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zimovet [89]
3 years ago
14

A field is to be fertilized at a cost of $0.08 per square yard. the rectangular part of the field is 95 yd long and the diameter

of each semicircle is 49 yd. Find the cost of fertilizing the field.
Mathematics
1 answer:
Alekssandra [29.7K]3 years ago
8 0
Assuming that this field is rectangular with 2 semi-circles (one on both ends of the field), we must first calculate the total area of the field. this is done by adding the area of the rectangular portion and the circular portion. This is done below:

Given:

Rectangular length = 95 yards
Rectangular width = semi-circle diameter = 49 yards

Area of rectangle = length * width = 95 * 49
Area of rectangle = 4655 yd^2

The area of the 2 semi-circles can be obtained by treating both as one circle.
Area of circle = pi * (d/2)^2 = 3.1416 * (49/2)^2
Area of circle = 1885.74 yd^2

Total area = 1885.74 + 4655 = 6540.74

We multiply the area by the cost of fertilizing:
Total cost = cost per sq. yd * total area 
Total cost = 0.08 * 6540.74
Total cost = $523.26
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Graph ΔABC and its image after a rotation of 180º about the origin.
ValentinkaMS [17]

Answer:

The vertices of image are A'(0,0), B'(-1,-5) and C'(4,-5). The graph of image and preimage is shown below.

Step-by-step explanation:

From the given figure it is noticed that the vertices of  triangle ABC are A(0,0), B(1,5) and C(-4,5).

If a figure rotated at 180º about the origin, then

P(x,y)\rightarrow P'(-x,-y)

The vertices of image are

A(0,0)\rightarrow A'(0,0)

B(1,5)\rightarrow B'(-1,-5)

C(-4,5)\rightarrow C'(4,-5)

Therefore the vertices of image are A'(0,0), B'(-1,-5) and C'(4,-5).

The graph of image and preimage is shown below.

3 0
3 years ago
Find the product. 7/10 X 3/6
tiny-mole [99]
7 . 3
--- × ------
10. . 6


= 21
--------
60

= 7/20
4 0
3 years ago
A college student took 4 courses last semester. His final grades, along with the credits each class is worth, are as follow: A (
NikAS [45]

Answer:

A college student took 4 courses last semester. His final grades, along with the credits each class is worth, are as follow: A (3), B (4), C (2), and D (3). The grading system assigns quality points as follows: A: 4; B: 3; C: 2; D: 1; and F: 0. Find the student’s GPA for this semester. Round your answer to the nearest thousandth.

another way is

This is a weighted average question. You are going to "weight" each course by the number of credits it is worth and then divide by the total number of credits. In other words, you are going to multiply each grade (A=4, B=3) by the number of credits attached to that grade. This will ensure that the courses that have more credits count more in the overall average. Then you are going to divide by the total number of credits to get the overall GPA.

So,

(3*4 + 4*3 + 2 *2 + 3*1)/(3+4+2+3) = GPA

Step-by-step explanation:

bran-list please

6 0
3 years ago
Use quadratic formula to find both solutions to the quadratic equation given below . 3x^2-5x-1=0
Rudiy27

Answer:

x1=\frac{5+\sqrt{37}} {6}

x2=\frac{5-\sqrt{37}} {6}

Step-by-step explanation:

we know that

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

3x^{2} -5x-1=0  

so

a=3\\b=-5\\c=-1

substitute in the formula

x=\frac{5(+/-)\sqrt{(-5)^{2}-4(3)(-1)}} {2(3)}

x=\frac{5(+/-)\sqrt{37}} {6}

x1=\frac{5+\sqrt{37}} {6}

x2=\frac{5-\sqrt{37}} {6}

8 0
3 years ago
Read 2 more answers
A triangle is in the shape of a 30-60-90 right triangle. If the hypotenuse is 11 cm long, find the length of the shorter leg.
soldier1979 [14.2K]
A 30 60 90 triangle has its shorter side equal to one half the hypotenuse.
So, if hypotenuse = 11 cm then shorter side = 5.5.

Source:
http://www.1728.org/trig2.htm


6 0
3 years ago
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