B it is linear Bc it increases constantly. There is a constant slope that doesn’t change
Step-by-step explanation:
it filled up half the circle (up to the center point) - if we had a full circle. but a little bit is cut off (below AB).
what we see is that the shaded area is the sum of the area of the triangle AOB and 2 equally sized circle segment areas left and right of AOB.
since we are dealing with a half-circle, we have 180° in total. 120° are taken by AOB, so, that leaves us with 180-120 = 60° for both circle segments (so, one has an angle of 30°).
and 2×30° = 1×60°, so we can calculate the area of one 60° segment instead of two 30° segments.
AOB is an isoceles triangle (the legs are equally long, and therefore also the 2 side angles are equal).
the area of this triangle AOB is
1/2 × a × b × sin(C) = 1/2 × 3 × 3 × sin(120) =
= 3.897114317... m²
a circle segment area of 60° is 60/360 = 1/6 of the full circle area (as a full circle = 360°).
so, it's area is
pi×r² × 1/6 = pi×3²/6 = pi×3/2 = 4.71238898... m²
so, the total area of the shaded area is
3.897114317... m² + 4.71238898... m² =
= 8.609503297... m²
Answer:
7(7x -4)
Step-by-step explanation:
49x-28
We can factor a 7 from each term
49x = 7*7x
28 = 7*4
7(7x -4)
Answer:
x = 7
Step by step explanation:
<em>minus</em><em> </em><em>x</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em>
6 = x -1
<em>add</em><em> </em><em>1</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em>
<em>x</em><em> </em><em>=</em><em> </em><em>7</em>
