Answer: The exact length of segment HC is sqrt(3) units
The approximate length is roughly 1.73205080756888 (round that however you need to)
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Work Shown:
Let x = length of HC
Since AH = 3*HC, this means AH = 3*x
Draw out the picture. This step is optional but helpful in my opinion. The drawing is attached below.
After adding in the altitude BH, we have three similar triangles. So we can form the proportion shown below to solve for x
HC/BH = BH/AH
HC/3 = 3/AH ... replace BH with 3
x/3 = 3/AH ... replace HC with x
x/3 = 3/(3x) ... replace AH with 3x
x/3 = 1/x ... reduce
x*x = 3*1 ... cross multiply
x^2 = 3
x = sqrt(3) ... which is shorthand for "square root"
HC = sqrt(3)
HC = 1.73205080756888 which is approximate
Answer:
Step-by-step explanation: its c. x+ 5
The average speed for the journey from Llanfyllin to Bala is 14 km/h
Average speed is the ratio of total distance travelled to total time taken. Also, average speed can be given as the average of the different speeds.
Let d₁ represent the distance from Llanfyllin to Vyrnwy and t represent the time taken, since the speed is 12 km/h, hence:
12 = d₁/t
d₁ = 12t
Let d₂ represent the distance from Vyrnwy to bala and t represent the time taken. Since they spent the same time, since the speed is 16 km/h, hence:
16 = d₂/t
d₂ = 16t
Average speed = total distance/total time
Average speed = (d₁ + d₂) / (t + t)
Average speed = (12t + 16t) / 2t
Average speed = 28t / 2t
Average speed = 14 km/h
Hence their average speed for the journey from Llanfyllin to Bala is 14 km/h.
Find out more at: brainly.com/question/23774048
From the problem, the vertex = (0, 0) and the focus = (0, 3)
From the attached graphic, the equation can be expressed as:
(x -h)^2 = 4p (y -k)
where (h, k) are the (x, y) values of the vertex (0, 0)
The "p" value is the difference between the "y" value of the focus and the "y" value of the vertex.
p = 3 -0
p = 3
So, we form the equation
(x -0)^2 = 4 * 3 (y -0)
x^2 = 12y
To put this in proper quadratic equation form, we divide both sides by 12
y = x^2 / 12
Source:
http://www.1728.org/quadr4.htm
