Answer:
Total area = 237.09 cm²
Step-by-step explanation:
Given question is incomplete; here is the complete question.
Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)
From the figure attached,
Area of the right triangle I = 
Area of ΔADC = 
= 
= 
= 
= 
= 30 cm²
Area of equilateral triangle II = 
Area of equilateral triangle II = 
= 
= 73.0925
≈ 73.09 cm²
Area of rectangle III = Length × width
= CF × CD
= 7 × 5
= 35 cm²
Area of trapezium EFGH = 
Since, GH = GJ + JK + KH
17 = 
12 = 
144 = (81 - x²) + (225 - x²) + 2
144 - 306 = -2x² + 
-81 = -x² + 
(x² - 81)² = (81 - x²)(225 - x²)
x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴
144x² - 11664 = 0
x² = 81
x = 9 cm
Now area of plot IV = 
= 99 cm²
Total Area of the land = 30 + 73.09 + 35 + 99
= 237.09 cm²
I believe it would be 1.5 cm squared because 7.5/5 = 5
Answer:
d = 1.1 units
Step-by-step explanation:
Since the x- coordinates of the 2 points are equal, both 3, then they lie on a vertical line. The distance between the 2 points is the difference in the y- coordinates, that is
d = | 2.4 - 1.3 | = | 1.1 | = 1.1 units
19 players.
2033-1225.5= 807.5
807.5/42.5=19
In order to vertically stretch the parent function, you have to make the absolute value of the coefficient of x^3 larger. meaning that the coefficient of x^3 has to be greater than one or less than negative one