Identify the segment bisector of JK
1 answer:
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Answer:
y = 16x/65
Step-by-step explanation:
Given:
Triangle ABE is similar to triangle ACD. AED and ABC are straight lines
EB and DC are parallel
The area of quadrilateral BCDE = xcm²
The area of triangle ABE = ycm²
Find attached the diagram from the above information.
In similar triangles, the ratio of their corresponding angles are equal.
Also, the ratio of the area of the two triangles = square of ratio of the corresponding sides of the two triangles.
Area ∆ACD/area of ∆ABE = (DC/EB)²
Area ∆ACD/area of ∆ABE = [(area of quadrilateral BCDE +
area of ∆ABE)]/(area of ∆ABE)
(x+y)/y = (DC/EB)²
(x+y)/y = (9/4)²
x+y = (81/16)y
x = (81/16)y - y
x = (81y - 16y)/16
x = 65y/16
Making y subject of formula
16x = 65y
y = 16x/65
An expression for y in terms of x:
y = 16x/65
Answer:
Multiply by ∛2 and translate the graph to left by 4 units.
Step-by-step explanation:
The initial function given is:
y = -∛(x - 4)
The transformed function is:
y = -∛(2x - 4)
Consider the initial function.
y = -∛(x - 4)
(Represented by Black line in the graph)
Multiply the function by ∛2. The function becomes:
y = -∛(x - 4) × ∛2
y = -∛(2)(x-4)
y = -∛(2x-8)
(Represented by Red line in the graph represents this function)
Translate the graph 4 units to the left by adding 4 to the x component:
y = -∛(2x-8+4)
y= -∛(2x - 4)
(Represented by Blue line in the graph)
