The cross product of the normal vectors of two planes result in a vector parallel to the line of intersection of the two planes.
Corresponding normal vectors of the planes are
<5,-1,-6> and <1,1,1>
We calculate the cross product as a determinant of (i,j,k) and the normal products
i j k
5 -1 -6
1 1 1
=(-1*1-(-6)*1)i -(5*1-(-6)1)j+(5*1-(-1*1))k
=5i-11j+6k
=<5,-11,6>
Check orthogonality with normal vectors using scalar products
(should equal zero if orthogonal)
<5,-11,6>.<5,-1,-6>=25+11-36=0
<5,-11,6>.<1,1,1>=5-11+6=0
Therefore <5,-11,6> is a vector parallel to the line of intersection of the two given planes.
Answer:
The standard form is 8 y ⁵ - 17 y⁴ + 6 y³ +2 y² - 11
The degree of given polynomial is '5'
the co-efficient of y⁴ is '-17'
Step-by-step explanation:
Given standard form 2 y²+ 6 y³-11-17 y⁴+8 y⁵
<em>The form ax² + b x + c is called the standard form of the quadratic expression of 'x'.This is second degree standard form of polynomial.</em>
<em>The form ax⁵ + b x⁴ + c x³ +d x² +ex +f is called the standard form of the quadratic expression of 'x'.This is fifth degree standard form of polynomial</em>
now Given polynomial is 2 y²+ 6 y³-11-17 y⁴+8 y⁵
The standard form is
8 y ⁵ - 17 y⁴ + 6 y³ +2 y² - 11
<u><em>Conclusion</em></u>:-
<em>The degree of given polynomial is '5'</em>
<em>The co-efficient of y⁴ is '-17'</em>
<em> </em>
Answer:
920ft^2
Step-by-step explanation:
20*17=340
15*20=300
60*2=120
8*20=160
340+300+120+160= 920 ft^2
147 = 3 * 7 * 7
245 = 5 * 7 * 7
so the gcf is 7*7 = 49
abd gcf of y^3 and y^2 is y^2
answer:- 49y^2
