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3 years ago

Finding x - cicrle theorems

Mathematics

2 answers:

Setler79 [48]3 years ago

4 0

Answer:

x = 30

Step-by-step explanation:

AB is a tangent to the circle so ∠ ABO = 90°

The sum of the 3 angles in a triangle = 180° , then

x + 2x + 90 = 180

3x + 90 = 180 ( subtract 90 from both sides )

3x = 90 ( divide both sides by 3 )

x = 30

ella [17]3 years ago

3 0

Answer: x = 30

OA cuts the circle (O) at C

because ∠ABC is the angle created by tangent rays and chord

=> ∠ABC = 1/2.∠AOB = 1/2.2x = x

ΔOBC has OB = OC (are both radii)

=> ΔOBC is the isosceles triangle at O

=> ∠OCB = (180° - ∠COB)/2 = (180° - 2x)/2 = 90° - x

because ∠OCB is the outer angle of ΔABC

=> ∠OCB = ∠ABC + ∠CAB

⇔ 90° - x = x + x

⇔ 90 = 3x

⇔ x = 30

Step-by-step explanation:

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Which equation describes a relationship such that x = 0 must be excluded from the domain?

zalisa [80]

Answer:

Example: f(x) = (1 + x + x^2) / x, x=0 is excluded from the domain, since division by zero is undefined.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

HELP with these questions

zlopas [31]

Step-by-step explanation:

transform the parent graph of f(x) = ln x into f(x) = - ln (x - 4) by shifting the parent graph 4 units to the right and reflecting over the x-axis

(???, 0): 0 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

0 = ln (x - 4)

$e^{0} = e^{ln (x - 4)}$

1 = x - 4

+4 +4

5 = x

(5, 0)

(???, 1): 1 = - ln (x - 4)

$\frac{0}{-1} = \frac{-ln (x - 4)}{-1}$

1 = ln (x - 4)

$e^{1} = e^{ln (x - 4)}$

e = x - 4

+4 +4

e + 4 = x

6.72 = x

(6.72, 1)

Domain: x - 4 > 0

+4 +4

x > 4

(4, ∞)

Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that

No vertical asymptotes

*************************************************************************

transform the parent graph of f(x) = 3ˣ into f(x) = - 3ˣ⁺⁵ by shifting the parent graph 5 units to the left and reflecting over the x-axis

Domain: there is no restriction on x so domain is all real number

(-∞, ∞)

Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0. the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.

(-∞, 0)

Y-intercept is when x = 0:

f(x) = - 3ˣ⁺⁵

= - 3⁰⁺⁵

= - 3⁵

= -243

Horizontal Asymptote: y = 0 (explanation above)