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Sergio [31]
3 years ago
8

Someone help me with this please?

Mathematics
2 answers:
MariettaO [177]3 years ago
7 0
The question's cut off. repost & ill see if i can help (:
WINSTONCH [101]3 years ago
6 0
I am sorry, but I think that the question is cut off.  Perhaps you can re-post it for us to see it completely?  I would be happy to help!  :)
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Step-by-step explanation:

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1000*0+321*15678*9(-4563)
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Step-by-step explanation:

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Two ships leave a harbor together, traveling on courses that have an angle of 135°40' between them. If they each travel 402 mile
mario62 [17]

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}

Taking last two ratio,

\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}

Putting the value of BC , AC ,∠A,∠B

\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}

\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}

         ≈734.106 miles

Therefore they are 734.106 miles apart.

3 0
3 years ago
A. 4+2(5)+(−3)<br> b. 4+2[5+(−3)]<br> PLEASE- I NEED HELP!!!
Ugo [173]
A = 11 and b = 8 I’m pretty sure brosky
3 0
3 years ago
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