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3 years ago

5

The line l passes through the point P = (1,-1,1) and has direction vector d = [2,3,1]. Determine whether l and the plane 4x-y+5z

= 0 are parallel, perpendicular, or neither.

1 answer:

suter [353]3 years ago

3 0

Hello :
the normal vector of the plane is : d' = [4,-1,5]...(vector perpendicular to the plane)<span>
d </span><span>⊥ d' because : (4)(2)+(-1)(3)+(5)(1)=0
</span>the line is perpendicular to the plane .

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The system of equations above is graphed below. Find the solution to the system.

Oksana_A [137]

Option D: $x=3,y=-2$ is the solution to the system of equations.

Explanation:

Given that the equations are $4x+y=10$ and $2x+y=4$

We need to determine the solution of the system of equations.

The solution can be determined by solving the two equations.

Let us solve the system of equations using substitution method.

Consider the equation $4x+y=10$ , the value of y is given by

$y=10-4x$

Substituting the value of $y=10-4x$ in the equation $2x+y=4$ , we get,

$2x+(10-4x)=4$

Simplifying, we get,

$2x+10-4x=4$

$-2x+10=4$

$-2x=-6$

$x=3$

Substituting the value of $x=3$ in $4x+y=10$ , we get,

$4(3)+y=10$

$12+y=10$

$y=-2$

Thus, the solution to the system of equations is $x=3,y=-2$

Hence, Option D is the correct answer.

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3 years ago

Write an equation of the line that passes through the given point and is (a) parallel and (b) perpendicular to the given line.

Harrizon [31]

Answer:

Equation of the line that passes through the given point and is (a) parallel is: $\mathbf{y=-2x+11}$

Equation of the line that passes through the given point and is (b) perpendicular is: $\mathbf{y=\frac{1}{2}x+1}$

Step-by-step explanation:

We need to Write an equation of the line that passes through the given point and is (a) parallel and (b) perpendicular to the given line.

First we will find slope of the line given in graph

The slope can be found using formula: $Slope=\frac{y_2-y_1}{x_2-x_1}$

We have points (1,6) and (2,2)

Slope is:

$Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{2-6}{2-1}\\Slope=\frac{-4}{2}\\Slope=-2$

Part a)

Write an equation of the line that passes through the given point and is (a) parallel

When the lines are parallel, they have same slope. So, slope of required line: m = -2

Using slope m =-2 and point(4,3) we can find y-intercept b

$y=mx+b\\3=-2(4)+b\\3=-8+b\\b=3+8\\b=11$

The equation of line will be:

$y=mx+b\\y=-2x+11$

Equation of the line that passes through the given point and is (a) parallel is: $\mathbf{y=-2x+11}$

Part b)

Write an equation of the line that passes through the given point and is (b) perpendicular

When the lines are perpendicular they have opposite slope. So, slope of required line: m = 1/2

Using slope m =1/2 and point(4,3) we can find y-intercept b

$y=mx+b\\3=\frac{1}{2}(4)+b\\3=2+b\\b=3-2\\b=1$

The equation of line will be:

$y=mx+b\\y=\frac{1}{2}x+1$

Equation of the line that passes through the given point and is (b) perpendicular is: $\mathbf{y=\frac{1}{2}x+1}$

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3 years ago

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attashe74 [19]

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The answer, r = t⁻³.

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n₅ = (t⁻¹)(t⁻³) = t⁻⁴
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The answers for the next three terms are as reflected above as n₄, n₅, and n₆, respectively.

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Ulleksa [173]

Step-by-step explanation:

let equation be y = mx + b.

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sub (1, 5):

5 = 2(1) + b

b = 3

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3 years ago