Answer:
15
17-2=15
Hope this helps please consider giving me brainliest :)
The answer is 12.9375 because if you add all the fractions and then divided it you get 12.9375
Answer:
y=2x-4
Step-by-step explanation:
To find an equation perpendicular to another one, you must find the negative inverse of the slope.
In this case, -1/2 has an inverse of -2, but because we're finding the negative inverse, the two negatives cancel out and we get 2, the slope of our perpendicular equation, which crosses the initial equation, making a right angle.
Answer:
y=3x+b
Step-by-step explanation:
finding slope:
slope= y2-y1/x2-x1
-2--5/3-2
3/1
slope is 3
i'm not sure how to find the y-intercept using this format but hope this kinda helped!
Answer:
The points are
and ![\bold{\left( -\cfrac{1}6\sqrt{\cfrac {15}{22}} ,-\cfrac 1{10}\sqrt{\cfrac {15}{22}},-\cfrac{2}3\sqrt{\cfrac {15}{22}}\right)}](https://tex.z-dn.net/?f=%5Cbold%7B%5Cleft%28%20%20-%5Ccfrac%7B1%7D6%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%20%2C-%5Ccfrac%201%7B10%7D%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%2C-%5Ccfrac%7B2%7D3%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%5Cright%29%7D)
Step-by-step explanation:
We can find the normal to both the surface and the plane, and set a relation between them since tangent plane and the given are parallel. Once we have that we can use the original surface to find the points on the surface that are tangent to the plane.
Normal vectors.
The normal vectors can be found using partial derivatives or gradient, thus we can call the surface:
![F(x,y,z)=3x^2+5y^2+3z^2-1=0](https://tex.z-dn.net/?f=F%28x%2Cy%2Cz%29%3D3x%5E2%2B5y%5E2%2B3z%5E2-1%3D0)
So its normal is
![\vec n_1 = \left< \cfrac{\partial F}{\partial x},\cfrac{\partial F}{\partial y},\cfrac{\partial F}{\partial z} \right>\\\vec n_1= \left< 6x,10y,6z\right>](https://tex.z-dn.net/?f=%5Cvec%20n_1%20%3D%20%5Cleft%3C%20%5Ccfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20x%7D%2C%5Ccfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D%2C%5Ccfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20z%7D%20%5Cright%3E%5C%5C%5Cvec%20n_1%3D%20%5Cleft%3C%206x%2C10y%2C6z%5Cright%3E)
And the normal vector to the plane is
![\vec n_2= < 1,1,4>](https://tex.z-dn.net/?f=%5Cvec%20n_2%3D%20%3C%201%2C1%2C4%3E)
Notice that the normal vector of the plane is just represented by the coefficients of x,y and z.
Parallel condition
The two planes are parallel only if their normal vectors are parallel as well, so we can write
![\vec n_1=t\vec n_2](https://tex.z-dn.net/?f=%5Cvec%20n_1%3Dt%5Cvec%20n_2)
Using the previously found normal vectors.
![\left< 6x,10y,6z\right>=t< 1,1,4>](https://tex.z-dn.net/?f=%5Cleft%3C%206x%2C10y%2C6z%5Cright%3E%3Dt%3C%201%2C1%2C4%3E)
So we get
![x= t/6\\y=t/10\\z=2t/3](https://tex.z-dn.net/?f=x%3D%20t%2F6%5C%5Cy%3Dt%2F10%5C%5Cz%3D2t%2F3)
Finding the points where we have the tangent plane.
We can plug the expressions we have found of x, y and z on the given surface equation.
![3(t/6)^2+5(t/10)^2+3(2t/3)^2=1](https://tex.z-dn.net/?f=3%28t%2F6%29%5E2%2B5%28t%2F10%29%5E2%2B3%282t%2F3%29%5E2%3D1)
And we can simplify and solve for t.
![\cfrac{t^2}{12}+\cfrac{t^2}{20}+\cfrac{4t^2}{3}=1](https://tex.z-dn.net/?f=%5Ccfrac%7Bt%5E2%7D%7B12%7D%2B%5Ccfrac%7Bt%5E2%7D%7B20%7D%2B%5Ccfrac%7B4t%5E2%7D%7B3%7D%3D1)
The LCD is 60 so we can multiply all terms in both sides by 60 to get
![5t^2+3t^2+80t^2=60\\88t^2=60](https://tex.z-dn.net/?f=5t%5E2%2B3t%5E2%2B80t%5E2%3D60%5C%5C88t%5E2%3D60)
So we get
![t^2 = \cfrac {15}{22}](https://tex.z-dn.net/?f=t%5E2%20%20%3D%20%5Ccfrac%20%7B15%7D%7B22%7D)
That give us the values of t
![t_{1,2} =\pm \sqrt{\cfrac {15}{22}}](https://tex.z-dn.net/?f=t_%7B1%2C2%7D%20%3D%5Cpm%20%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D)
Lastly at
we have:
![x= \cfrac{1}6\sqrt{\cfrac {15}{22}} \\y=\cfrac 1{10}\sqrt{\cfrac {15}{22}}\\z=\cfrac{2}3\sqrt{\cfrac {15}{22}}](https://tex.z-dn.net/?f=x%3D%20%5Ccfrac%7B1%7D6%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%20%5C%5Cy%3D%5Ccfrac%201%7B10%7D%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%5C%5Cz%3D%5Ccfrac%7B2%7D3%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D)
So one point is
![\boxed{\left( \cfrac{1}6\sqrt{\cfrac {15}{22}} ,\cfrac 1{10}\sqrt{\cfrac {15}{22}},\cfrac{2}3\sqrt{\cfrac {15}{22}}\right)}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cleft%28%20%20%5Ccfrac%7B1%7D6%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%20%2C%5Ccfrac%201%7B10%7D%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%2C%5Ccfrac%7B2%7D3%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%5Cright%29%7D)
And for the value of ![t_2=-\sqrt{\cfrac {15}{22}}](https://tex.z-dn.net/?f=t_2%3D-%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D)
We have
![x= -\cfrac{1}6\sqrt{\cfrac {15}{22}} \\y=-\cfrac 1{10}\sqrt{\cfrac {15}{22}}\\z=-\cfrac{2}3\sqrt{\cfrac {15}{22}}](https://tex.z-dn.net/?f=x%3D%20-%5Ccfrac%7B1%7D6%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%20%5C%5Cy%3D-%5Ccfrac%201%7B10%7D%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%5C%5Cz%3D-%5Ccfrac%7B2%7D3%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D)
So the second point is
![\boxed{\left( - \cfrac{1}6\sqrt{\cfrac {15}{22}} ,-\cfrac 1{10}\sqrt{\cfrac {15}{22}},-\cfrac{2}3\sqrt{\cfrac {15}{22}}\right)}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cleft%28%20-%20%5Ccfrac%7B1%7D6%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%20%2C-%5Ccfrac%201%7B10%7D%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%2C-%5Ccfrac%7B2%7D3%5Csqrt%7B%5Ccfrac%20%7B15%7D%7B22%7D%7D%5Cright%29%7D)