9514 1404 393
Answer:
3) D
4) D
5) B
6) B
Step-by-step explanation:
3) A "Pythagorean triple" is a set of 3 integers that could be the sides of a right triangle. The only triangle shown with integer side lengths is choice D.
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5) The only triangle for which the square of the hypotenuse is the sum of the squares of the other two sides is choice D.
11² +2² = (5√5)²
121 + 4 = 125
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6) The lengths of sides of each right triangle are 2/3 of those of choice D in problem 4. That is, they are a 7-24-25 triangle, multiplied by 2. That means the height is 24·2 = 48, and the area is ...
A = 1/2bh
A = 1/2(28 m)(48 m) = 672 m² . . . . matches choice B
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7) The side lengths of a 30-60-90 triangle have the ratios 1 : √3 : 2. That is, the short leg is half the hypotenuse, a fact stated in A and D. Those true statements make it clear that statement B is false.
(343.98)x400,000/(413.4)
137,592,000/413.4
332,830.188679
Answer:
Step-by-step explanation:
Given that angle A is in IV quadrant
So A/2 would be in II quadrant.
sin A = -1/3
cos A =
(cos A is positive since in IV quadrant)
Using this we can find cos A/2
Answer:
g ≈ 34.7 in
Step-by-step explanation:
The law of sines is useful for this:
f/sin(F) = g/sin(G)
Multiplying by sin(G), we have ...
g = f·sin(G)/sin(F) = (61 in)·sin(34°)/sin(79°)
g ≈ 34.7 in
<h3>
Answer: (4,2)</h3>
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Explanation:
C is at (0,0). Ignore the other points.
Reflecting over y = 1 lands the point on (0,2) because we move 1 unit up to arrive at the line of reflection, and then we keep going one more unit (same direction) to complete the full reflection transformation. I'll call this point P.
Then we reflect point P over the line x = 2 to arrive at the location Q = (4,2). Note how we moved 2 units to the right to get to the line of reflection, and then keep moving the same direction 2 more units, then we have applied the operation of "reflect over the line x = 2"
So we have started at C = (0,0), moved to P = (0,2) and then finally arrived at the destination Q = (4,2). This is the location of C' as well.
All of this is shown in the diagram below.