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kifflom [539]
2 years ago
13

if alpha and beta are zeros of the polynomial x² - 6 X + k find k such that (alpha+beta)²-2alha beta =40​

Mathematics
1 answer:
lidiya [134]2 years ago
8 0

Answer:

k = - 2

Step-by-step explanation:

Given α and β are the zeros of x² - 6x + k = 0 , with

a = 1, b = - 6 and c = k , then

α + β = - \frac{b}{a} = - \frac{-6}{1} = 6

αβ = \frac{c}{a} = \frac{k}{1} = k

Then solving

(α + β)² - 2αβ = 40

6² - 2k = 40

36 - 2k = 40 ( subtract 36 from both sides )

- 2k = 4 ( divide both sides by - 2 )

k = - 2

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X² - 4x - 77 = 0

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x + 7 = 0

x - 11 = 0

x = -7
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3 years ago
A quadrilateral has vertices at $(0,1)$, $(3,4)$, $(4,3)$ and $(3,0)$. Its perimeter can be expressed in the form $a\sqrt2+b\sqr
seraphim [82]

Answer:

a + b = 12

Step-by-step explanation:

Given

Quadrilateral;

Vertices of (0,1), (3,4) (4,3) and (3,0)

Perimeter = a\sqrt{2} + b\sqrt{10}

Required

a + b

Let the vertices be represented with A,B,C,D such as

A = (0,1); B = (3,4); C = (4,3) and D = (3,0)

To calculate the actual perimeter, we need to first calculate the distance between the points;

Such that:

AB represents distance between point A and B

BC represents distance between point B and C

CD represents distance between point C and D

DA represents distance between point D and A

Calculating AB

Here, we consider A = (0,1); B = (3,4);

Distance is calculated as;

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

(x_1,y_1) = A(0,1)

(x_2,y_2) = B(3,4)

Substitute these values in the formula above

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

AB = \sqrt{(0 - 3)^2 + (1 - 4)^2}

AB = \sqrt{( - 3)^2 + (-3)^2}

AB = \sqrt{9+ 9}

AB = \sqrt{18}

AB = \sqrt{9*2}

AB = \sqrt{9}*\sqrt{2}

AB = 3\sqrt{2}

Calculating BC

Here, we consider B = (3,4); C = (4,3)

Here,

(x_1,y_1) = B (3,4)

(x_2,y_2) = C(4,3)

Substitute these values in the formula above

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

BC = \sqrt{(3 - 4)^2 + (4 - 3)^2}

BC = \sqrt{(-1)^2 + (1)^2}

BC = \sqrt{1 + 1}

BC = \sqrt{2}

Calculating CD

Here, we consider C = (4,3); D = (3,0)

Here,

(x_1,y_1) = C(4,3)

(x_2,y_2) = D (3,0)

Substitute these values in the formula above

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

CD = \sqrt{(4 - 3)^2 + (3 - 0)^2}

CD = \sqrt{(1)^2 + (3)^2}

CD = \sqrt{1 + 9}

CD = \sqrt{10}

Lastly;

Calculating DA

Here, we consider C = (4,3); D = (3,0)

Here,

(x_1,y_1) = D (3,0)

(x_2,y_2) = A (0,1)

Substitute these values in the formula above

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

DA = \sqrt{(3 - 0)^2 + (0 - 1)^2}

DA = \sqrt{(3)^2 + (- 1)^2}

DA = \sqrt{9 +  1}

DA = \sqrt{10}

The addition of the values of distances AB, BC, CD and DA gives the perimeter of the quadrilateral

Perimeter = 3\sqrt{2} + \sqrt{2} + \sqrt{10} + \sqrt{10}

Perimeter = 4\sqrt{2} + 2\sqrt{10}

Recall that

Perimeter = a\sqrt{2} + b\sqrt{10}

This implies that

a\sqrt{2} + b\sqrt{10} = 4\sqrt{2} + 2\sqrt{10}

By comparison

a\sqrt{2} = 4\sqrt{2}

Divide both sides by \sqrt{2}

a = 4

By comparison

b\sqrt{10} = 2\sqrt{10}

Divide both sides by \sqrt{10}

b = 2

Hence,

a + b = 2 + 10

a + b = 12

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A method for determining whether a critical point is a minimum. True or false?
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Answer:

We can find if a critical point is a local minimum or maximum by looking at the second derivatives.

Step-by-step explanation:

If you take the first derivative, you will find the slope at the given point, which if it is a minimum or a maximum will be 0.

Then we take the second derivative. If that number is a positive number, then we have a local minimum. If it is a negative number, then it is a local maximum.

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If Sunil earns $3672 per month, what is his annual salary?
snow_tiger [21]

Answer:

$44064

Step-by-step explanation:

Sunil earns $3672 every month. To find out his annual salary, we will need to multiply this monthly salary by the number of months in a year; 12.

3672 * 12 = $44064

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3 years ago
I don’t understand.<br> online school is hard
Minchanka [31]
I think you got to multiply the numbers hopefully I helped
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3 years ago
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