The rectangle below has an area of 30k3 + 6k2.
1 answer:
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Answer:
Segment EH and segment E prime H prime both pass through the center of dilation (A)
The complete question related to this found on brainly (ID:16812154) is stated below:
Triangle EFG is dilated by a scale factor of 3 centered at (0, 1) to create triangle E'F'G'. Which statement is true about the dilation?
E(0,5) F(1,1) G(-2,1) H(0,1)
a) segment EH and segment E prime H prime both pass through the center of dilation.
b)The slope of segment EF is the same as the slope of segment E prime H prime.
c) segment E prime G prime will overlap segment EG. segment
d) EH ≅ segment E prime H prime.
Step-by-step explanation:
∆EFG = Original image
∆EFG is dilated to give ∆E'F'G'
∆E'F'G' = New image
Scale factor = 3
Center of dilation = (0,1) = H(0,1)
Coordinates ∆EFG : E(0,5) F(1,1) G(-2,1)
To determine the statement that is true about the dilation from the options,
First we would make a diagram on the coordinates of ∆EFG and center of dilation (H).
Find attached the diagram.
Length GF = 3unit
Length G'F' = 3×scale factor = 9unit
Length EH = 4unit
Length E'H' = 4×scale factor = 12unit
Then we would move 3units to the left on same line from G to get the coordinate of G'(mark the point).
Also move 3units to the right on same line from F to get the coordinate of F'(mark the point).
Both of these give length of G'F' = 9unit
Now move 8units to the top from E to get the coordinate of E'(mark the point).
From this you get length E'H' = 12unit
Draw lines connecting the three points to get ∆E'F'G'
See diagram for better understanding
From the diagram, EH and E'H' both pass through (0,1). The other options are wrong.
Therefore, Segment EH and segment E prime H prime both pass through the center of dilation (A)
Xy=10
x+y=29
subtract x from both sides for second equation
y=-x-29
subsitute in second equaiton
x(-x-29)=10
-x^2-29x=10
add x^2+29x to both sides
x^2+29x+10=0
quadratic formula
if you have
ax^2+bx+c=0,
x=
so
x^2+29x+10=0
a=1
b=29
c=10
x=
x=
x=
x=
x=
or 
those are the 2 numbers
aprox=-28.65 and -0.349
x+y=29
subtract x from both sides for second equation
y=-x-29
subsitute in second equaiton
x(-x-29)=10
-x^2-29x=10
add x^2+29x to both sides
x^2+29x+10=0
quadratic formula
if you have
ax^2+bx+c=0,
x=
so
x^2+29x+10=0
a=1
b=29
c=10
x=
x=
x=
x=
x=
those are the 2 numbers
aprox=-28.65 and -0.349