Answer:
Therefore they are 734.106 miles apart.
Step-by-step explanation:
Given that ,
Two ships have a harbor together. The angle between two ships is 135°40'. Each of two ships travel 402 miles.
It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.
Let ∠B= 135°40', and AB = 402 miles , BC = 402 miles
Then the distance between the ships = AC
We know
The sum of all angles = 180°
⇒∠A+∠B+∠C=180°
⇒∠A+135°40'+∠C=180°
⇒2∠A= 180°- 135°40' [ since ∠A=∠C]
⇒2∠A=44°60'
⇒∠A= 22°30'
Again we know that,

Taking last two ratio,

Putting the value of BC , AC ,∠A,∠B


≈734.106 miles
Therefore they are 734.106 miles apart.
Answer:
3.75
Step-by-step explanation:
5 1/4÷1/3
so,
= 5 1/4×3/1
= 5 × 0.75
= 3.75
9514 1404 393
Answer:
5
Step-by-step explanation:
To solve the inequality, we can do the following.
4.55g + 8.50 ≤ 35 . . . given
4.55g ≤ 26.50 . . . . . . subtract 8.50 from both sides
g ≤ 5.82... . . . . . . . . . . divide both sides by 4.55
The number of games must be an integer value, so the maximum number of games they can bowl is 5.
Answer: 32/35
Step-by-step explanation:
got it correct