Can anyone show me the steps in order solve this problem
1 answer:
M∠A = 42° [isosceles triangle]
m∠B = 180 - 42 - 42 = 96° [in a triangle, the three interior angles always add to 180°]
m∠C = 180 - 96 - (x+12)
m∠C = 84 - x - 12
m∠C = 72 - x
2x+9 + 3x-1 + 72-x = 180
4x + 80 = 180
4x = 180 - 80
4x = 100
x = 100/4
x = 25
m∠C = 72 - x = 72 - 25 = 47°
<span>I hope this helped</span>
m∠B = 180 - 42 - 42 = 96° [in a triangle, the three interior angles always add to 180°]
m∠C = 180 - 96 - (x+12)
m∠C = 84 - x - 12
m∠C = 72 - x
2x+9 + 3x-1 + 72-x = 180
4x + 80 = 180
4x = 180 - 80
4x = 100
x = 100/4
x = 25
m∠C = 72 - x = 72 - 25 = 47°
<span>I hope this helped</span>
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Refer to the diagram shown below.
When x = 30 ft, the cable is at 15 ft, therefore y(30) = 15.
That is,
a(30 - h)² + k = 15 (1)
Also, because the distance between the supports is 90 ft, therefore
y(0) = 6 ft, and y(90) = 6 ft
That is,
a(-h)² + k = 6 (2)
a(90 - h)² + k = 6 (3)
From (2) and (3), obtain
a(90 - h)² = ah²
90² - 180h + h² = h²
180h = 90²
h = 45 ft.
From (1) and (2), obtain
225a + k = 15
2025a + k = 6
Therefore
1800a = -9
a = - 0.005
k = 15 - 225(-0.005) = 16.125 ft
Answer:
The equation for the cable is
y = - 0.005(x - 45)² + 16.125
A graph of the solution verifies that the solution is correct.
When x = 30 ft, the cable is at 15 ft, therefore y(30) = 15.
That is,
a(30 - h)² + k = 15 (1)
Also, because the distance between the supports is 90 ft, therefore
y(0) = 6 ft, and y(90) = 6 ft
That is,
a(-h)² + k = 6 (2)
a(90 - h)² + k = 6 (3)
From (2) and (3), obtain
a(90 - h)² = ah²
90² - 180h + h² = h²
180h = 90²
h = 45 ft.
From (1) and (2), obtain
225a + k = 15
2025a + k = 6
Therefore
1800a = -9
a = - 0.005
k = 15 - 225(-0.005) = 16.125 ft
Answer:
The equation for the cable is
y = - 0.005(x - 45)² + 16.125
A graph of the solution verifies that the solution is correct.
Answer:
51.81 cm^2
Step-by-step explanation:
Find The Half Of 33 which is 16.5
Use The Formula For Area
A= x r^2
A= 3.14 x 16.5
A= 51.81 cm^2