Find the surface area of the regular hexagonal prism.
1 answer:
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Answer:
A. (0, 1) and (2, -2)
B. Slope (m) = -³/2
C. y + 2 = -³/2(x - 2)
D.
E.
Step-by-step explanation:
A. Two points on the line from the graph are: (0, 1) and (2, -2)
B. The slope can be calculated using two points, (0, 1) and (2, -2):
Slope (m) = -³/2
C. Equation in point-slope form is represented as y - b = m(x - a). Where,
(a, b) = any point on the graph.
m = slope.
Substitute (a, b) = (2, -2), and m = -³/2 into the point-slope equation, y - b = m(x - a).
Thus:
y - (-2) = -³/2(x - 2)
y + 2 = -³/2(x - 2)
D. Equation in slope-intercept form, can be written as y = mx + b.
Thus, using the equation in (C), rewrite to get the equation in slope-intercept form.
y + 2 = -³/2(x - 2)
2(y + 2) = -3(x - 2)
2y + 4 = -3x + 6
2y = -3x + 6 - 4
2y = -3x + 2
y = -3x/2 + 2/2
E. Convert the equation in (D) to standard form:
Answer: The answers are (a) 40 cm and (b)
Step-by-step explanation: The calculations are as follows:
(a) See the figure (a). As given in the question, A circle with centre 'O' circumscribes a triangle ABC with BC = 20 cm and ∠BAC = 30°. We need to find the diameter DC of the circle.
Let us draw BD. Now, ∠BAC and ∠BDC are angles on the same arc BC, so we have
∠BAC = ∠BDC = 30°.
Also, ∠CBD = 90°, since it stands on the diameter DC. So, ΔBCD will be a right angled triangle.
We can write
Thus, the diameter of the circle = 40 cm.
(b) See the figure (b).
As given in the question, A circle with centre 'O'' circumscribes a triangle DEF with EF = 4.6 inches and diameter GF = 12.4 in.. We need to find the angle EDF.
Let us draw GE. Now, ∠EGF and ∠EDF are angles on the same arc EF, so we have
∠EGF = ∠EDF = ?
Also, ∠GEF = 90°, since it stands on the diameter GF. So, ΔGEF will be a right angled triangle.
We can write
Thus,
