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strojnjashka [21]
3 years ago
11

Write the polynomial in standard form. Then find its degree and the leading coefficient.

Mathematics
1 answer:
Contact [7]3 years ago
3 0

Answer:

6c^5+4c−301, assuming that you meant 6c^5  

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Quadrilateral ABCD has coordinates A (3,5), B (5,2), C (8,4), D (6, 7).
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B. Square, because all four sides are congruent and adjacent sides are perpendicular  

Step-by-step explanation:

A graph of your figure is below.

The figure is a square because

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Determine the equation of the line that is perpendicular to the lines r​(t) equalsleft angle6​t,1 plus3 ​t,4tright angle and R​(
Lynna [10]

\vec r(t)=\langle6t,1+3t,4t\rangle

\vec R(s)=\langle2+s,-8+3s,-12+4s\rangle

Take the derivatives of each to get the tangent vectors:

\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=\langle6,3,4\rangle

\dfrac{\mathrm d\vec R(s)}{\mathrm ds}=\langle1,3,4\rangle

Take the cross product of the tangent vectors to get a vector that is normal to both lines:

\langle6,3,4\rangle\times\langle1,3,4\rangle=\langle0,-20,15\rangle

The two given lines intersect when \vec r(t)=\vec R(s):

\langle6t,1+3t,4t\rangle=\langle2+s,-8+3s,-12+4s\rangle\implies t=1,s=4

that is, at the point (6, 4, 4).

The line perpendicular to both of the given lines through the origin is obtained by scaling the normal vector found earlier by \tau\in\Bbb R; translate this line by adding the vector \langle6,4,4\rangle to get the line we want,

\vec\rho(\tau)=\langle6,4,4\rangle+\langle0,-20,15\rangle\tau

\boxed{\vec\rho(\tau)=\langle6,4-20\tau,4+15\tau\rangle}

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