When the base of a triangle is fixed at 2.218 millimeters, the area will be 0.1098 mm².
<h3>What will the area be?</h3>
It should be noted that the area of a triangle is calculated as:
Area of triangle = (1/2)×base×height
= (1/2) × 2.218 mm × 0.099 mm
= 1.109 mm × 0.099 mm
= 0.109791 mm²
≈ 0.1098 mm²
Therefore, the <em>area</em> is 0.1098 mm².
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The base of a triangle is fixed at 2.218 millimeters. Determine the number of
significant figures of the area of the triangle with each given height of 0.099mm.
The given equations are:
5x - 2y = 88
3x + 4y = 58
Multiplying the 1st equation by 2, we get the new set of equations as:
10x - 4y = 176
3x + 4y = 58
Adding the two equations, we get:
10x - 4y + 3x + 4y = 176 + 58
13x =234
x = 18
Using the value of x in 1st equation, we get:
5(18) - 2y = 88
- 2y = 88 -5(18)
-2y = -2
y = 1
So, the solution of the equation is (18, 1)
So, 9 out of 10 students prefer class, so, obviously, it would not be 100. It would be 150. So in conclusion, 50 students prefer lunch over math class, whilst 150 other students prefer math class over lunch.
The equation of a line starting from two points is:

From the first point you get: x1 = -1, y1 = -2
From the second point you get: x2 = 3, y2 = 10
Replace x1, y1, x2, y2 in the equation of the line and you get:



From this you get the equation of your line: