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timofeeve [1]
3 years ago
10

The line segment joining the points ( 3,-4) and (1,2) is trisected at the points P and Q . If the coordinates of P and Q are (p

, -2 ) and ( 5/3 , q ) respectively, find the values of p and q.
Mathematics
1 answer:
sdas [7]3 years ago
3 0

Answer:

P = 7/3 and q = 0

Step-by-step explanation:

The given parameters are;

The endpoints of the segment are, (3, -4) and (1, 2)

The points that trisect the given points are P and Q with coordinates (p, -2) and (5/3, q) respectively

Therefore, we have;

The first point of the trisection cuts 1/3 of the length from one point and the second point of the trisection cuts 2/3 of the length from the same point

The coordinates of P or Q = (3 - (3 - 1)/3, -4 - (-4 - 2)/3) = (7/3 , -2)

Therefore, given that the y-coordinate value of the derived point coincides with the y-coordinate value of the point P, (p, -2) and there is only one point with x = -2 on the line, we have that the coordinate of the point P is (p, -2) = (7/3 , -2)

∴ P = 7/3

Similarly we have the second point of the trisection, Q, given as follows;

We are already given the x-coordinate value of the point Q as the 5/3 in (5/3, q)

Point Q = (5/3, q) = (3 - 2×(3 - 1)/3, -4 - 2×(-4 - 2)/3) = (5/3 , 0)

Point Q = (5/3, q) = (5/3 , 0)

∴ q = 0.

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For each direct variation find the constant of variation then find the value of Y when X equals -0.5. Y equals 2 when X equals 3
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For direct variation y varies directly with x, then we use equation y = kx

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Please help!!!
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Answer:

c. 2a^3x^4-3a^3x^3+a^4x^2

d. 63x^5y^3-30x^2y^4-7x^3y^2

Step-by-step explanation:

For this exercise you need to remember the following:

 1. The multiplication of signs:

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2. The Product of powers property, which states that:

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d. (6.3x^3y-3y^2-0.7x)(10x^2y^2)=\\\\=(6.3x^3y)(10x^2y^2)+(-3y^2)(10x^2y^2)+(-0.7x)(10x^2y^2)=63x^5y^3-30x^2y^4-7x^3y^2

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