Answer:
396cm I think.
Step-by-step explanation:
10 x (27 - 8) + 9 x 24
CPI in 1960 = 29.6
CPI in 1980 = 82.4
Growth (%) in CPI = 
× 100
⇒ Growth (%) in CPI = 
× 100
⇒ Growth (%) in CPI = 178.3783.. %
⇒ Growth (%) in CPI = 178.4%
Now, we know that price of frozen chicken pie in 1960 = $0.28
Also, we know that over 1960 to 1980, frozen chicken pie price grew at the same rate as CPI
Hence, Price of Chicken Pie in 1980 = Price of Chicken Pie in 1960 × (1 + Growth in CPI)
⇒ Price of Chicken Pie in 1980 = 0.28 × (1 + 178.4%)
⇒ Price of Chicken Pie in 1980 = 0.28 + (0.28 × 178.4%)
⇒ Price of Chicken Pie in 1980 = 0.28 + 0.49952
⇒ Price of Chicken Pie in 1980 = 0.28 + 0.50
⇒ Price of Chicken Pie in 1980 = $0.78
Hence, price of chicken pie in 1980 would be ~$0.78
B = 16
x^2 + 16x + 64 can be factored down to (x + 8)^2.
We know that
<span>Figures can be proven similar if one, or more, similarity transformations (reflections, translations, rotations, dilations) can be found that map one figure onto another.
In this problem to prove circle 1 and circle 2 are similar, a translation and a scale factor (from a dilation) will be found to map one circle onto another.
</span>we have that
<span>Circle 1 is centered at (4,3) and has a radius of 5 centimeters
</span><span> Circle 2 is centered at (6,-2) and has a radius of 15 centimeters
</span>
step 1
<span>Move the center of the circle 1 onto the center of the circle 2
</span>the transformation has the following rule
(x,y)--------> (x+2,y-5)
so
(4,3)------> (4+2,3-5)-----> (6,-2)
so
center circle 1 is now equal to center circle 2
<span>The circles are now concentric (they have the same center)
</span>
step 2
A dilation is needed to increase the size of circle 1<span> to coincide with circle 2
</span>
scale factor=radius circle 2/radius circle 1-----> 15/5----> 3
radius circle 1 will be=5*scale factor-----> 5*3-----> 15 cm
radius circle 1 is now equal to radius circle 2
A translation, followed by a dilation<span> will map one circle onto the other, thus proving that the circles are similar</span>
