<h3>Answer:</h3>
56 m²
<h3>Explanation:</h3>
The altitude from point B to segment CP is the same for ∆BMP as for ∆BMC. Since both have the same base length (MP = MC), both have the same area, 21 m². Hence the area of ∆CPB is (21+21) m² = 42 m².
The altitude from point C to segment AB is the same for ∆CPA as for ∆CPB, so the areas of those triangles will have the same proportion as the base segments AP and BP. That is, ...
... AACPA : ACPB = PA : PB = 1 : 3
The ratio of ACPB to the total is then ...
... ACPB : (ACPA +ACPB) = 3 : (1+3) = 3 : 4
The area of ∆ABC is the total of the areas of the smaller triangles CPA and CPB, so we have
... ACPB : AABC = 3 : 4
... AABC/42 m² = 4/3 . . . . . rearranging slightly and substituting for ACPB
... AABC = (42 m²)×(4/3) . . . . multiply by the denominator
... AABC = 56 m²
...so M must equal K. Since K and M would be across from each other on the completed rectangle, and because of how rectangles work, M would have to be equal to K. I hope this helped!
Answer:
c
Step-by-step explanation:
the domain is every possible input value and the range is every possible output value. since the outputs are all positive, the range is equal to all real numbers. and since the graph starts at 1,000, the domain is greater than or equal to 1,000
So the probablity is desired outcomes divided by total possible outcomes
so assume duncan gets an orange on the first try, that would be 4/9
then carlos picks another fruit, also assuming that duncan got an orange so 5/8
then these are dependant on each other so we multiply
4/9 times 5/8=20/72=10/36=5/18