Answer:
Note that orthogonal to the plane means perpendicular to the plane.
Step-by-step explanation:
-1x+3y-3z=1 can also be written as -1x+3y-3z=0
The direction vector of the plane -1x+3y-3z-1=0 is (-1,3,-3).
Let us find a point on this line for which the vector from this point to (0,0,5) is perpendicular to the given line. The point is x-0,y-0 and z-0 respectively
Therefore, the vector equation is given as:
-1(x-0) + 3(y-0) + -3(z-5) = 0
-x + 3y + (-3z+15) = 0
-x + 3y -3z + 15 = 0
Multiply through by - to get a positive x coordinate to give
x - 3y + 3z - 15 = 0
i. Let t be the line tangent at point J. We know that a tangent line at a point on a circle, is perpendicular to the diameter comprising that certain point. So t is perpendicular to JL
let l be the tangent line through L. Then l is perpendicular to JL ii. So t and l are 2 different lines, both perpendicular to line JL.
2 lines perpendicular to a third line, are parallel to each other, so the tangents t and l are parallel to each other.
Remark. Draw a picture to check the
Answer:good for him
Step-by-step explanation:
The polynomial p(x)=x^3+7x^2-36p(x)=x 3 +7x 2 −36p, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 7, x, square
Iteru [2.4K]
Answer:
(x-2)(x+3)(x+6)
Step-by-step explanation:
Given the polynomial function p(x)=x^3+7x^2-36
We are to write it as a product of its linear factor
Assuming the value of x that will make the polynomial p(x) to be zero
Let x = 2
P(2) = 2³+7(2)²-36
P(2) = 8+7(4)-36
P(2) = 8+28-36
P(2) = 0
Since p(2) = 0 hence x-2 is one of the linear factors
Also assume x = -3
P(-3) = (-3)³+7(-3)²-36
P(-3) = -27+7(9)-36
P(-3) = -27+63-36
P(-3) = 36-36
P(-3) = 0
Since p(-3) = 0, hence x+3 is also a factor
The two linear pair are (x-2)(x+3)
(x-2)(x+3) = x²+3x-2x-6
(x-2)(x+3) = x²+x-6
To get the third linear function, we will divide x^3+7x^2-36 by x²+x-6 as shown in the attachment.
x^3+7x^2-36/x²+x-6 = x+6
Hence the third linear factor is x+6
x^3+7x^2-36 = (x-2)(x+3)(x+6)
