Any one can help me with problem 4?
1 answer:
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<span>B) the bisectors of angles D, E, and F
</span>
In an inscribed circle of a triangle, all angle bisectors will pass through the center of the circle.
Pls. see attachment.
1st attachment is Triangle DEF. 2nd attachment is how inscribed circle relates to the triangle it is inscribed in.
</span>
In an inscribed circle of a triangle, all angle bisectors will pass through the center of the circle.
Pls. see attachment.
1st attachment is Triangle DEF. 2nd attachment is how inscribed circle relates to the triangle it is inscribed in.
Answer:
1.9cm
Step-by-step explanation:
The density d of a material is related to its mass m and volume V as follows;
d = ------------------(i)
The material in question here is the lead ball.
Now, from known experiment;
the density of lead is <em>11.34g/cm³</em>
From the question, the weight/mass of the lead ball is <em>326g</em>
Substitute these values into equation (i) as follows;
11.34 =
V =
V = 28.75cm³
Now, since the ball is of course spherical, we can get the radius by using the following relation from the volume of a sphere;
V = [V = volume, r = radius]
V = 28.75cm³
=> 28.75 =
=> 3 x 28.75 = 4 π r³
=> 86.25 = 4 π r³
=> 21.5625 = π r³ [Take π = 3.142]
=> 21.5625 = (3.142) r³ [divide both sides by 3.142]
=> 6.86 = r³ [Take the cube root of both sides]
=> ∛6.86 = ∛r³
=> 1.90 = r
Therefore, the radius is 1.9cm to the nearest tenth