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3 years ago

5

In circle O, central angle AOB measures StartFraction pi Over 3 EndFraction radians What is the length of arc AB? 6π cm 12π cm 1

8π cm 36π cm

2 answers:

Radda [10]3 years ago

8 0

Answer:

6 pi

Step-by-step explanation:

edg 2020

Nina [5.8K]3 years ago

7 0

Answer:

a

Step-by-step explanation:

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Which are solutions of x2=-11+4?

Dovator [93]

For this case we have a quadratic equation of the form:

$ax ^ 2 + bx + c = 0\\x ^ 2 + 11x-4 = 0$

Where:

$a = 1\\b = 11\\c = -4$

We find the roots:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}\\x = \frac {-11 \pm \sqrt {11 ^ 2-4 (1) (- 4)}} {2 (1)}\\x = \frac {-11 \pm \sqrt {121 + 16}} {2}\\x = \frac {-11 \pm \sqrt {137}} {2}$

The roots are:

$x_ {1} = \frac {-11+ \sqrt {137}} {2}\\x_ {2} = \frac {-11- \sqrt {137}} {2}$

Answer:

$x_ {1} = \frac {-11+ \sqrt {137}} {2}\\x_ {2} = \frac {-11- \sqrt {137}} {2}$

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3 years ago

Read 2 more answers

Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.

Olin [163]

Answer:

The correct option is d

$f(1) = -5$

$f(2) = 11$

The correct option is d

The correct option is c

the correct option is b

Step-by-step explanation:

The given equation is

$f(x) = x^4 + x -7 =0$

The give interval is $(1,2)$

Now differentiating the equation

$f'(x) = 4x^3 +7 > 0$

Therefore the equation is positive at the given interval

Now at x= 1

$f(1) = (1)^4 + 1 -7 =-5$

Now at x= 2

$f(2) = (2)^4 + 2 -7 =11$

Now at the interval (1,2)

$f(1) < 0 < f(2)$

i.e

$-5 < 0 < 11$

this tell us that there is a value z within 1,2 and

f(z) = 0

Which implies that there is a root within (1,2) according to the intermediate value theorem

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3 years ago

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At first, she had +2y in the upper and -5y in the lower.

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