The value of a family's home, in Camrose AB, is given by the following exponential function f(x), where x is the number of years
1 answer:
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The correct question is
<span>Peyton is a sprinter who can run the 40-yard dash in 4.5 seconds. He converts his speed into miles per hour, as shown below
</span>Which ratio is incorrectly written to convert his speed?
the picture in the attached figure
we know that
The term (5280 ft/1 mi) is incorrect
"Feet"<span> needs to be on the bottom to cancel with the previous term.
</span>"Mile"<span> needs to be on top in the numerator so that the answer can be expressed in "miles per hour"
</span>
the correct term is (1 mi/5280 ft)
<span>Peyton is a sprinter who can run the 40-yard dash in 4.5 seconds. He converts his speed into miles per hour, as shown below
</span>Which ratio is incorrectly written to convert his speed?
the picture in the attached figure
we know that
The term (5280 ft/1 mi) is incorrect
"Feet"<span> needs to be on the bottom to cancel with the previous term.
</span>"Mile"<span> needs to be on top in the numerator so that the answer can be expressed in "miles per hour"
</span>
the correct term is (1 mi/5280 ft)
Problem 1)
AC is only perpendicular to EF if angle ADE is 90 degrees
(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88
Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle
Triangle AED is acute (all 3 angles are less than 90 degrees)
So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF
-------------------------------------------------------------
Problem 2)
a) The center is (2,-3)
The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2
---------------------
b) The radius is 3 and the diameter is 6
From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2
where
h = 2
k = -3
r = 3
so, radius = r = 3
diameter = d = 2*r = 2*3 = 6
---------------------
c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.
Some points on the circle are
A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)
Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.
AC is only perpendicular to EF if angle ADE is 90 degrees
(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88
Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle
Triangle AED is acute (all 3 angles are less than 90 degrees)
So because angle ADE is NOT 90 degrees, this means AC is NOT perpendicular to EF
-------------------------------------------------------------
Problem 2)
a) The center is (2,-3)
The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2
---------------------
b) The radius is 3 and the diameter is 6
From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2
where
h = 2
k = -3
r = 3
so, radius = r = 3
diameter = d = 2*r = 2*3 = 6
---------------------
c) The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.
Some points on the circle are
A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)
Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.