Matrix :
1 4 4
5 2 2
1 5 5
=> Determinant =
1 * (2*5 - 2*5) - 4 * (5*5 - 1*2) + 4 * ( 5*5 - 2*1)
= 1*(10 - 10) - 4*(25 - 2) + 4*(25 -2) = 1*0 + 4* 23 + 4*(25 - 2) = 0 - 4*23 + 4*23 = 0
Answer: 0
Answer:
Step-by-step explanation:
<u>The area is:</u>
- 22*4 + 2*(1/2)*4*4 = 88 + 16 = 104 cm²
Correct choice is A
Bro just go on Symbolab and use the calculator
Lets find 2 points, one obvious one is (0,0), another point is (12,4)

y = mx + b
m is the slope
we got m = 1/3
We can eliminate answer choices C and D.
y = 1/3x + b
we need to solve for b
Plug in the point (0,0) and solve for b
0 = 1/3(0) + b
0 = 0 + b
b = 0
The equation of the line is y = 1/3x + 0, or
y = 1/3xYour answer is
B.
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