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8_murik_8 [283]
3 years ago
7

For problems 29 - 31 the graph of a quadratic function y=ax^2 + bx + c is shown. Tell whether the discriminant of ax^2 + bx + c

= 0 is positive, negative, or zero.

Mathematics
1 answer:
Elan Coil [88]3 years ago
3 0

Answer:

29) discriminant  is positive

30) discriminant  is 0

31) discriminant  is negative

Step-by-step explanation:

the graph of a quadratic function y=ax^2 + bx + c is shown. Tell whether the discriminant of ax^2 + bx + c = 0 is positive, negative, or zero.

In the graph of question number 29 we can see that the graph intersects the x axis at two points

so the equation has 2 solutions.

When the equation has two solution then the discriminant is positive

In the graph of question number 30 we can see that the graph intersects the x axis at only one point

so the equation has only  1 solution.

When the equation has only one solution then the discriminant is equal to 0

In the graph of question number 30 we can see that the graph does not intersects the x axis  

so the equation has 2 imaginary solutions.

When the equation has two imaginary solutions then the discriminant is negative

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Y_Kistochka [10]

Hello from MrBillDoesMath!

Answer:   8 (Pi - sqrt(3))

Discussion:

The area of the shaded region is that of the semicircle minus the area of the triangle..


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    Where R^2 is the square of the radius of the circle. In our case, R ( = OC)

     = 4 so the semicircle area is

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Area of triangle.

   First of all, angle ACB is a right angle ( i.e. 90 degrees).

     * This is the Theorem of Thales from elementary Plane Geometry. *

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 But CB = 4 (given) and AB = 4*2 = 8 ( the diameter is twice the radius).

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    Hence AC = sqrt(48) = sqrt (16*3) = 4 * sqrt(3)

We are almost done! The area of the triangle is given by

   (1/2) b * h = (1/2)  BC * AC = (1/2) 4 * (4 * sqrt(3)) =  8 sqrt(3)

We conclude the area area of the shaded part is

  8 PI - 8 sqrt(3)   = 8 (Pi - sqrt(3))

Note that sqrt(3) is approx  1.7 so (PI - sqrt(3)) is a positive number, as it better well be!


Thank you,

MrB

6 0
3 years ago
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