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3 years ago

7

the rocks near the shore between two lighthouses at point a and b make the waters unsafe. The measure of AXB is 300. Waters insi

de this arc are unsafe. Suppose you are a navigator on a ship at sea. How can you use the lighthouses to keep the ship in safe waters?

1 answer:

exis [7]3 years ago

6 0

You can, use the lighthouses to know where you are going in the dark night.

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Change 192 liters to kiloliters ______kl

Nikitich [7]

<span>192L= <span>0.1920000kl
</span>You just multiply the liters by </span><span>0.001 kiloliters, and you get the correct answer in kiloliters. </span><span>

</span>

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3 years ago

How to find the perimeter of the triangle

Sliva [168]

Answer:

A. 26.2

Step-by-step explanation:

To find the perimeter of the triangle, you have to find the distances of all three lines and add them up.

<u>Line AB</u>

Let's start off by finding the distance of line AB.

We will use the formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point A is (-2,5) and Point B is (4,-3).

To substitute the values, it will get to $d = \sqrt{(4--2)^{2}+(-3-5)^{2}}$ which in other words is $d = \sqrt{(4+2)^{2}+(-3-5)^{2}}$ .

Now we have to solve the parenthesis to get $d = \sqrt{(6)^{2}+(-8)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{36+64}$ .

Now we have to simplify the square root to $d = \sqrt{100}$ . In other words, that is $d = 10$ .

Line AB = 10

<u>Line BC</u>

Now let's find the distance of line BC.

We will use the same formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point B is (4,-3) and Point C is (0,-6).

To substitute the values, it will get to $d = \sqrt{(0-4)^{2}+(-6--3)^{2}}$ which in other words is $d = \sqrt{(0-4)^{2}+(-6+3)^{2}}$ .

Now we have to solve the parentheses to get $d = \sqrt{(-4)^{2}+(-3)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{16+9}$ .

Now we have to simplify the square root to $d = \sqrt{25}$ . In other words, that is $d = 5$ .

Line BC = 5.

<u>Line AC</u>

Now let's fine the distance of line AC.

We will use the same formula, $d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ .

Point A is (-2,5) and Point C is (0,-6).

To substitute the values, it will get to $d = \sqrt{(0--2)^{2}+(-6-5)^{2}}$ which in other words is $d = \sqrt{(0+2)^{2}+(-6-5)^{2}}$ .

Now we have to solve the parentheses to get $d = \sqrt{(2)^{2}+(-11)^{2}}$ .

Now we have to solve the exponents which would get to $d = \sqrt{4+121}$ .

Now we have to simplify the square root to $d = \sqrt{125}$ . In other words, that is $d = 5\sqrt{5}$ . To round that, $d = 11.2$ .

Line AC = 11.2.

<u>Perimeter of Triangle ABC</u>

Perimeter of Triangle ABC = Line AB + Line BC + Line AC.

Perimeter of Triangle ABC = 10 + 5 + 11.2

Perimeter of Triangle ABC = 26.2

Hope this helped! If not, please let me know <3

3 0

3 years ago

You have 24 homework problems for math class.you finish 1/2 of them during some extra time in class. If you complete 4 more prob

vodomira [7]

Answer:

who gets 24 homework problems...8/24 if you simplify by dividing 8 on the num. and den. then it would be 1/3

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4 years ago

What is the area of a circle with 11m and 9m?

Naya [18.7K]

Answer:

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Step-by-step explanation:

multiply the two numbers

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3 years ago

How to do two step equations

Leokris [45]

Simplify using the inverse of addition or subtraction.

Simplify further by using the inverse of multiplication or division.

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3 years ago