Nxnxnxbbxnxndnnsnsn nsnsnsndnnsndnndndnsndnndndnndnd
Answer:
Step-by-step explanation:
We are given a circle with a partially shaded region. First, we need to determine the area of the whole circle. To do this, we need the measurement of the radius of the circle:
Use the Pythagorean theorem to solve for the other leg of the right triangle inside the circle:
5^2 = 3^2 + x^2
x = 4
The radius is 4 + 1 cm = 5 cm
So the area of the circle is A = pi*r^2
A = 3.14 * (5)^2
A = 25pi cm^2
To solve for the area of the shaded region:
Ashaded = Acircle - Atriangles
we need to solve for the area of the triangles:
A = 1/2 *b*h
A = 1/2 *6 * 5
A = 15 cm^2
Atriangles = 2 * 15
Atriangles = 30 cm^2
Ashaded = 25pi - 30
40√3+432/7 is the answer
Hope this helps :D
Absolute value means the distance from 0.
|-0.45| =0.45
|-0.0045|=0.0045
0.45>0.0045
So
|-0.45|>|-0.0045|
<h2>JK = 18m</h2><h2>_______________</h2>
<u>Step-by-step explanation:</u>
ΔJLK = ΔNLM ( <em>v</em><em>e</em><em>r</em><em>t</em><em>i</em><em>c</em><em>a</em><em>l</em><em>l</em><em>y</em><em> </em><em>o</em><em>p</em><em>p</em><em>o</em><em>s</em><em>i</em><em>t</em><em>e</em><em> </em>Δ)
ΔJKL = ΔNML ( <em>e</em><em>a</em><em>c</em><em>h</em><em> </em><em>9</em><em>0</em><em>°</em><em> </em><em>)</em>
so,
triangle JKL = triangle NML (<em>b</em><em>y</em><em> </em><em>A</em><em>A</em><em> </em><em>s</em><em>i</em><em>m</em><em>i</em><em>l</em><em>a</em><em>r</em><em>i</em><em>t</em><em>y</em><em>)</em>
JK / KL = NM / ML
JK / 21m = 42m / 49m
JK = 42 × 21 ÷ 49
JK = 18m
<h2>_______________</h2><h2>FOLLOW ME</h2>
