{2}\\\\y_{max}=12ft+115.5584ft-16.1\frac{ft}{s^{2}}*(3.5887s^{2})\\\\y_{max}=127.5584ft-57.7780ft=69.7804ft

Hence, we have that the maximum height before the landing will be 69.7804 feet.

Have a nice day!

3 0
3 years ago
2. What is the equation in slope-intercept form of the line that passes through thepoints (-4, 47) and (2, -16)?O21 979y=-2*+ 21
trapecia [35]

hello

the points given are (-4, 47) and (2, -16)

let's find the intercept of this equation

\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ x_2=2 \\ y_2=-16 \\ y_1=47 \\ x_1=-4 \end{gathered}\begin{gathered} m=\frac{-16-47}{2-(-4)} \\ m=-\frac{21}{2} \\ slope=-\frac{21}{2} \end{gathered}

now, since we know the value of the slope, we can use that in the standard equation on a straight line

the standard equation of a straight line is given as

\begin{gathered} y=mx+c \\ m=\text{slope} \\ c=\text{intercept} \end{gathered}

we can pick any of the points and solve for intercept

let's use (2, -16)

\begin{gathered} x=2 \\ y=-16 \\ y=mx+c \\ m=-\frac{21}{2} \\ -16=-\frac{21}{2}(2)+c \\ -16=-21+c \\ \text{collect like terms} \\ c=-16+21 \\ c=5 \end{gathered}

now we know the value of intercept (c) = 5 and the slope (m) = 21/2

let's use this to write equation of the straight line

\begin{gathered} y=mx+c \\ y=-\frac{21}{2}x+5 \end{gathered}

from the calculations above, the equation of the straight line is given as y = -21/2x + 5

8 0
1 year ago
Riley's mom has asked him to go to the
Ierofanga [76]

Answer:

4 dollars

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
It is almost the end of the grading period.I have 730 out of 880 points so far in my math class which means I currently have a B
Murljashka [212]
There is at least 90
4 0
3 years ago
FIRST TO ANSWER GET BRAINLEST.
Lostsunrise [7]

Answer:

I believe the answer is B sorry if I'm wrong

3 0
3 years ago
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