Graph the image of this figure after a dilation with a scale factor of 1/2 centered at the point (−3, 0).
1 answer:
The vertices of the image of dilation are (4, 2), (-1, 2) and (3, 4)
<h3>How to graph the dilated image?</h3>The given parameters are:
Center, (a, b) = (-3, 0)
Scale factor, k = 1/2
The vertices of the figure are
(5,4), (-5, 4) and (3,8)
The rule of dilation is represented as:
(x, y) ⇒ (k(x -a), k(y -b))
So, we have:
Point (5,4)
(x, y) ⇒ (1/2(5 + 3), 1/2(4 -0))
(x, y) ⇒ (4, 2)
Point (-5,4)
(x, y) ⇒ (1/2(-5 + 3), 1/2(4 -0))
(x, y) ⇒ (-1, 2)
Point (3,8)
(x, y) ⇒ (1/2(3 + 3), 1/2(8 -0))
(x, y) ⇒ (3, 4)
So, the vertices of the image of dilation are (4, 2), (-1, 2) and (3, 4)
See attachment for the image
Read more about dilation at:
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I am sorry for my bad hand writing
Answer:
Therefore the concentration of salt in the incoming brine is 1.73 g/L.
Step-by-step explanation:
Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.
Let the concentration of salt be a gram/L
Let the amount salt in the tank at any time t be Q(t).
Incoming rate = (a g/L)×(1 L/min)
=a g/min
The concentration of salt in the tank at any time t is = g/L
Outgoing rate =
Integrating both sides
[ where c arbitrary constant]
Initial condition when t= 20 , Q(t)= 15 gram
Therefore ,
.......(1)
In the starting time t=0 and Q(t)=0
Putting t=0 and Q(t)=0 in equation (1) we get
Therefore the concentration of salt in the incoming brine is 1.73 g/L