Answer:
The distance A’C’ is 4.47 units
Step-by-step explanation:
Before we go on, we need to get the appropriate transformation
Mathematically, we have a 90 degrees clockwise rotation yielding the following;
(x,y) to (-y,x)
A is (-4,1)
C is (-2,5)
By transforming, we have
A’( -1,-4)
C’ (-5,-2)
To get the magnitude of the line segment, we are going to use the distance formula between points
We have this as;
D = √(x2-x1)^2 + (y2-y1)^2
D = √(-5-(-1))^2 + (-2-(-4))^2
D = √(-4)^2 + (2)^2
D = √(16 + 4)
D = √20
D = 4.47 units
Answer:
-1.375
Step-by-step explanation:
Answer:
The equation of the quadratic graph is f(x)= - (1/8) (x-3)^2 + 3 (second option)
Step-by-step explanation:
Focus: F=(3,1)=(xf, yf)→xf=3, yf=1
Directrix: y=5 (horizontal line), then the axis of the parabola is vertical, and the equation has the form:
f(x)=[1 / (4p)] (x-h)^2+k
where Vertex: V=(h,k)
The directix y=5 must intercept the axis of the parabola at the point (3,5), and the vertex is the midpoint between this point and the focus:
Vertex is the midpoint between (3,5) and (3,1):
h=(3+3)/2→h=6/2→h=3
k=(5+1)/2→k=6/2→k=3
Vertex: V=(h,k)→V=(3,3)
p=yf-k→p=1-3→p=-2
Replacing the values in the equation:
f(x)= [ 1 / (4(-2)) ] (x-3)^2 + 3
f(x)=[ 1 / (-8) ] (x-3)^2 + 3
f(x)= - (1/8) (x-3)^2 + 3
The amount of heat released when 12.0g of helium gas condense at 2.17 K is; -250 J
The latent heat of vapourization of a substance is the amount of heat required to effect a change of state of the substance from liquid to gaseous state.
However, since we are required to determine heat released when the helium gas condenses.
The heat of condensation per gram is; -21 J/g.
Therefore, for 12grams, the heat of condensation released is; 12 × -21 = -252 J.
Approximately, -250J.
Read more on latent heat:
brainly.com/question/19863536
It’s either 11.25 or 27.75. Next time please show the answer choices too!
