This equation can be used when comparing ages.
An example to illustrate this:
Assume that adding 6 to 3 times the age of Jack will give us the age of his grandfather.
When translating this into equations, assuming that the age of jack is "a" and the age of his grandfather is "b", we will find that:
b = 6 + 3a
You haven't shared the given line, so all I can do here is to invent a line and then show you how to write the equation of a new line which is parallel to mine and which has an x-intercept of 4.
My invented line: y = (2/3)x + 3
The new line MUST have the same slope: m = 2/3.
Then y = mx + b becomes y = (2/3)x + b. Find the x-intercept by setting y = 0 and solving for x: (2/3)x = 0 - b. Now replace x with 4 and find b:
-b = (2/3)(4) = 8/3. Then b = -8/3, and the new line is
y = (2/3)x - 8/3.
Answer:
7.)
Therefore FT is 33 unit.
8.)
Therefore SU is 98 unit.
Step-by-step explanation:
7.)
Given:
ΔFUT ~ ΔFHG
FU = 39
FH = 130
FG = 110
To Find:
FT = ?
Solution:
ΔFUT ~ ΔFHG ............Given
If two triangles are similar then their sides are in proportion.
Substituting the values we get
Therefore FT is 33 unit.
8.)
Given:
ΔSTU ~ ΔFED
ST= 63
FE = 9
FD = 14
To Find:
SU = ?
Solution:
ΔSTU ~ ΔFED ............Given
If two triangles are similar then their sides are in proportion.
Substituting the values we get
Therefore SU is 98 unit.
F(x) = 18-x^2 is a parabola having vertex at (0, 18) and opening downwards.
g(x) = 2x^2-9 is a parabola having vertex at (0, -9) and opening upwards.
By symmetry, let the x-coordinates of the vertices of rectangle be x and -x => its width is 2x.
Height of the rectangle is y1 + y2, where y1 is the y-coordinate of the vertex on the parabola f and y2 is that of g.
=> Area, A
= 2x (y1 - y2)
= 2x (18 - x^2 - 2x^2 + 9)
= 2x (27 - 3x^2)
= 54x - 6x^3
For area to be maximum, dA/dx = 0 and d²A/dx² < 0
=> 54 - 18x^2 = 0
=> x = √3 (note: x = - √3 gives the x-coordinate of vertex in second and third quadrants)
d²A/dx² = - 36x < 0 for x = √3
=> maximum area
= 54(√3) - 6(√3)^3
= 54√3 - 18√3
= 36√3.
