r
sin
θ
=
−
3
Explanation:
Imagine we have a point
P
with Rectangular (also called Cartesian) coordinates
(
x
,
y
)
and Polar coordinates
(
r
,
θ
)
.
The following diagram will help us visualise the situation better:
https://keisan.casio.com/exec/system/1223526375
https://keisan.casio.com/exec/system/1223526375
We can see that a right triangle is formed with sides
x
,
y
and
r
, as well as an angle
θ
.
We have to find the relation between the Cartesian and Polar coordinates, respectively.
By Pythagora's theorem, we get the result
r
2
=
x
2
+
y
2
The only properties we can say about
θ
are its trigonometric functions:
sin
θ
=
y
/
r
⇒
y
=
r
sin
θ
cos
θ
=
x
/
r
⇒
x
=
r
cos
θ
So we have the following relations:
⎧
⎪
⎨
⎪
⎩
r
2
=
x
2
+
y
2
y
=
r
sin
θ
x
=
r
cos
θ
Now, we can see that saying
y
=
−
3
in the Rectangular system is equivalent to say
r
sin
θ
=
−
3
Answer link
Jim G.
May 19, 2018
r
=
−
3
sin
θ
Explanation:
to convert from
cartesian to polar
∙
x
x
=
r
cos
θ
and
y
=
r
sin
θ
⇒
r
sin
θ
=
−
3
⇒
r
=
−
3
sin
θ
Answer:
C
Step-by-step explanation:
Answer:
Answer Choice A
Step-by-step explanation:
I took this assignment
(please give brainliest)
Answer:
The answer is C pay attention in class next time
Answer:
Step-by-step explanation:
Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?
Let p(x)=kpx+dp and q(x)=kqx+dq than
f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7
p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7
(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)
So you want:
−2kpk2q=0
and
kpkq=−1
and
2kpd2p−3kpdq+7=0
Now I amfraid this doesn’t work as −2kpk2q=0 that either kp or kq is zero but than their product can’t be anything but 0 not −1 .
Answer: there are no such linear functions.
