1.) 15 × 2 = 30
2.) 20 × 2 = 40
3.) A number between 30 and 40 or 70
I hope that that was what you were asking.
<span>1. X = -2 or 3
2. X = -5 or 3
3. X = -2.5 or 3
4. X = -4 or 2
5. X = 3 or -3
6. X = -4 or 2
I am assuming that you're looking for the intersections between the two equations for each problem. The general approach to each of the given problems is to solve both equations for y (only need to do this with problems 4 through 6 since you've already been given the equations solved for y with problems 1 through 3). After you have two equations solved for y, simply set them equal to each other and then manipulate until you have a quadratic equation of the form:
Ax^2 + Bx + C = 0
After you've gotten your quadratic equation, just find the roots to the equation and you'll know both X values that will result in the same Y value as the equations you've been given for each problem. I'm personally using the quadratic formula for getting the desired roots, but you can also factor manually. So let's do it.
1. y = x+2, y = x^2 - 4
Set the equations equal to each other
x + 2 = x^2 - 4
2 = x^2 - x - 4
0 = x^2 - x - 6
Using the quadratic formula with A=1, B=-1, C=-6, you get the solutions -2 and 3.
2. y = x^2 + 3x - 1, y = x+14
Same thing, set the equations equal to each other.
x^2 + 3x - 1 = x + 14
x^2 + 2x - 1 = 14
x^2 + 2x - 15 = 0
Use the quadratic formula with A=1, B=2, C=-15. Roots are -5 and 3.
3. y = 2x^2 + x - 7, y = 2x + 8
Set the equations equal to each other again.
2x^2 + x - 7 = 2x + 8
2x^2 - x - 7 = 8
2x^2 - x - 15 = 0
Quadratic formula with A=2, B=-1, C=-15, gives you the roots of -2.5 and 3
4. y = x(x + 3), y - x = 8
A little more complicated. Solve the second equation for y
y - x = 8
y = x + 8
Multiply out the 1st equation
y = x(x + 3)
y = x^2 + 3x
Now set the equations equal to each other
x + 8 = x^2 + 3x
8 = x^2 + 2x
0 = x^2 + 2x - 8
And use the quadratic formula with A=1, B=2, C=-8. Roots are -4, 2
5. y = -3x^2 - 2x + 5, y + 2x + 22 = 0
Solve the 2nd equation for y
y + 2x + 22 = 0
y + 22 = -2x
y = -2x - 22
Set equal to 1st equation
-2x - 22 = -3x^2 - 2x + 5
-22 = -3x^2 + 5
0 = -3x^2 + 27
Use the quadratic formula with A=-3, B=0, C=27, giving roots of 3 and -3
6. y + 6 = 2x^2 + x, y + 3x = 10
Solve the 1st equation for y
y + 6 = 2x^2 + x
y = 2x^2 + x - 6
Solve the 2nd equation for y
y + 3x = 10
y = -3x + 10
Set the solved equations equal to each other
2x^2 + x - 6 = -3x + 10
2x^2 + 4x - 6 = + 10
2x^2 + 4x - 16 = 0
Use the quadratic formula with A=2, B=4, C=-16, getting roots of -4 and 2.</span>
Just think of 4/8. Four is half of 8, right? therefore, it is 1/2. 4/8 reduced is 1/2
Answer:
Well, when we have a point in rectangular coordinates (x, y) and we want to write it in polar coordinates (r, θ) the rule we use is:
r = √(x^2 + y^2)
θ = Atg(Iy/xI) + 90°*(n - 1)
where n is number of the quadrant where our point is.
For example, if we are on the second quadrant, we use n = 2.
Now, if instead, we have the point (r, θ) and we want to rewrite it in rectangular coordinates, then the transformation is:
x = r*cos(θ)
y = r*sin(θ)
Now we have the point (- 2, -(5pi)/4) because the first part is negative, this number can not be in polar coordinates (r can not be negative) then we have:
x = -2
y = -(5*pi)/4
Using the first relation, we can find that:
r = √( (-2)^2 + (-5*3.14/4)^2) = 4.4
(rememer that the point (- 2, -(5pi)/4) is on the third quadrant, then we will use n = 3.)
θ = Atg(-(5*pi)/4/-2) + 90°*(3 - 1)= Atg((5*pi)/2) + 180° = 82.7° + 180° = 262.7°
Then the point that represents (- 2, - (5pi)/4) in polar coordinates is the point:
(4.4, 262.7°)
If instead of degrees, the angle part is written in radians, we have:
180° = 3.14 radians.
262.7° = x radians.
then:
x/3.14 = (262.7)/180
x = 3.14*(262.7)/180 = 4.58 radians.
then the point will be:
(4.4, 4.58 radians)
Answer:
there is the answer. ❤ hope it helps