All categories

3 years ago

Pls help find shaded area

Mathematics

1 answer:

Pavlova-9 [17]3 years ago

6 0

<h3>Exact area = 12pi - 9*sqrt(3) square inches</h3><h3>Approximate area = 22.1106545749577 square inches</h3>

==========================================================

Work Shown:

Assuming the chord is horizontal along with the diameter, then this means the two smallest minor arcs are both 30 degrees each. This leaves the northern most arc to be 180-30-30 = 120 degrees in measure

A = area of northern sector (you can think of it as a pizza slize)

A = (angle/360)*pi*r^2

A = (120/360)*pi*6^2

A = 12pi

B = area of triangle

B = (1/2)*side1*side2*sin(included angle)

B = (1/2)*6*6*sin(120)

B = 18*sqrt(3)/2

B = 9*sqrt(3)

C = area of shaded region

C = A - B

C = 12pi - 9*sqrt(3) .... exact area

C = 22.1106545749577 ... use a calculator to get the approximate area

You might be interested in

PLEASE HELP ME!!!!! Factor completely b^4 − b^3 + b − 1

Afina-wow [57]

7 0

3 years ago

Gronala costs $4.50 per pound. Assorted nuts cost $7.00 per pound. how many pounds of granola and nuts are needed to make 7 poun

yulyashka [42]

Let G be the granola we need (in pound)

Let N be the nut we need (in pound)

Trail mix cost $5.00 per pound---> cost for 7 pound = 5 x 7 = $ 35

4.5G+7N= 35

G+ N = 7

Therefore G = 5.6 N=1.4, we would need 5.6 pound granola & 1.4 pound nut

7 0

3 years ago

Read 2 more answers

Which of the following is true of the values of x and y in the diagram below?

belka [17]

Answer: y/x = 1

Step-by-step explanation:

It was on quizlet

8 0

3 years ago

If <img src="https://tex.z-dn.net/?f=f%28x%29%3Dx%5E2-2" id="TexFormula1" title="f(x)=x^2-2" alt="f(x)=x^2-2" align="absmiddle"

jek_recluse [69]

Answer:

$\boxed{f(6) = 34}$

Step-by-step explanation:

Composition of functions occurs when we have two functions normally written similar or exactly like f(x) & g(x) - you can have any coefficients to the (x), but the most commonly seen are f(x) and g(x). They are written as either f(g(x)) or (f o g)(x). Because our composition is written as $(f \circ g)(2)$ , we are replacing the x values in the g(x) function with 2 and simplifying the expression.

$g(2) = 3(2)$

$g(2) = 6$

Now, because we are composing the functions, this value we have solved for now replaces the x-values in the f(x) function. So, f(x) becomes f(6), and we use the same manner as above to simplify.

$f(6) = (6)^2-2$