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3 years ago

Find the arc length of the semicircle

Mathematics

2 answers:

sukhopar [10]3 years ago

4 0

Answer:

18.8496 units (rounded o1ff to four decimal places)

Step-by-step explanation:

Arc length = 2 × π × r × $\frac{central-angle}{360}$

r (radius) = 12 units ÷ 2 = 6 units

Central angle = 180°

For our semicircle, the arc length = 2 × π × 6 × $\frac{180}{360}$ = 18.8495559215 units

Or 18.8496 units (rounded o1ff to four decimal places)

Sergeeva-Olga [200]3 years ago

4 0

Answer:

6pi

Step-by-step explanation:

hop it helps

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3 years ago

The maximum value of the function: f(x)= -5 x ^2 +30x-200 is?

VARVARA [1.3K]

Answer:

$\displaystyle - 155$

Step-by-step explanation:

we are given a quadratic function

$\displaystyle f(x) = - 5 {x}^{2} + 30x - 200$

we want to figure out the minimum value of the function

to do so we need to figure out the minimum value of x in the case we can consider the following formula:

$\displaystyle x _{ \rm min} = \frac{ - b}{2a}$

the given function is in the standard form i.e

$\displaystyle f(x) = a {x}^{2} + bx + c$

so we acquire:

b=30
a=-5

thus substitute:

$\displaystyle x _{ \rm min} = \frac{ - 30}{2. - 5}$

simplify multiplication:

$\displaystyle x _{ \rm min} = \frac{ - 30}{ - 10}$

simply division:

$\displaystyle x _{ \rm min} = 3$

plug in the value of minimum x to the given function:

$\displaystyle f (3)= - 5 {(3)}^{2} + 30.3 - 200$

simplify square:

$\displaystyle f (3)= - 5 {(9)}^{} + 30.3 - 200$

simplify multiplication:

$\displaystyle f (3)= - 45 + 90- 200$

simplify:

$\displaystyle f (3)= - 155$

hence,

the minimum value of the function is -155

5 0

3 years ago