A set of angles has a ratio of 5 to 2. If the larger angle measures 105 degrees then what is the measure of the smaller angle

The diagram is not available, so I included a supplementary figure.

Response:

The segments ST and UT are equal in length.

Detailed explanation:

As seen in the additional figure

∵ ST and UT act as tangents to circle K at points S and U respectively

∵ SK and UK are the radii of circle K

- A tangent is perpendicular to the radius at the point where it touches the circle.

Thus, ST ⊥ KS at point S

Thus, m∠KST = 90°

Thus, UT ⊥ KU at point U

Thus, m∠KUT = 90°

Therefore, m∠KST = m∠KUT

In triangles KST and KUT

∵ KS = KU because they are both radii

∵ m∠KST = m∠KUT confirms the equality

∵ KT is common to both triangles

- Therefore, the triangles are congruent according to the HL criterion

∴ Δ KST ≅ KUT as per the HL criterion

- Thus, from the congruency result

∴ ST = UT

The segments ST and UT are equal in length.

B I hope this was helpful, though I'm not entirely certain.

Answer:

Given that the frog jumps every 10 seconds

   (using digits from a random number table)

• It requires 7 jumps with 2 in the reverse direction (either left or right) for the frog to get off the board in 60 seconds.
• Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
• Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.

Step-by-step explanation:

A frog positioned right at the center of a 5ft long board is 2.5 ft away from either edge.

Every 10 seconds, the frog jumps left or right.

If the frog's jumps are LLRLRL, it will remain on the board at the leftmost square.

If it jumps as LLRLL, it will jump off the board after fifty seconds.

Given that the frog jumps every 10 seconds

   (using digits from a random number table)

• It requires 7 jumps with 2 in reverse direction (either left or right) for the frog to get off the board in 60 seconds.
• Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
• Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.

Out of 100 surveyed individuals, 35 work out in the morning, 45 in the afternoon, and 20 at night.

Jim is accurate.....
The ratio of morning exercisers to the whole group is 35/100.
The ratio of afternoon exercisers is 45/100.
The ratio for night exercisers is 20/100.

The average speed for his entire journey from York to Blackpool is about 61.41 km/h.

Here’s a breakdown of how we arrive at this:

The distance he travelled from York to Leeds is 45 km,

and the speed during that section was 54 km/h.

Therefore, the time taken to travel from York to Leeds is 45/54 hours (since Time = Distance/Speed).

Next, the distance from Leeds to Blackpool is 42 km,

and the time for that leg of the journey is 35 minutes, which is 35/60 hours.

This leads to the total duration for his trip as

hours.

The cumulative distance covered equals 45 + 42 = 87 km.

Thus, his average speed is calculated as: