A lead ball weighs 326 grams. Find the radius of the ball to the nearest tenth of a
centimeter.

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 11.34g/cm³

From the question, the weight/mass of the lead ball is 326g

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

he spend 5.1 minutes chatting with his friends

step-by-step explanation:

2/5+1/2=9/10

6/1-9/10=5 1/10

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

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