First way is create an equation like this:
and calculating it.
Second way is to find 37% of each number and then add it
![0.37*27 \frac{3}{5}](https://tex.z-dn.net/?f=0.37%2A27%20%5Cfrac%7B3%7D%7B5%7D%20)
+
![0.37* \frac{15}{9}](https://tex.z-dn.net/?f=0.37%2A%20%5Cfrac%7B15%7D%7B9%7D%20)
In both cases the result is 10.829
Answer:
i think its b
Step-by-step explanation:
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,
![\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}](https://tex.z-dn.net/?f=%5Cfrac%7BAB%7D%7Bsin%5Cangle%20C%7D%3D%5Cfrac%7BBC%7D%7Bsin%20%5Cangle%20A%7D%3D%5Cfrac%7BAC%7D%7Bsin%20%5Cangle%20B%7D)
Taking last two ratio,
![\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}](https://tex.z-dn.net/?f=%5Cfrac%7BBC%7D%7Bsin%20%5Cangle%20A%7D%3D%5Cfrac%7BAC%7D%7Bsin%20%5Cangle%20B%7D)
Putting the value of BC , AC ,∠A,∠B
![\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}](https://tex.z-dn.net/?f=%5Cfrac%7B402%7D%7Bsin%2022%5E%5Ccirc30%27%7D%3D%5Cfrac%7BAC%7D%7Bsin%20135%5E%5Ccirc40%27%7D)
![\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}](https://tex.z-dn.net/?f=%5CRightarrow%20AC%3D%5Cfrac%7B402%20%5Ctimes%20sin135%5E%5Ccirc40%27%7D%7Bsin%2022%5E%5Ccirc30%27%7D)
≈734.106 miles
Therefore they are 734.106 miles apart.
0.99999696 is your answer I am pretty sure.
Sorry if my handwriting sloppy.
I showed all of it step by step.