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OverLord2011 [107]
3 years ago
10

Solve for v 5v^2 +21v=-4

Mathematics
2 answers:
andrey2020 [161]3 years ago
8 0

0.000000000000000000000000000000000000

aksik [14]3 years ago
6 0

Answer

=-15

v=-4

Step-by-step explanation:

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Consider the planes 4x+3y+4z=1 and 4x+4z=0. (A) Find the unique point P on the y−axis which is on both planes. ( 0 equation edit
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Step-by-step explanation:

A) From the order of the exercise we already know that the intersection points lies on the Y-axis, so its coordinates are P(0;y;0). In order to find it, we only need to substitute the equation 4x+4z=0 into the equation 4x+3y+4z=1. Then,

1=4x+3y+4z = 3y + (4x+4z)= 3y+0.

From the expression above it is easy to obtain that y=1/3, and the intersection point is P(0;1/3;0).

B) To obtain the parallel vector to both planes we use the cross product of the normal vector of the planes.

\left[\begin{array}{ccc}i&j&k\\4&3&4\\4&0&4\end{array}\right] = 12i-0j+12k

As we want a unit vector, we must calculate the modulus of u:

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C) In order to obtain the vector equation of the intersection line of both planes, we just need to put together the above results.

r = \frac{1}{3}j +\lambda \left( \frac{\sqrt{2}}{2}i- \frac{\sqrt{2}}{2}k \right)

where \lambda is a real number.

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