All categories

3 years ago

Find the parametric equation of the line r(t), 0 ≤ t ≤ 1, between P = (2, − 3, 1) and Q = (− 2, 1, 7).

Mathematics

1 answer:

alisha [4.7K]3 years ago

7 0

Answer:

x=-2+3t , y=-9-10t

Step-by-step explanation:

I'm going to start by solving the first equation for t....which will give (x-x1)/(x2-x1) =t.

You can now substitute this in the second equation to get....y=y1 + (y2-y1)(x-x1)/(x2-x1)

Rearranging, you get y-y1 = (y2-y1) / (x2-x1) * (x-x1) which is equivalent to y-y1= m(x-x1) which is the point-slope formula for a line.

So since t was varying between 0 and 1,

plugging in t=0 you get x=x1 and y = y1 givin you the coordinate (x1,y1)

similarly, plugging in t=1 leads to (x2,y2)

So this is a representation of a segment connecting these 2 points.

B) just plug in the points into your given formula from part a to get...

x=-2 +(1--2)t and y = 9 +(-1-9)t

or x=-2+3t , y=-9-10t

You might be interested in

Simplify the following polynomial

Amiraneli [1.4K]

Answer:

-3x^4 - 13x^3 + 14x - 7

Step-by-step explanation:

(5x^4 – 9x^3 + 7x – 1) + (-8x^4 + 4x^2 – 3x + 2) - (-4x^3 + 5x - 1)(2x – 7)

simplify multiplied terms

(-4x^3 + 5x - 1)(2x – 7)

(-4x^3+10x-8)

group like terms together

(5x^4-8x^4) + (-9x^3-4x^3) + (7x-3x+10x) + (-1+2-8)

simplify grouped terms

-3x^4 - 13x^3 + 14x - 7

3 0

3 years ago

Find the values of k so that each remainder is three. 10. (x^2+ 5x + 7) = (x + k)

Goryan [66]

Answer:

$k=1\text{ or } k=4$

Step-by-step explanation:

We can use the Polynomial Remainder Theorem. It states that if we divide a polynomial P(x) by a binomial in the form (x - a), then our remainder will be P(a).

We are dividing:

$(x^2+5x+7)\div(x+k)$

So, a polynomial by a binomial factor.

Our factor is (x + k) or (x - (-k)). Using the form (x - a), our a = -k.

We want our remainder to be 3. So, P(a)=P(-k)=3.

Therefore:

$(-k)^2+5(-k)+7=3$

Simplify:

$k^2-5k+7=3$

Solve for k. Subtract 3 from both sides:

$k^2-5k+4=0$

Factor:

$(k-1)(k-4)=0$

Zero Product Property:

$k-1=0\text{ or } k-4=0$

Solve:

$k=1\text{ or } k=4$

So, either of the two expressions:

$(x^2+5x+7)\div(x+1)\text{ or } (x^2+5x+7)\div(x+4)$

Will yield 3 as the remainder.

5 0

3 years ago

I need help with this

Fynjy0 [20]

Answer:

a stationary sells 8pencil for the cost price of 10pencil find his gain percent

5 0

3 years ago

The expression when m=6 and n=4

OverLord2011 [107]

Step 1: plug in the numbers. 7(6)-4
Step 2 multiply 7x6 42-4

7 0

3 years ago

Use the graph to determine the domain and range of the relation, and state whether the relation is a function.

never [62]

Answer:

the domain is { 0, 1, 2, 3, 4... to positive infinity}

The range is {-1, -3, 0, -4 and so on}

this is not a function

Step-by-step explanation:

domain is a set of all x values, so just write the x values based on the graph

range is a set of all y values, so just write y values based on the graph

not function because it doesnt satisfy with the vertical line test.

7 0

3 years ago