Answer:
35/3x +200 = # of calories
Step-by-step explanation:
35/3 = calories burnt in 1 minute
x = minutes exercised for
add 200 to account for calories eaten
35/3x +200 = y
Let
. Then
Meanwhile, since
, you have
.
It follows that
,
, and
, so that the quadratic fit for
that satisfies the given points is
Note that this is just the second-order Taylor polynomial for
about
.
Answer:
A AND D ARE CORRECT.
Step-by-step explanation:
IT IS A NUMBER LINE SO IT IS JUST BASIC ADDITION AND SUBTRACTION.
Problem 10
<h3>Answer: (x+5)^2+(y-3)^2=25</h3>
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Work Shown:
All we're doing really is replacing the original 'y' with 'y-3' to shift the center up 3 units. This shifts every point on the circle the same as well.
This new circle has center (-5,3) and radius 5. The old center was (-5,0). The radius stayed the same.
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Problem 11
<h3>Answer: (x-8)^2+(y-15)^2 = 9</h3>
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Work Shown:
x^2+y^2-16x-30y+280 = 0
(x^2-16x)+(y^2-30y)+280 = 0
(x^2-16x+64)-64+(y^2-30y+225)-225+280 = 0
(x-8)^2+(y-15)^2-9 = 0
(x-8)^2+(y-15)^2 = 9
This circle has center (8,15) and radius 3.
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Problem 12
<h3>Answer: (x+6)^2+(y-11)^2 = 20</h3>
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Work Shown:
x^2+y^2+12x-22y+137 = 0
(x^2+12x)+(y^2-22y)+137 = 0
(x^2+12x+36)-36+(y^2-22y+121)-121+137 = 0
(x+6)^2+(y-11)^2-20 = 0
(x+6)^2+(y-11)^2 = 20
This circle has center (-6,11) and radius sqrt(20) = 2*sqrt(5).
