Lori is making 3 mugs of hot chocolate. Each mug holds 1/2 a pint. How many cups of hot chocolate does Lori pour?
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<span>Given that the railroad rail is 6 kilometers before expansion.
When the railroad expands, it makes an isosceles triangle at the center of the rail.
The base of the triangle is 6 kilometers while each leg of the triangle is given by (6 + 10/100,000) / 2 = 6.0001 / 2 = 3.00005 kilometers.
[recall that 100,000 centimeters gives 1 kilometers, thus 10 centimetres gives 10/100,000 kilometers]
Taking one half of the isoceles triangle, we have a right triangle with one of the leg as 6 / 2 = 3 kilometers and the hypothenus = 3.00005 kilometers.
Thus, using the Pythagoras theorem, we find the other leg (height) of the right triangle as follows.
Let the other leg of the right triangle be h, then
Thus, the center of the rail will rise 0.01732058 kilometers above the ground.
But, there are 1,000 meters in 1 kilometers, therefore, </span><span><span>the center of the rail will rise 0.01732058 x 1,000 = 17.32 meters above the ground.</span></span>
When the railroad expands, it makes an isosceles triangle at the center of the rail.
The base of the triangle is 6 kilometers while each leg of the triangle is given by (6 + 10/100,000) / 2 = 6.0001 / 2 = 3.00005 kilometers.
[recall that 100,000 centimeters gives 1 kilometers, thus 10 centimetres gives 10/100,000 kilometers]
Taking one half of the isoceles triangle, we have a right triangle with one of the leg as 6 / 2 = 3 kilometers and the hypothenus = 3.00005 kilometers.
Thus, using the Pythagoras theorem, we find the other leg (height) of the right triangle as follows.
Let the other leg of the right triangle be h, then
Thus, the center of the rail will rise 0.01732058 kilometers above the ground.
But, there are 1,000 meters in 1 kilometers, therefore, </span><span><span>the center of the rail will rise 0.01732058 x 1,000 = 17.32 meters above the ground.</span></span>
Answer:
1) Please find the attached drawing of the archway created with MS Excel
2) The y-intercept is (0, 24)
The x-intercepts are (-3, 0), and (8, 0)
3) The width of the archway at its base is 11
The height of the archway at its highest point = 30.25
Step-by-step explanation:
The given function representing the archway is y = -x² + 5·x + 24
1) Please find attached the required drawing of the archway created with MS Excel
2) The y-intercept is given by the point where x = 0
Therefore, we have, the y-value at the y-intercept = -0² + 5×0 + 24 = 24
The y-intercept = (0, 24)
The x-intercept is given by the point where y = 0
Therefore, the x-values at the x-intercept are found using the following equation;
0 = -x² + 5·x + 24
x² - 5·x - 24 = 0
By inspection, we have;
x² - 8·x + 3·x - 24 = 0
x·(x - 8) + 3·(x - 8) = 0
∴ (x + 3) × (x - 8) = 0
Either (x + 3) = 0, and x = -3, or (x - 8) = 0, and x = 8
Therefore, the x-intercepts are (-3, 0), and (8, 0)
3) The width of the archway at its base = The distance between the x-values at the two x-intercepts
∴ The width of the archway at its base = 8 - (-3) = 11
The highest point of the arch is given by the vertex of the parabola, y = a·x² + b·x + c, which has the x-value of the vertex = -b/(2·a)
∴ The x-value of the vertex of the given parabola, y = -x² + 5·x + 24, is x = -5/(2×(-1)) = 2.5
Therefore;
The y-value of the vertex, is y = -(2.5)² + 5×2.5 + 24 = 30.25 = The height of the archway at its highest point
∴ The height of the archway at its highest point = 30.25.
