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bonufazy [111]
3 years ago
10

X-3 and 2x-1 are factors of 2x^3-px^2-2qx+q. Find the values of p and q.

Mathematics
1 answer:
Nonamiya [84]3 years ago
8 0

Answer:

p =1

q = 9

Step-by-step explanation:

f(x) = 2x³ - px² + 2qx + q

(x - 3) is a factor of f(x)

⇒f(3)  = 0

2(3)³ - p*3² - 2q*3 +q = 0

2*27 - 9p - 6q + q = 0

    54 - 9p - 5q = 0

            -9p - 5q = -54 -------------------(I)

(2x - 1) is a factor of f(x)

2x - 1 = 0

      2x = 1

        \st x = \dfrac{1}{2}

f(1/2) = 0

2*(\dfrac{1}{2})^{3}-p*(\dfrac{1}{2})^{2}-2q*\dfrac{1}{2}+q=0\\\\2*\dfrac{1}{8}-p*\dfrac{1}{4}-q+q = 0\\\\\dfrac{1}{4}-\dfrac{1}{4}p =0\\\\[Multiply the entire equation by 4]\\\\4*\dfrac{1}{4}-4*\dfrac{1}{4}p=0\\\\

1 - p = 0

    -p = -1

p = 1

Substitute p =1 in equation (I)

-9*1 - 5q = -54

 -9 - 5q = -54

      -5q = -54 + 9

      -5q = -45

           q = -45/(-5)

q = 9

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On a coordinate plane, a circle has a center at (0, 0). Point (3, 0) lies on the circle.
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Answer:

The correct option is;

No, the distance from (0, 0) to (2, √6) is not 3 units

Step-by-step explanation:

The given parameters of the question are as follows;

Circle center (h, k) = (0, 0)

Point on the circle (x, y) = (3, 0)

We are required to verify whether point (2, √6) lie on the circle

We note that the radius of the circle is given by the equation of the circle as follows;

Distance \, formula = \sqrt{\left (x_{2}-x_{1}  \right )^{2} + \left (y_{2}-y_{1}  \right )^{2}}

Distance² = (x - h)² + (y - k)² = r² which gives;

(3 - 0)² + (0 - 0)² = 3²

Hence r² = 3² and r = 3 units

We check the distance of the point (2, √6) from the center of the circle (0, 0) as follows;

\sqrt{\left (x_{2}-x_{1}  \right )^{2} + \left (y_{2}-y_{1}  \right )^{2}} = Distance

Therefore;

(2 - 0)² + (√6 - 0)² = 2² + √6² = 4 + 6 = 10 = √10²

\sqrt{\left (2-0 \right )^{2} + \left (\sqrt{6} -0 \right )^{2}} = 10

Which gives the distance of the point (2, √6) from the center of the circle (0, 0) = √10

Hence the distance from the circle center (0, 0) to (2, √6) is not √10 which s more than 3 units hence the point  (2, √6), does not lie on the circle.

4 0
3 years ago
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Step-by-step explanation:

Since the set

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<em>{F(1,0), F(0,1)}  </em>

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F(1,0) = a+b = (2i-2j+k)+(i+2j+k) = 3i+2k

F(0,1) = a+c = (2i-2j+k)+(2i+j+2k) = 4i-j+3k

Now, we just have to find the equation of the plane that contains the vectors 3i+2k and 4i-j+3k

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Let x,y be the two numbers.

Given that one number is 8 greater than another.

Let x be the smaller number ans y be the greater number.

That is y=x+8. Let this be the first equation.

And also given that product of the two numbers is 84.

That is x × y = 84, let us plugin y=x+8 here.

x × (x+8) = 84

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That is x= 4 or -12.

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Hence two positive numbers corresponding to given conditions are 4,12.

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Step-by-step explanation:

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