We have that
<span>Circle 1: center (8, 5) and radius 6
</span><span>Circle 2: center (−2, 1) and radius 2
we know that
the equation of a circle is
(x-h)</span>²+(y-k)²=r²
for the circle 1---------> (x-8)²+(y-5)²=36
for the circle 2---------> (x+2)²+(y-1)²=4
using a graph tool
see the attached figure
Part A)<span>What transformations can be applied to Circle 1 to prove that the circles are similar?
we know that
r1/r2---------> 6/2------> 3
</span><span>
to prove that the circle 1 and circle 2 are similar, the radius of circle 1 </span>must be divided by 3 and translate the center of the circle 1 (10) units to the left and (4) units down
<span>
the answer part A) is
</span>
the radius of circle 1 must be divided by 3 and translate the center of the circle 1 (10) units to the left and (4) units down
Part B) <span>What scale factor does the dilation from Circle 1 to Circle 2 have?
the answer Part B) is
the scale factor is (3/1)</span>
Answer:
Step-by-step explanation:
Let x represent the number of 3-lb bags purchased. Then the total purchase was ...
$2(8 -x) +$5.50(x) = $37
16 +3.50x = 37 . . . . . . . . . divide by $, collect terms
3.50x = 21 . . . . . . . . . . . . . subtract 16
21/3.50 = x = 6 . . . . . . . . divide by the coefficient of x
You bought 6 3-lb bags of peanuts and 2 1-lb bags.
By inspection, it's clear that the sequence must converge to

because

when

is arbitrarily large.
Now, for the limit as

to be equal to

is to say that for any

, there exists some

such that whenever

, it follows that

From this inequality, we get




As we're considering

, we can omit the first inequality.
We can then see that choosing

will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that

.
Yes <span>-(12/-17) is equal to 12/17
I hope this helps.
Have a awesome day. :)</span>