y varies jointly as x and z, and y = 32 m, x = 6 m, and z = 8 m. What is the value of y when x = 5 m and z = 12 m
2 answers:
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The polynomial function whose real zeros are in -1, 1, 3 and whose degree is 3 is
Step-by-step explanation:
We need to find a polynomial function whose real zeros are in -1, 1, 3 and whose degree is 3.
If -1, 1 and 3 are real zeros, it can be written as:
x= -1, x= 1, and x = 3
or x+1=0, x-1=0 and x-3=0
Finding polynomial by multiply these factors:
So, The polynomial function whose real zeros are in -1, 1, 3 and whose degree is 3 is
Keywords: Real zeros of Polynomials
Learn more about Real zeros of Polynomials at:
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Answer:
Step-by-step explanation:
1. r=3
To convert from polar coordinates (r,θ) to rectangular coordinates (x,y), we use the following equations
x=rCosθ
Cosθ=x/r
y=rSinθ
Sinθ=y/r
Also, r²=x²+y²
When r =3,
Then we have
r=3,
x²+y²=3
All points with r = 3 are at
distance 3 from the origin, so r = 3 describes the circle of radius 3, with center at the origin (0,0).
Option C
2. r=2sinθ
When r=2sinθ
To convert from polar coordinates (r,θ) to rectangular coordinates (x,y), we use the following equations
x=rCosθ
Cosθ=x/r
y=rSinθ
Sinθ=y/r
Also, r²=x²+y²
Now, apply the given information
r=2sinθ
Since Sinθ=y/r
r=2y/r
Cross multiply
r²=2y
x²+y²=2y
x²+y²-2y =0
x²+ (y-1)² -1 =0
x²+(y-1)²=1²
Then,
It is a circle with center (0,1) and radius 1.
Because the sine is periodic, we know that we will get the entire curve for values of θ in [0, 2π). As θ runs from 0 to π/2, r increases
from 0 to 2. Then as θ continues to π, r decreases again to 0. When θ runs from π to
2π, r is negative, and it is not hard to see that the first part of the curve is simply traced
out again, so in fact we get the whole curve for values of θ in [0, π). Thus, check attachment for curve, Now, this suggests that the curve could possibly be a circle,
and if it is, it would have to be the circle x² + (y − 1)² = 1. Having made this guess, we can easily check it. First we substitute for x and y to get (r cos θ)² + (r sin θ − 1)² = 1;
expanding and simplifying does indeed turn this into r = 2 sin θ.
Option E
A circle of radius 1 with center on the x-axis
3. θ=π/4
This is a point on the circle
p=(r, θ)
So, r=0
x = rCosπ/4=rCos45
x=r√2 /2
and
y = rsin(π/4) = rSin45
y=r√2 /2
This makes it very
easy to convert equations from rectangular to polar coordinates.
x²+y²= (r√2/2)²+(r√2/2)²
x²+y²= 2r²/4 + 2r²/4
x²+y²= r²
Since r=0
x²+y²= 0
x²=y²
Therefore, x=y
Option D
The straight line y=x
4. r=4θ
Here the distance from the origin
exactly matches the angle, so a bit of thought makes it clear that when θ ≥ 0 we get the spiral of Archimedes ( check attachment ) When θ < 0, r is also negative, and so the full graph is the right hand picture in the figure. Check attachment
The correct option is B
None of the above
5. r=2Cosθ
To convert from polar coordinates (r,θ) to rectangular coordinates (x,y), we use the following equations
x=rCosθ
Cosθ=x/r
y=rSinθ
Sinθ=y/r
Also, r²=x²+y²
Now, apply the given information
r=2cosθ
Since cosθ=x/r
r=2x/r
Cross multiply
r²=2x
x²+y²=2x
x²+y²-2x=0
(x-1)² + y² -1 =0
(x-1)² + y² = 1²
Then,
It is a circle with center (1,0) and radius 1.
A circle of radius 1 with center on the y-axis
Option A is correct
