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lisov135 [29]
3 years ago
14

Anyone know wether gcse are techer assessed or wht in england this year

Mathematics
1 answer:
miskamm [114]3 years ago
6 0

Answer:

wait what is ur question

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Complete the given diagram by dragging expressions to each leg of the triangle. Then, correctly complete the equation to derive
Nimfa-mama [501]

The equation to derive the distance d is \sqrt{(x2-x1)^2+(y2-y1)^2}. The lengths of the other legs of the given triangle are (y2 - y1) and (x2 - x1).

<h3>What is the formula for calculating the distance between two points?</h3>

Consider the two points (x1, y1) and (x2, y2)

The formula used for calculating the distance between the two points is

distance = \sqrt{(x2-x1)^2+(y2-y1)^2}

<h3>Calculation:</h3>

Given that,

The triangle in the graph has vertices (x1, y1), (x2, y2), and (x2, y1)

Since this triangle makes 90°, it is a right-angled triangle.

Hypotenuse = (x1, y1) to (x2, y2), Adjacent = (x1, y1) to (x2,y1), and Opposite = (x2, y1) to (x2, y2).

Consider the length of the hypotenuse = d

So, using the distance formula, the length of the hypotenuse(d) is,

d = \sqrt{(x2-x1)^2+(y2-y1)^2}

And the lengths of the other two legs of the given triangle are,

Length of the adjacent side: (x1, y1) to (x2,y1)

= \sqrt{(x2-x1)^2+(y1-y1)^2}

= \sqrt{(x2-x1)^2+0}

= (x2-x1)

Length of the opposite side: (x2, y1) to (x2, y2)

= \sqrt{(x2-x2)^2+(y2-y1)^2}

= \sqrt{0+(y2-y1)^2}

= (y2-y1)

Therefore, the derived distances for the given triangle are:

d=\sqrt{(x2-x1)^2+(y2-y1)^2}, (x2 - x1), and (y2 - y1).

Learn more about the distance between two points here:

brainly.com/question/661229

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4 0
2 years ago
What is the slope intercept form of y + 1 = -2 (x-1)?
Mars2501 [29]
Y+1 = -2x+2 (distribute)
Y=-2x + 1 (isolate y)
4 0
3 years ago
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Kierra recorded her daily exercise activity for a month in this frequency table. Which problem shows how to solve for the follow
sineoko [7]

Answer:

i accidentaly put awnser and i dont know how to get out of here

Step-by-step explanation:

6 0
3 years ago
Find the distance and the midpoint of the line segment with endpoints (-1,4) and (-5,3)
Marina CMI [18]

Answer:

see explanation

Step-by-step explanation:

Calculate the distance (d) using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 1,4) and (x₂, y₂ ) = (- 5, 3)

d = \sqrt{(-5+1)^2+(3-4)^2}

  = \sqrt{(-4)^2+(-1)^2}

  = \sqrt{16+1} = \sqrt{17} ≈4.12 ( to 2 dec. places )

To find the midpoint use the midpoint formula

[0.5(x₁ + x₂ ), 0.5(y₁ + y₂ ) ]

Using the same points as above then

midpoint = [0.5(- 1- 5), 0.5(4 + 3 ) ] = [0.5(- 6), 0.5(7) ] = (- 3, 3.5 )

4 0
3 years ago
PLEASE HELP !! ILL GIVE BRAINLIEST *EXTRA POINTS*.. <br> IM GIVING 50 POINTS !! DONT SKIP :((.
8090 [49]

Answer:

poroportional because he will earn 3.50 the same number for every shoot and positve

Step-by-step explanation:

3 0
3 years ago
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