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Anvisha [2.4K]
3 years ago
14

PLZ HELP, DUE TONIGHT!! Find the values of a and b such that f(x) is continuous at x=1

Mathematics
1 answer:
Sedaia [141]3 years ago
6 0

Answer:

  • a = 2, b = -4

Step-by-step explanation:

<u>To make the given piecewise function continuous we need:</u>

  • ax² - b = 6 and 5ax + b = 6 at x = 1

<u>It gives us the system:</u>

  • a(1²) - b = 6 ⇒ a - b = 6
  • 5a(1) + b = 6 ⇒ 5a + b = 6

<u>Sum of the two:</u>

  • a - b + 5a + b = 6 + 6
  • 6a = 12
  • a = 2

<u>Then b is:</u>

  • 2 - b = 6
  • b = -4
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XY is a diameter of a circle and Z is a point on the circle such that ZY=6. If the area of the triangle XYZ is 18 square root 3
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<h2>Answer:</h2>

4π

<h2>Step-by-step explanation:</h2>

As shown in the diagram, triangle XYZ is a right triangle. Therefore, its area (A) is given by:

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Where;

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<em>Substitute these values into equation(i) and solve as follows:</em>

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Triangle XOZ is isosceles, therefore the following are true;

(i) |OZ| = |OX|

(ii) XZO = ZXO = 30°

(iii) XOZ + XZO + ZXO = 180°   [sum of angles in a triangle]

=>  XOZ + 30° + 30° = 180°

=>  XOZ + 60° = 180°

=>  XOZ = 180° - 60°

=>  XOZ = 120°

Therefore we can calculate the radius |OZ| of the circle using sine rule as follows;

\frac{sin|XOZ|}{XZ} = \frac{sin|ZXO|}{OZ}

\frac{sin120}{6\sqrt{3} } = \frac{sin 30}{OZ}

\frac{\sqrt{3} /2}{6\sqrt{3} } = \frac{1/2}{|OZ|}

\frac{1}{12}  = \frac{1}{2|OZ|}

\frac{1}{6} = \frac{1}{|OZ|}

|OZ| = 6

The radius of the circle is therefore 6.

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The length(L) of an arc is given by;

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r = radius of the circle.

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θ = ZOX = 120°

r = |OZ| = 6

Substitute these values into equation (ii) as follows;

L = 120/360 x 2π x 6

L = 4π

Therefore the length of the arc XZ is 4π

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