Answer:
D) 3.8 cm
Step-by-step explanation:
There are several ways this problem can be solved. Maybe the easiest is to use the Law of Cosines to find angle BAC. Then trig functions can be used to find the length of the chord.
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In triangle BAC, the Law of Cosines tells us ...
a² = b² +c² -2bc·cos(A)
A = arccos((b² +c² -a²)/(2bc)) = arccos((8² +6² -3²)/(2·8·6)) = arccos(91/96)
A ≈ 18.573°
The measure of half the chord is AB times the sine of this angle:
BD = 2(AB·sin(A)) ≈ 3.82222
The length of the common chord is about 3.8 cm.
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<em>Additional comment</em>
Another solution can be found using Heron's formula to find the area of triangle ABC. From that, its altitude can be found.
Area ABC = √(s(s-a)(s-b)(s-c)) . . . . where s=(a+b+c)/2
s=(3+8+6)/2 = 8.5
A = √(8.5(8.5 -3)(8.5 -8)(8.5 -6)) = √54.4375 ≈ 7.64444
The altitude of triangle ABC to segment AC is given by ...
A = 1/2bh
h = 2A/b = 2(7.64444)/8 = 1.911111
BD = 2h = 3.822222
Answer:
0.360 % or 40%
Step-by-step explanation:
Area circle = 50.27 (50)
Area square = 32.3 (32)
Area circle minus area square = 50.27-32.3= 17.97 (18)
18/50 = 9/25 = 0.36
<h3>Option B</h3><h3>At 2 second and 1.75 second, the object be at a height of 56 feet</h3>
<em><u>Solution:</u></em>
Given that,
<em><u>The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation:</u></em>

<em><u>At what times will the object be at a height of 56 feet</u></em>
<em><u>Substitute h = 56</u></em>

Solve the above equation by quadratic formula


Thus, at 2 second and 1.75 second, the object be at a height of 56 feet
Answer:
9sin (3)and f,(0)=4,AND f(0)=2