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Misha Larkins [42]
1 year ago
7

A center pivot irrigation system provides water to a sector shaped field, find the area of the field if 0=140 degrees and r= 40

yd
Mathematics
1 answer:
Anastasy [175]1 year ago
4 0

The area of the sector is approximately 1954.77 square yards.

A circular sector, also known as a circle sector or a disc sector (symbol:), is a piece of a disc (a closed region bordered by a circle) encompassed by two radii and an arc. The smaller area is known as the minor sector, and the bigger is known as the major sector.

Any point on the circle outside of the sector can be connected to the arc's endpoints to produce half of the central angle.

We know that the area of a circle is πr² and the area of the sector of a circle with angle θ is given by θ/360 πr² .

Area of the sector

= 140°/360°  × π × 40²

= 1954.7687... square yards.

≈ 1954.77 square yards.

Therefore the area of the sector is approximately 1452.113 square yards.

To learn more about sector visit:

brainly.com/question/16989821

#SPJ1

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Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term x^2, which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form f(x)=ax^2+bx+c is given by the expression: x_v=\frac{-b}{2a}

Since in our case a=1 and b=5, we get that the x-position of the vertex is: x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the x_v=-\frac{5}{2}. Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.

When evaluating the function at these points, we notice that two of them render zero (which indicates they are the actual roots of the equation):

f(-1) = (-1)^2+5(-1)+4= 1-5+4 = 0\\f(-4)=(-4)^2+5(-4)_4=16-20+4=0

The actual graph we can complete with this info is shown in the image attached, where the actual roots (x-axis crossings) are pictured in red.

Then, the two roots are: x = -1 and x = -4.

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