Enter the mixed number as an improper fraction. 2 1/2 plz

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

6 0

3 years ago

Two passenger train, A and B, 450 km apart, star to move toward each other at the same time and meet after 2hours.

likoan [24]

Let $v$ be the speed of train A, and let's set the origin in the initial position of train A. The equations of motion are

$\begin{cases}s_A(t) = vt\\s_B(t) = -\dfrac{8}{7}vt+450\end{cases}$

where $s_A,\ s_B$ are the positions of trains A and B respectively, and t is the time in hours.

The two trains meet if and only if $s_A=s_B$ , and we know that this happens after two hours, i.e. at $t=2$

$\begin{cases}s_A(2) = 2v\\s_B(2) = -\dfrac{16}{7}v+450\end{cases}\implies 2v = -\dfrac{16}{7}v+450$

Solving this equation for v we have

$2v = -\dfrac{16}{7}v+450 \iff \dfrac{30}{7}v=450 \iff v=\dfrac{450\cdot 7}{30} = 105$

So, train A is travelling at 105 km/h. This implies that train B travels at

$105\cdot \dfrac{8}{7} = 15\cdot 8=120 \text{ km/h}$

5 0

3 years ago

A formula for the perimeter of a rectangle is P=2(l+w), where P is the perimeter, l is the length, and w is the width.

NNADVOKAT [17]

1408 because you need to multiply it to get the perimeter

7 0

3 years ago