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The base of a rectangular prism has a length of 13 inches and a width of 1over2. The volume of the prism is less than 65 cubic i
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<h3>1. Inscribe a circle ΔJKL. Identify the point of concurrency that is the center of the circle you drew. </h3>
<h3>2. Circumscribe a circle ΔFGH. Identify the point of concurrency that is the center of the circle you drew. </h3>
To inscribe a circle in the attached triangle, we must follow the following steps. The resultant figure is the first one below.
1. Draw the angle bisector of angle JKL. The straight line in the green one.
2. Draw the angle bisector of angle KLJ. The straight line in the blue one.
3. Draw point D at these two lines.
4. Draw a line through point D perpendicular to side JK. The straight line in the red one.
5. Mark point G at the red line and side JK.
6. Draw a circle with center at D (point of concurrency) and a radius of DG.
<h3>2. Circumscribe a circle ΔFGH. Identify the point of concurrency that is the center of the circle you drew. </h3>
To circumscribe a circle in the attached triangle, we must follow the following steps:
1. Draw the perpendicular bisector of the side FG. The straight line in the red one. The resultant figure is the second one below.
2. Draw the perpendicular bisector of the side GH. The straight line in the blue one.
3. The center of the circle is the intersection of these two lines
4. Draw a circle with center at D (point of concurrency).
<h3>3. Construct the two lines tangents to circle</h3>A tangent to a circle is a straight line that touches the circle at an only point. If a point A is outside a circle, we can draw two lines that pass through this point and are tangent to the circle at two different points each. So this is shown in the third figure. Lines red and blue are tangent to the circle at two different points.
To solve this, we have to find the volume of the cylinder first. The formula to be used is 
Given:V= ?r= 6cmh= 10cm
Solution:
V= (3.14)(6cm)
x 10cmV= (3.14)(
) x 10cmV= (
) x 10cmV= 1130.4cm^3
Finding the volume of the cylinder, we can now solve what the weight of the oil is. Using the formula of density, Density = mass/volume, we can derive a formula to get the weight.
Given:Density = 0.857 gm/cm^3Volume = 1130.4 cm^3
Solution:weight = density x volumew= (0.857 gm/cm^3) (1130.4cm^3)w= 968.7528 gm
The weight of the oil is 968.75 gm.
Given:V= ?r= 6cmh= 10cm
Solution:
V= (3.14)(6cm)
Finding the volume of the cylinder, we can now solve what the weight of the oil is. Using the formula of density, Density = mass/volume, we can derive a formula to get the weight.
Given:Density = 0.857 gm/cm^3Volume = 1130.4 cm^3
Solution:weight = density x volumew= (0.857 gm/cm^3) (1130.4cm^3)w= 968.7528 gm
The weight of the oil is 968.75 gm.