The grade point averages of a large population of college students are approximately normally distributed with a mean of 2.4 and
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Answer:
The answer is below
Step-by-step explanation:
A tangent is a line which touches a circle at one point. The angle between a tangent and a circle is 90°.
The two tangent theorem states that tangent that meet at the same point are equal in length.
Therefore:
CD = c + d
but c = 5.1, hence:
CD = 5.1 + d
12.1 = 5.1 + d
d = 7
AD = a + d; but AD = 12.2, d = 7
12.2 = a + 7
a = 5.2
AB = a + b; but b = 3.3
AB = 5.2 + 3.3 = 8.5
BC = b + c = 3.3 + 5.1 = 8.4
Perimeter of ABCD = AB + BC + CD + AD = 8.5 + 8.4 + 12.1 + 12.2 = 41.2
Vertex form is
y=a(x-h)^2+k
vertex is (h,k)
axis of symmetry is x=4, therfor h=4
y=a(x-4)^2+k
we have some points
(3,-2) and (6,-26)
input and solve for a and k
(3,-2)
-2=a(3-4)^2+k
-2=a(-1)^2+k
-2=a(1)+k
-2=a+k
(6,-26)
-26=a(6-4)^2+k
-26=a(2)^2+k
-26=a(4)+k
-26=4a+k
we have
-2=a+k
-26=4a+k
multiply first equation by -1 and add to second
2=-a-k
<u>-26=4a+k +</u>
-24=3a+0k
-24=3a
divide both sides by 3
-8=a
-2=a+k
-2=-8+k
add 8 to both sides
6=k
the equation is
y=a(x-h)^2+k
vertex is (h,k)
axis of symmetry is x=4, therfor h=4
y=a(x-4)^2+k
we have some points
(3,-2) and (6,-26)
input and solve for a and k
(3,-2)
-2=a(3-4)^2+k
-2=a(-1)^2+k
-2=a(1)+k
-2=a+k
(6,-26)
-26=a(6-4)^2+k
-26=a(2)^2+k
-26=a(4)+k
-26=4a+k
we have
-2=a+k
-26=4a+k
multiply first equation by -1 and add to second
2=-a-k
<u>-26=4a+k +</u>
-24=3a+0k
-24=3a
divide both sides by 3
-8=a
-2=a+k
-2=-8+k
add 8 to both sides
6=k
the equation is