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aniked [119]
3 years ago
9

1.) Find the distance between the points at (2, 3) and (-4,6)

Mathematics
1 answer:
Elza [17]3 years ago
7 0

Answer:

6.7082

Step-by-step explanation:

Distance Formula=√(x2−x1)^2+(y2−y1)^2

                                √(-4-2)^2+(6-3)^2

                                 -4 - 2= -6 squared is -36

                                 6-3= 3 squared is 9

                                 add 36 and 9 which = 45

                                 √45 = 6.7082

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Can someone help me with this please??
Evgen [1.6K]

Answer:

z=20.8

x=16.97

y=12

Step-by-step explanation:

The triangle on the right is a 30,60,90 triangle, a special right triangle with simple rules to find the side lengths. sides A,B, and C where C is the hypotenuse (24 in your diagram). The short side A is= 1/2 of side C (12 in your diagram). The long side B is= side A*√3 (20.8 in your diagram).

The triangle on the left is another special right triangle called a 45,45,90. The rules for this triangle are C= A*√2, and B=A. Because of the rules in the 30,60,90 triangle, we know that side A is 12. This means that side B (y)=12. Side C (x)= 16.97.

Hope this helped! Sorry if my explanation is confusing, but the rules for special right triangles can be easily found online and then calculated :)

4 0
4 years ago
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vovikov84 [41]
I cannot do that unfortunately
3 0
3 years ago
The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?
Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

A(-6, 2)) = (x_1, y_1)

B(2, 6) = (x_2, y_2)

AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}

AB = \sqrt{(8)^2 + (4)^2}

AB = \sqrt{64 + 16} = \sqrt{80}

AB = 8.9 (nearest tenth)

Distance between B(2, 6) and C(7, 1):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

B(2, 6) = (x_1, y_1)

C(7, 1) = (x_2, y_2)

BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}

BC = \sqrt{(5)^2 + (-5)^2}

BC = \sqrt{25 + 25} = \sqrt{50}

BC = 7.1 (nearest tenth)

Distance between C(7, 1) and D(3, -5):

CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

C(7, 1) = (x_1, y_1)

D(3, -5) = (x_2, y_2)

CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}

CD = \sqrt{(-4)^2 + (-6)^2}

CD = \sqrt{16 + 36} = \sqrt{52}

CD = 7.2 (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Let,

D(3, -5) = (x_1, y_1)

A(-6, 2) = (x_2, y_2)

DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}

DA = \sqrt{(-9)^2 + (7)^2}

DA = \sqrt{81 + 49} = \sqrt{130}

DA = 11.4 (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0
3 years ago
Emergency helpn please i dk wut they mean
galben [10]

Answer:

the first one is correct

Step-by-step explanation:

4 0
3 years ago
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Tasya [4]
Hey You!

250 + 350 = 600

2,000 - 600 = 1,400

1,400 / 17.95 = 77.9944289694

The greatest amount of students the school can take is 77.
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