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3 years ago

Suppose you know that a line passes through the point (0,1).it's slope is 1. Find another point on the line.

Mathematics

1 answer:

tamaranim1 [39]3 years ago

4 0

I think it will probably be (0,2)

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A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now

algol [13]

The answers to the bearing problems are listed below:

How far North is the ship now ___________ 38.82 km
How far East is the ship now ____________ 144.89 km
How far is the ship from its starting point_____241.48 km
On what bearing is it now from its original position____035°

<h3>Meaning of bearing.</h3>

Bearing can defined as branch of mathematics that describes the accurate location of an object at any point in time.

<h3>Analysis</h3>

The answers to the problem from 1-3 can be gotten by using sin and cos to find the missing sides.

The final value can be gotten using the cosine rule

4). bearing from original position = $cos^{-1}$ ( $a^{2}$ + $b^{2}$ ﹣ $c^{2}$ ) / $2ab$ = 35°

In conclusion, the answers for each is given in the list above.

Learn more about Bearing: brainly.com/question/24142612

#SPJ1

8 0

1 year ago

In Exercises 26 and 27 write an equation of the line with the given slope and y-intercept.

Neko [114]

26. y= -7/3+ 0 or just y= -7/3

27. y= -10

7 0

2 years ago

What is happening to the graph when the x-values are between 4 and 8

VashaNatasha [74]

The numbers are going down

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2 years ago

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can you please explain to me how 570ml = oz. I'm having a hard time figuring it out. I'm having a problem with converting.

Vladimir79 [104]

19.2740122 fluid ounces = 570 milliliters. I'm positive this really is the answer. :)

8 0

2 years ago

7+3√5/3+√5 - 7-3√5/3-√5 = a + b√5

daser333 [38]

<h3>Given:-</h3>

$\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\$

The value of a and b

<h3>Solution:-</h3>

$\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\$

$\\ \sf \implies\frac{( \: 7 + 3 \sqrt{5} \: \: ( 3 - \sqrt{5}) \: \: - 7 - 3 \sqrt{5} \: \: ( 3 + \sqrt{5}) \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{( \: 21 - 7 \sqrt{5} \: + 9 \sqrt{5} - 15) \: \: - ( \: 21 + 7 \sqrt{5} \: - 9 \sqrt{5} + 15)\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{( \: 6 + 2 \sqrt{5} ) \: \: - ( \: 6 - 2 \sqrt{5} )\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 6 + 2 \sqrt{5} \: \: - \: \: 6 - 2 \sqrt{5} \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{3 {}^{2} - {\sqrt{5} }^{2} } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 9 - 5 \: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 4 \: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: \cancel{4 } \sqrt{5} \: \: }{ \: \: \: \: \cancel{4 }\: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies \: \sqrt{5} = a + \sqrt{5} \: b\\$

we can also write it as ;

$\\ \sf \implies \: 0 + \sqrt{5} = a + \sqrt{5} \: b\\$

★<u> </u><u>Henceforth, the value of a and b are</u> :

→ a = 0

→ b = 1

6 0

2 years ago

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