Answer:
(3,-2)
Step-by-step explanation:
Given equations of line
3x-2y=13
2y+x+1=0
=> x = -1 -2y
Point of intersection will coordinates where both equation have same value of (x,y)
top get that we have to solve the both equations by using method of substitution of simultaneous equation.
using this value of x in 3x-2y=13, we have
3(-1-2y) -2y = 13
=> -3 -6y-2y = 13
=> -8y = 13+3 = 16
=> y = 16/-8 = -2
x = -1 - 2y = -1 -2(-2) = -1+4= 3
Thus, point of intersection of line is (3,-2)
4
Is that right answer go for what you think who every you are
Answer:
![r = \±\sqrt{14](https://tex.z-dn.net/?f=r%20%3D%20%5C%C2%B1%5Csqrt%7B14)
![Product = -14](https://tex.z-dn.net/?f=Product%20%3D%20-14)
Step-by-step explanation:
Given
![\frac{1}{2x} = \frac{r - x}{7}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2x%7D%20%3D%20%5Cfrac%7Br%20-%20x%7D%7B7%7D)
Required
Find all product of real values that satisfy the equation
![\frac{1}{2x} = \frac{r - x}{7}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2x%7D%20%3D%20%5Cfrac%7Br%20-%20x%7D%7B7%7D)
Cross multiply:
![2x(r - x) = 7 * 1](https://tex.z-dn.net/?f=2x%28r%20-%20x%29%20%3D%207%20%2A%201)
![2xr - 2x^2 = 7](https://tex.z-dn.net/?f=2xr%20-%202x%5E2%20%3D%207)
Subtract 7 from both sides
![2xr - 2x^2 -7= 7 -7](https://tex.z-dn.net/?f=2xr%20-%202x%5E2%20-7%3D%207%20-7)
![2xr - 2x^2 -7= 0](https://tex.z-dn.net/?f=2xr%20-%202x%5E2%20-7%3D%200)
Reorder
![- 2x^2+ 2xr -7= 0](https://tex.z-dn.net/?f=-%202x%5E2%2B%202xr%20%20-7%3D%200)
Multiply through by -1
![2x^2 - 2xr +7= 0](https://tex.z-dn.net/?f=2x%5E2%20-%202xr%20%2B7%3D%200)
The above represents a quadratic equation and as such could take either of the following conditions.
(1) No real roots:
This possibility does not apply in this case as such, would not be considered.
(2) One real root
This is true if
![b^2 - 4ac = 0](https://tex.z-dn.net/?f=b%5E2%20-%204ac%20%3D%200)
For a quadratic equation
![ax^2 + bx + c = 0](https://tex.z-dn.net/?f=ax%5E2%20%2B%20bx%20%2B%20c%20%3D%200)
By comparison with ![2x^2 - 2xr +7= 0](https://tex.z-dn.net/?f=2x%5E2%20-%202xr%20%2B7%3D%200)
![a = 2](https://tex.z-dn.net/?f=a%20%3D%202)
![b = -2r](https://tex.z-dn.net/?f=b%20%3D%20-2r)
![c =7](https://tex.z-dn.net/?f=c%20%3D7)
Substitute these values in ![b^2 - 4ac = 0](https://tex.z-dn.net/?f=b%5E2%20-%204ac%20%3D%200)
![(-2r)^2 - 4 * 2 * 7 = 0](https://tex.z-dn.net/?f=%28-2r%29%5E2%20-%204%20%2A%202%20%2A%207%20%3D%200)
![4r^2 - 56 = 0](https://tex.z-dn.net/?f=4r%5E2%20-%2056%20%3D%200)
Add 56 to both sides
![4r^2 - 56 + 56= 0 + 56](https://tex.z-dn.net/?f=4r%5E2%20-%2056%20%2B%2056%3D%200%20%2B%2056)
![4r^2 = 56](https://tex.z-dn.net/?f=4r%5E2%20%3D%2056)
Divide through by 4
![r^2 = 14](https://tex.z-dn.net/?f=r%5E2%20%3D%2014)
Take square roots
![\sqrt{r^2} = \±\sqrt{14](https://tex.z-dn.net/?f=%5Csqrt%7Br%5E2%7D%20%3D%20%5C%C2%B1%5Csqrt%7B14)
![r = \±\sqrt{14](https://tex.z-dn.net/?f=r%20%3D%20%5C%C2%B1%5Csqrt%7B14)
Hence, the possible values of r are:
or ![-\sqrt{14](https://tex.z-dn.net/?f=-%5Csqrt%7B14)
and the product is:
![Product = \sqrt{14} * -\sqrt{14}](https://tex.z-dn.net/?f=Product%20%3D%20%5Csqrt%7B14%7D%20%2A%20-%5Csqrt%7B14%7D)
![Product = -14](https://tex.z-dn.net/?f=Product%20%3D%20-14)
Answer:
36×3/4
we will simplify it in four's table
9×3=27
Step-by-step explanation:
hope it help;)
The bearing of the plane is approximately 178.037°. ![\blacksquare](https://tex.z-dn.net/?f=%5Cblacksquare)
<h3>Procedure - Determination of the bearing of the plane</h3><h3 />
Let suppose that <em>bearing</em> angles are in the following <em>standard</em> position, whose vector formula is:
(1)
Where:
- Magnitude of the vector, in miles per hour.
- Direction of the vector, in degrees.
That is, the line of reference is the
semiaxis.
The <em>resulting</em> vector (
), in miles per hour, is the sum of airspeed of the airplane (
), in miles per hour, and the speed of the wind (
), in miles per hour, that is:
(2)
If we know that
,
,
and
, then the resulting vector is:
![\vec v = 239 \cdot (\sin 180^{\circ}, \cos 180^{\circ}) + 10\cdot (\sin 53^{\circ}, \cos 53^{\circ})](https://tex.z-dn.net/?f=%5Cvec%20v%20%3D%20239%20%5Ccdot%20%28%5Csin%20180%5E%7B%5Ccirc%7D%2C%20%5Ccos%20180%5E%7B%5Ccirc%7D%29%20%2B%2010%5Ccdot%20%28%5Csin%2053%5E%7B%5Ccirc%7D%2C%20%5Ccos%2053%5E%7B%5Ccirc%7D%29)
![\vec v = (7.986, -232.981) \,\left[\frac{mi}{h} \right]](https://tex.z-dn.net/?f=%5Cvec%20v%20%3D%20%287.986%2C%20-232.981%29%20%5C%2C%5Cleft%5B%5Cfrac%7Bmi%7D%7Bh%7D%20%5Cright%5D)
Now we determine the bearing of the plane (
), in degrees, by the following <em>trigonometric</em> expression:
(3)
![\theta = \tan^{-1}\left(-\frac{7.986}{232.981} \right)](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%5Cleft%28-%5Cfrac%7B7.986%7D%7B232.981%7D%20%5Cright%29)
![\theta \approx 178.037^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%5Capprox%20178.037%5E%7B%5Ccirc%7D)
The bearing of the plane is approximately 178.037°. ![\blacksquare](https://tex.z-dn.net/?f=%5Cblacksquare)
To learn more on bearing, we kindly invite to check this verified question: brainly.com/question/10649078