Find the distance from the point (4, 1, −1) to the plane x − 2y + 2z + 3 = 0

Mathematics

2 answers:

diamong [38]4 years ago

8 0

I believe the answer is -17????

g100num [7]4 years ago

5 0

<span>Distance from point (4,1,-1) to plane x-2y+2z+3 = 0 is 1. The formula for the distance from a point to a plane is d = abs(Ax + By + Cz + D) / sqrt(A^2 + B^2 + C^2) where d = distance (x,y,z) = coordinates of point A,B,C,D = coefficients of equation for the plane For this problem, the values for A,B,C,and D are A=1, B=-2, C=2, and D=3. So substituting the known values into the equation, we get: d = abs(1*x - 2*y + 2*z + 3) / sqrt(1^2 + (-2)^2 + 2^2) d = abs(1*4 - 2*1 + 2*(-1) + 3) / sqrt(1 + 4 + 4) d = abs(4 - 2 - 2 + 3) / sqrt(9) d = abs(3) / 3 d = 1</span>

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Show an example of commutative property of addition

vaieri [72.5K]

For example, if we use 5+3=8.
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Therefore, 3+5=5+3 because at the end, they both have equal value aka both end up with 8 using the same numbers. (goes by the rule a+b=b+a)

5 0

3 years ago

Read 2 more answers

Find the differential of each function. (a) y = tan( 5t ) dy = Correct: Your answer is correct. (b) y = 5 − v2 5 + v2

Trava [24]

Answer:

$(a)$ $dy = 5sec^2(5t) \ dt$

$(a) \ dy = \frac{-20v}{(5+v^2)^2} \ dt$

Step-by-step explanation:

Given;

(a) y = tan(5t)

$(b) \ y = \frac{5-v^2}{5+v^2}$

Solving for (a)

y = tan(5t)

let u = 5t

⇒y = tan(u)

du/dt = 5

dy/du = sec²u

$\frac{dy}{dt} =\frac{dy}{du} *\frac{du}{dt} \\\\\frac{dy}{dt} = sec^2(u)*5\\\\\frac{dy}{dt} = 5sec^2(u)\\\\\frac{dy}{dt} = 5sec^2(5t)$

$dy = 5sec^2(5t) \ dt$

Solving for b;

let u = 5 - v²

du/dv = -2v

let v = 5+ v²

dv/du = 2v

$\frac{dy}{dv} = \frac{vdu - udv}{v^2} \\\\\frac{dy}{dv} = \frac{-2v(5+v^2) - 2v(5-v^2)}{(5+v^2)^2}\\\\\frac{dy}{dv} = \frac{-10v-2v^3-10v+2v^3}{(5+v^2)^2}\\\\$