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miv72 [106K]
3 years ago
9

PLEASE HELP!! I WILL AWARD BRAINLIEST AND 50 POINTS!!!!!!!!!!!!!!!!!!!!!! For the given picture: a) List all of the triangles th

at are already drawn in the picture without changing it. b) For each triangle, list the exterior angle(s) that are already drawn in the picture without changing it. c) For each of the exterior angles in part (b), identify its remote interior angles in the corresponding triangle(s).

Mathematics
1 answer:
PIT_PIT [208]3 years ago
6 0

Answer:

There are 4 triangles. For the exterior and remote interior, I’m naming the angles with three letters.  If your teacher just wants one, it would be the middle one.  (I assume they’ll want all three, since the letters just name points, not angles.)

1) A) Name of triangle: ABO

B) Exterior AngleS: EAO, BOC and AOD

C) Remote Interior Angles: For EAO, it’s ABO and AOB.  For BOC, it’s BAO and ABO. For AOD, it’s BAO and ABO

2) A) Name of triangle: ABC

B) Exterior Angles:EAC

C) Remote Interior Angles: ABO and BCO

3) A) Name of triangle: DOC

B) Exterior Angles: ODH, DOA, COB, and OCG

C) Remote Interior Angles: For ODH, it’s DOC and COD.  For DOA, it’s ODC and DCO. For COB, it’s ODC and DCO.  For OCG, it has CDO and DOC

4) A) Name Of triangle: COB

B) Exterior Angles: AOB, DOC, and OBF

C) Remote Interior Angles: For AOB, it’s OBC and OCB. For DOC, it’s OBC and OCB.  For OBF, it’s BOC and OCB

Step-by-step explanation:

There are 4 triangles. For the exterior and remote interior, I’m naming the angles with three letters.  If your teacher just wants one, it would be the middle one.  (I assume they’ll want all three, since the letters just name points, not angles.)

1) A) Name of triangle: ABO

B) Exterior AngleS: EAO, BOC and AOD

C) Remote Interior Angles: For EAO, it’s ABO and AOB.  For BOC, it’s BAO and ABO. For AOD, it’s BAO and ABO

2) A) Name of triangle: ABC

B) Exterior Angles:EAC

C) Remote Interior Angles: ABO and BCO

3) A) Name of triangle: DOC

B) Exterior Angles: ODH, DOA, COB, and OCG

C) Remote Interior Angles: For ODH, it’s DOC and COD.  For DOA, it’s ODC and DCO. For COB, it’s ODC and DCO.  For OCG, it has CDO and DOC

4) A) Name Of triangle: COB

B) Exterior Angles: AOB, DOC, and OBF

C) Remote Interior Angles: For AOB, it’s OBC and OCB. For DOC, it’s OBC and OCB.  For OBF, it’s BOC and OCB

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The question is incomplete. The complete question is :

The volume of a right circular cone with radius r and height h is V = pir^2h/3. a. Approximate the change in the volume of the cone when the radius changes from r = 5.9 to r = 6.8 and the height changes from h = 4.00 to h = 3.96.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.47 to r = 6.45 and the height changes from h = 10.0 to h = 9.92.

a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)

b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

Solution :

Given :

The volume of the right circular cone with a radius r and height h is

$V=\frac{1}{3} \pi r^2 h$

$dV = d\left(\frac{1}{3} \pi r^2 h\right)$

$dV = \frac{1}{3} \pi h \times d(r^2)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

a). The radius is changed from r = 5.9 to r = 6.8 and the height is changed from h = 4 to h = 3.96

So, r = 5.9  and dr = 6.8 - 5.9 = 0.9

     h = 4  and dh = 3.96 - 4 = -0.04

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (5.9)(4)(0.9)+\frac{1}{3} \pi (5.9)^2 (-0.04)$

$dV=44.484951 - 1.458117$

$dV=43.03$

Therefore, the approximate change in volume is dV = 43.03 cubic units.

b).  The radius is changed from r = 6.47 to r = 6.45 and the height is changed from h = 10 to h = 9.92

So, r = 6.47  and dr = 6.45 - 6.47 = -0.02

     h = 10  and dh = 9.92 - 10 = -0.08

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (6.47)(10)(-0.02)+\frac{1}{3} \pi (6.47)^2 (-0.08)$

$dV=-2.710147-3.506930$

$dV= -6.22$

Hence, the approximate change in volume is dV = -6.22 cubic units

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