Answer:
47 or 50 will be your answer, my apologies if it's wrong
Step-by-step explanation:
Measure 5 (property taxes) Ballot Measure 5, an initiative passed by Oregon voters in November 1990, fundamentally changed Oregon’s property tax and public school funding systems. Voter approval of Measure 5, and of Measure 47 in 1996 and Measure 50 in 1997, demonstrated the great force of anti-tax fervor in that decade.
Answer:
Angle 1 and 2 are adjacent .
For the answer to the question above,
<span>r = 1 + cos θ
x = r cos θ
x = ( 1 + cos θ) cos θ
x = cos θ + cos^2 θ
dx/dθ = -sin θ + 2 cos θ (-sin θ)
dx/dθ = -sin θ - 2 cos θ sin θ
y = r sin θ
y = (1 + cos θ) sin θ
y = sin θ + cos θ sin θ
dy/dθ = cos θ - sin^2 θ + cos^2 θ
dy/dx = (dy/dθ) / (dx/dθ)
dy/dx = (cos θ - sin^2 θ + cos^2 θ)/ (-sin θ - 2 cos θ sin θ)
For horizontal tangent line, dy/dθ = 0
cos θ - sin^2 θ + cos^2 θ = 0
cos θ - (1-cos^2 θ) + cos^2 θ = 0
cos θ -1 + 2 cos^2 θ = 0
2 cos^2 θ + cos θ -1 = 0
Let y = cos θ
2y^2+y-1=0
2y^2+2y-y-1=0
2y(y+1)-1(y+1)=0
(y+1)(2y-1)=0
y=-1
y=1/2
cos θ =-1
θ = π
cos θ =1/2
θ = π/3 , 5π/3
θ = π/3 , π, 5π/3
when θ = π/3, r = 3/2
when θ = π, r = 0
when θ = 5π/3 , r = 3/2
(3/2, π/3) and (3/2, 5π/3) give horizontal tangent lines
</span>---------------------------------------------------------------------------------
For horizontal tangent line, dx/dθ = 0
<span>-sin θ - 2 cos θ sin θ = 0 </span>
<span>-sin θ (1+ 2 cos θ ) = 0 </span>
<span>sin θ = 0 </span>
<span>θ = 0, π </span>
<span>(1+ 2 cos θ ) =0 </span>
<span>cos θ =-1/2 </span>
<span>θ = 2π/3 </span>
<span>θ = 4π/3 </span>
<span>θ = 0, 2π/3 ,π, 4π/3 </span>
<span>when θ = 0, r=2 </span>
<span>when θ = 2π/3, r=1/2 </span>
<span>when θ = π, r=0 </span>
<span>when θ = 4π/3 , r=1/2 </span>
<span>(2,0) , (1/2, 2π/3) , (0, π), (1/2, 4π/3) </span>
<span>At (2,0) there is a vertical tangent line</span>
Answer:
(5 , 11)
Step-by-step explanation:
![Equation\ 1\\\\\y=3x-4\\\\Equation\ 2\\\\\y=\frac{2}{5}x+9](https://tex.z-dn.net/?f=Equation%5C%201%5C%5C%5C%5C%5Cy%3D3x-4%5C%5C%5C%5CEquation%5C%202%5C%5C%5C%5C%5Cy%3D%5Cfrac%7B2%7D%7B5%7Dx%2B9)
There are three ways to solve the systems of linear equations
1) Elimination method - where we add/subtract the equations to eliminate the variable
2) Substitution method - where we substitute one variable in the form of another variable
3) Graphing method - where we sketch the graphs of both the equations and if they intersect that point would be the solution and if they don't there is no solution
We use the second method the Substitution method which is the easiest in my opinion.
Here we got two equations so lets just substitute equation 1 with equation 2 meaning insert equation 1 into equation 2 like here:
![3x-4=\frac{2}{5}x+9](https://tex.z-dn.net/?f=3x-4%3D%5Cfrac%7B2%7D%7B5%7Dx%2B9)
as you can see here we eliminated the variable y by substituting y in terms of x.
Now we just solve the equation algebraically
![3x-4=\frac{2}{5}x+9\\3x-\frac{2}{5}x=9+4\\\frac{15x-2x}{5} =13\\\\\frac{13x}{5}=13\\\\13x=65\\x=65/13\\x=5\\](https://tex.z-dn.net/?f=3x-4%3D%5Cfrac%7B2%7D%7B5%7Dx%2B9%5C%5C3x-%5Cfrac%7B2%7D%7B5%7Dx%3D9%2B4%5C%5C%5Cfrac%7B15x-2x%7D%7B5%7D%20%3D13%5C%5C%5C%5C%5Cfrac%7B13x%7D%7B5%7D%3D13%5C%5C%5C%5C13x%3D65%5C%5Cx%3D65%2F13%5C%5Cx%3D5%5C%5C)
The value of x is 5 now to find the value of y we insert the value of x in any two equations lets insert the value of x into equation 1
![y=3x-4\\y=3(5)-4\\y=15-4\\y=11](https://tex.z-dn.net/?f=y%3D3x-4%5C%5Cy%3D3%285%29-4%5C%5Cy%3D15-4%5C%5Cy%3D11)
so the solution of the system of linear equations is (5 , 11)