3:5 my answer has to be 20 characters so im just typing the rest of this to answer your question.
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
Let volume of 4% solution = x and of 14% be y
them x + y = 65........(1) also using the actual weights of salt:-
0.04x + 0.14y = 65*0.06 = 3.9
multiplying the above equation by -25 we have
-x - 3.5y = -97.5 adding this to equation (1)
--2.5y = -32.5
y= 13
and x = 65 - 13 = 52
so we need 13 mls of the 14%, and 52 mls of the 4% solution.
The answer is D. when an exponent on the outside you multiply the exponents
162 is not a perfect square therefore you have to find the largest perfect square that can go into 162 which would be 81.