Answer:
Andre is not correct; each card is .0054 inches thick
Step-by-step explanation:
Since the stack is 2.16 inches and there's 400 cards, divide 2.16 by 400 to get .0054 inches per card
<span>Suppose a Cartesian plane where the positive x-axis represents 0°. An angle measuring 92° would have a terminal side that lies in Quadrant II. This is because quadrants are counted counterclockwise from the positive x-axis. An angle of 92° would have a terminal side just past the positive y-axis, therefore being in quadrant II.</span>
X-6y=32 Eqn(1)
X-4y=24. Eqn(2)
From eqn(1)
X=32+6y. Eqn(3)
Put Eqn 3 to Eqn 2
(32+6y)-4y=24
32+6y-4y=24
32+2y=24
2y=24-32
Y=-8/2
Y =-4
Put y into Eqn 3
X=32+6y
X=32+6(-4)
X=32-24
X=8
The answer is B. It is an acute triangle because of 22^2 + 51^2 > 54^2. To identify the type of triangle referred to the problem, we use the Pythagorean Theorem which involves squaring the lengths of all sides of the triangle. If the hypotenuse’s square is less than the two smallest sides' squares added together, than it is an acute triangle. However, if the hypotenuse’s square is larger, then it is an obtuse triangle.
54 is the hypotenuse since it is the longest side.
Following the Pythagorean Theorem, a^2 + b^2 = c^2:
54^2=2916
51^2=2601
22^2=484
It becomes 2601+484 = 2916.
When the two shorter sides are added, the sum is greater than the square of the hypotenuse which is 2916. Thus, the formed triangle is acute.