Sweet................
a. (0,4500) initial height
b. take any to points and find slope which is rate in ft per min and is also negative in this case
c. (3,0) hits ground after 3 minutes
d. after b, use y-y1=m (x-x1) for any point (x1,y1)
ur welcome. plz follow
Answer:
2x^2 - 16x + 30
Step-by-step explanation:
Here, we want to get the composite function;
p•q(x)
All we have to do here is to replace the x value in p(x) by the totality of q(x)
we have this as ;
= 2(x-3)^2 - 4(x-3)
= 2(x^2 -6x + 9) - 4(x-3)
= 2x^2 -12x + 18 -4x + 12
= 2x^2 - 16x + 30
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4.) $1.60 per notebook
Answer:
Step-by-step explanation:
Any linear equation can be written as y = mx + b
The point will determine b
The slope (m) is found by this formula for a perpendicular line
m1 * m2 = - 1
m1 = -1/m2
m1 = -1/(-2/3)
m1 = 3/2
So our equation looks like
y = 3/2 x + b
Now we use the point (4,-8)
-8 = 3/2 (4) + b
-8 = 3(2) + b
-8 = 6 + b
b = - 14
The answer is
y = 4/2 - 14
Answer:
Differentiation will give you the gradient for the tangent at any point, and you use the product rule whenever a function can be thought of as two functions multiplied together.
If
f
(
x
)
=
g
(
x
)
×
h
(
x
)
then
f
'
(
x
)
=
g
'
(
x
)
h
(
x
)
+
g
(
x
)
h
'
(
x
)
so if
y
=
x
×
sin
x
then
d
y
d
x
=
1
×
sin
x
+
x
×
cos
x
=
sin
x
+
x
cos
x
We know that
x
=
π
2
, so the gradient is
m
=
sin
(
π
2
)
+
π
2
cos
(
π
2
)
=
1
+
π
2
×
0
=
1
Therefore, we can say that
y
=
m
x
+
c
y
=
(
1
)
x
+
c
y
=
x
+
c
So all we really need to find now is the value for
c
, the
y
intercept. We do this by working out a point
(
x
,
y
)
on the graph. We are already given that
x
=
π
2
, so
y
=
x
sin
x
=
π
2
sin
(
π
2
)
=
π
2
×
1
=
π
2
∴
(
x
,
y
)
=
(
π
2
,
π
2
)
Now we substitute this into the equation we already have for the tangent,
y
=
x
+
c
,
(
x
,
y
)
=
(
π
2
,
π
2
)
π
2
=
π
2
+
c
c
=
π
2
−
π
2
=
0
∴
y
=
x
+
c
=
x
+
(
0
)
=
x
which means the tangent to the curve
y
=
x
sin
x
at
(
π
2
,
π
2
)
is simply
y
=
x
.
