Find the equation for the plane through the point Upper P 0equals(2 comma 5 comma 9 )and normal to the vector nequals5 Bold i plus 8 Bold j plus 2 Bold k.

5x+8y+2z = 68

Step-by-step explanation:

Given the point P0 = (2, 5, 9), to Find an equation for a plane through that point and normal to the vector n = 5i+8j+2k the following steps must be followed:

The equation for the plane passing through the point is expressed as;

a(x-x0)+b(y-y0)+c(z-z0) = 0 where;

(x0, y0, z0) is the point on the plane and (a,b,c) is the normal vector n.

Given the point (2, 5, 9) and normal vector n =(5, 8, 2)

x0 = 2, y0 = 5, z0 = 9, a = 5, b = 8 and c= 2.

Substituting this values into the equation of the plane above will give;

5(x-2)+8(y-5)+2(z-9) = 0

On expansion:

5x-10+8y-40+2z-18 = 0

5x+8y+2z-10-40-18 = 0

5x+8y+2z-68 = 0

The required equation of the plane is 5x+8y+2z = 68

i think the answer is b.

step-by-step explanation:

not 100% sure though.

the correct answer option is (2x-3)(2x+3).

step-by-step explanation:

we are given the following expression and we are to factorize it:

we are to find such factors that when multiplied they give a result of 36 (9 x 4 = 36) and when added they give a result of 0.

the two such factors of 36 are negative six and postive six.

factorizing the given expression to get:

therefore, the correct factorization of is (2x-3)(2x+3).