Answer:
Professor 1 as he gave the maximum number of A's .
its A.........dont u have a calculator
You know that three points A,B,CA,B,C (two vectors A⃗ BA→B, A⃗ CA→C) form a plane. If you want to show the fourth one DD is on the same plane, you have to show that it forms, with any of the other point already belonging to the plane, a vector belonging to the plane (for instance A⃗ DA→D).
Since the cross product of two vectors is normal to the plane formed by the two vectors (A⃗ B×A⃗ CA→B×A→C is normal to the plane ABCABC), you just have to prove your last vector A⃗ DA→D is normal to this cross product, hence the triple product that should be equal to 00:
A⃗ D⋅(A⃗ B×AC)=0
9514 1404 393
Answer:
a) 10(3x+2) ≤ 140; x ≤ 4
b) no
Step-by-step explanation:
a) The area is the product of length and width. We want that product to be no greater than 140:
10(3x +2) ≤ 140 . . . . your inequality
We can solve this by dividing by 10 first.
3x +2 ≤ 14
3x ≤ 12 . . . . . . subtract 2
x ≤ 4 . . . . . . . . divide by 3; the solution to the inequality
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b) If the length were 15, the area would be 10×15 = 150, which is greater than 140. The length cannot be 15 if the area is at most 140.
Answer:
A=18
B=78
Step-by-step explanation:
