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dybincka [34]
3 years ago
14

The point (21,32) is on a line with slope 1.5.

Mathematics
1 answer:
Oxana [17]3 years ago
7 0
We can use the equation: y=mx+b
x represents the x-value, y represents the y-value, m represents slope, and b represent the y-intercept.

We must plug in what we know in order to find the value of b.
32 = 1.5*21 + b
32 = 31.5 + b
So b = 0.5

Now we know our equation is y = 1.5x + 0.5

We can find any other point on the line by plugging in any x-value. Let's try 5.
y = 1.5*5 + 0.5
So y =8 and our point is (5,8)
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3 years ago
Two ships leave a harbor together, traveling on courses that have an angle of 135°40' between them. If they each travel 402 mile
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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'}

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