86. D. -10
87. H. -6, -3, 0, 2
88. Quadrant 1
Answer:
The speed of the jet is 347 mph and the speed of the wind is 18 mph.
Step-by-step explanation:
We have the following:
x = the speed of the jet in still air.
y = the speed of the wind
we know that the speed is equal to:
v = d / t
therefore the distance would be:
d = v * t
if we replace with the information of the exercise we have:
3 * (x + y) = 1095
3 * (x - y) = 987
we must solve this system of equations, add both equations and we are left:
3 * x + 3 * y = 1095
3 * x - 3 * y = 987
3 * x + 3 * y + 3 * x - 3 * y = 1095 + 987
6 * x = 2082
x = 2082/6 = 347
now to calculate y, we replace:
3 * (347 + y) = 1095
1041 + 3 * y = 1095
3 * y = 1095 - 1041
y = 54/3 = 18
The speed of the jet is 347 mph and the speed of the wind is 18 mph.
Answer:
f
(
x
)
=
3
cos
(
2
x
)
+
2
EXPLANATION:
We are able to use the transformation formula
f
(
x
)
=
a
⋅
cos
(
x
−
h
b
)
+
k
. You start with
f
(
x
)
=
cos
(
x
)
and replace
a
with the desired amplitude,
h
with the desired horizontal shift, and
k
with the desired vertical shift. This leaves out the
b
-value. A regular cosine function has a period of
2
π
. If you want a period of
π
, since that is one half of the original period, you need to replace your
b
with a
1
2
.
This is about how it would work out.
f
(
x
)
=
3
⋅
cos
(
x
−
0
1
2
)
+
2
From there you simplify your equation giving you
f
(
x
)
=
3
cos
(
2
x
)
+
2
Answer:
There is no graph
Step-by-step explanation:
