Find the surface area of a triangular prism that has the following dimensions. Use the formula SA = 2 B + Ph . B = 84 mm 2 P = 5
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Answer:
Axis is a vertical line at x = 2
Vertex is (2, -1)
y-intercept is (0, 3)
Solutions are x = 1 and x = 3
Step-by-step explanation:
To draw the graph of the quadratic equation you must find at least 5 points lie on the graph by choose values of x and find their values of y
Let us do that
Use x = -1, 0, 1, 2, 3, 4, 5
∵ y = x² - 4x + 3
∵ x = -1
∴ y = (-1)² - 4(-1) + 3 = 1 + 4 + 3 = 8
→ Plot point (-1, 8)
∵ x = 0
∴ y = (0)² - 4(0) + 3 = 0 + 0 + 3 = 3
→ Plot point (0, 3)
∵ x = 1
∴ y = (1)² - 4(1) + 3 = 1 - 4 + 3 = 0
→ Plot point (1, 0)
∵ x = 2
∴ y = (2)² - 4(2) + 3 = 4 - 8 + 3 = -1
→ Plot point (2, -1)
∵ x = 3
∴ y = (3)² - 4(3) + 3 = 9 - 12 + 3 = 0
→ Plot point (3, 0)
∵ x = 4
∴ y = (4)² - 4(4) + 3 = 16 - 16 + 3 = 3
→ Plot point (4, 3)
∵ x = 5
∴ y = (5)² - 4(5) + 3 = 25 - 20 + 3 = 8
→ Plot point (5, 8)
→ Join all the points to form the parabola
From the graph
∵ The axis of symmetry is the vertical line passes through the vertex point
∵ x-coordinate of the vertex point is 2
∴ Axis is a vertical line at x = 2
∵ The coordinates of the vertex point of the parabola are (2, -1)
∴ Vertex is (2, -1)
∵ The parabola intersects the y-axis at point (0, 3)
∴ y-intercept is (0, 3)
∵ x² - 4x + 3 = 0
∵ The solutions of the equation are the values of x at y = 0
→ That means the intersection points of the parabola and the x-axis
∵ The parabola intersects the x-axis at points (1, 0) and (3, 0)
∴ Solutions are x = 1 and x = 3
The equation p=4+3y is maximised at (4,5) and the maximised value is 31.
Given equation p=4x+3y and constraints x>0,x<4, y<5.
We have to maximise the equation p=4x+3y with constraints x>0,x<4,y<5.
To find out the points we have to make graphs of the constraints first.
From the graph we have found that points are (0,0),(0,5),(4,0),(4,5).
The above points are common points of graphs of all constraints.
The value of equations are as under:
At (0,0) , p=4*0+3*0=0
At (0,5), p=4*0+3*5=15
At (4,0),p=4*4+3*0=16
At (4,5),p=4*4+3*5=16+15=31
Maximum value of p is 31 and that is for (4,5).
Hence the equation is maximised at (4,5) and the maximum value is 31.
Learn more about equation at brainly.com/question/2972832
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