I know for a fact that the reference angle for 3pi/2 is pi/2 so the first one is A, and if I'm not mistaken, the second one should be undefined. I'm not sure about the second one.
4d + 5i = 57
i = 2d - 11
4d + 5(2d - 11) = 57
4d + 10d - 55 = 57
14d = 57 + 55
14d = 112
d = 112/14
d = 8 <== Daphne's age
i = 2d - 11
i = 2(8) - 11
i = 16 - 11
i = 5 <==Isaac's age
Answer:
4
Step-by-step explanation:
Step 1: find (r- s) or r(x) - s(x)
r(x) - s(x) = 3x - 1 - (2x + 1)
r(x) - s(x) = 3x - 1 - 2x - 1 (distribute the -1 to 2x and 1)
r(x) - s(x) = x - 2 (combine like terms, 3x + (-2x) = x, -1 + (-1) = -2)
so r(x) - s(x) = x - 2, or (r - s)(x) = x - 2
Step 2: Plug in 6 to 'x' and find (r - s)(x)
(r - s)(6) = 6 - 2 = 4
<span>18.374 </span><span>r</span><span>ounded to the nearest tenth is </span><span>18.400.</span>
<span>Because 7 is after 5 so we will round to the next tenth......18.400
Hope I helped:P</span>
L
=
∫
t
f
t
i
√
(
d
x
d
t
)
2
+
(
d
y
d
t
)
2
d
t
. Since
x
and
y
are perpendicular, it's not difficult to see why this computes the arclength.
It isn't very different from the arclength of a regular function:
L
=
∫
b
a
√
1
+
(
d
y
d
x
)
2
d
x
. If you need the derivation of the parametric formula, please ask it as a separate question.
We find the 2 derivatives:
d
x
d
t
=
3
−
3
t
2
d
y
d
t
=
6
t
And we substitute these into the integral:
L
=
∫
√
3
0
√
(
3
−
3
t
2
)
2
+
(
6
t
)
2
d
t
And solve:
=
∫
√
3
0
√
9
−
18
t
2
+
9
t
4
+
36
t
2
d
t
=
∫
√
3
0
√
9
+
18
t
2
+
9
t
4
d
t
=
∫
√
3
0
√
(
3
+
3
t
2
)
2
d
t
=
∫
√
3
0
(
3
+
3
t
2
)
d
t
=
3
t
+
t
3
∣
∣
√
3
0
=
3
√
3
+
3
√
3
=6The arclength of a parametric curve can be found using the formula:
L
=
∫
t
f
t
i
√
(
d
x
d
t
)
2
+
(
d
y
d
t
)
2
d
t
. Since
x
and
y
are perpendicular, it's not difficult to see why this computes the arclength.
It isn't very different from the arclength of a regular function:
L
=
∫
b
a
√
1
+
(
d
y
d
x
)
2
d
x
. If you need the derivation of the parametric formula, please ask it as a separate question.
We find the 2 derivatives:
d
x
d
t
=
3
−
3
t
2
d
y
d
t
=
6
t
And we substitute these into the integral:
L
=
∫
√
3
0
√
(
3
−
3
t
2
)
2
+
(
6
t
)
2
d
t
And solve:
=
∫
√
3
0
√
9
−
18
t
2
+
9
t
4
+
36
t
2
d
t
=
∫
√
3
0
√
9
+
18
t
2
+
9
t
4
d
t
=
∫
√
3
0
√
(
3
+
3
t
2
)
2
d
t
=
∫
√
3
0
(
3
+
3
t
2
)
d
t
=
3
t
+
t
3
∣
∣
√
3
0
=
3
√
3
+
3
√
3
=
6
√
3
Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.
Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.
