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MrRa [10]
3 years ago
9

In the coordinate plane, the point A, 4− 2 is translated to the point A′, 1− 5. Under the same translation, the points B, 7− 4 a

nd C, 2− 5 are translated to B′ and C′, respectively. What are the coordinates of B′ and C′?
Mathematics
1 answer:
Inga [223]3 years ago
5 0

Answer:

The coordinates of B' and C' are B'(x,y) = (4, -7) and C'(x,y) = (-1, -8), respectively.

Step-by-step explanation:

From the Linear Algebra, we define the translation of a given point as:

O'(x,y) = O(x,y) + T(x,y) (1)

Where:

O(x,y) - Original point, dimensionless.

T(x,y) - Translation vector, dimensionless.

O'(x,y) - Translated point, dimensionless.

If we know that A'(x,y) = (1, -5) and A(x,y) = (4,-2), then the translation vector is:

T(x,y) = A'(x,y)-A(x,y) (2)

T(x,y) = (1,-5)-(4,-2)

T(x,y) = (-3,-3)

If we know that B(x,y) = (7,-4), C(x,y) = (2,-5) and T(x,y) = (-3,-3), then the translated points are, respectively:

B'(x,y) = B(x,y)+T(x,y) (3)

B'(x,y) = (7,-4) +(-3,-3)

B'(x,y) = (4, -7)

C'(x,y) = C(x,y) +T(x,y)

C'(x,y) = (2,-5) + (-3,-3)

C'(x,y) = (-1, -8)

The coordinates of B' and C' are B'(x,y) = (4, -7) and C'(x,y) = (-1, -8), respectively.

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Anna71 [15]

Answer:

The answer is

<h2>9.4 units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

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

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-6, -5) and (2, 0)

The distance between them is

d =  \sqrt{ ({ - 6 - 2})^{2} +  ({ - 5 - 0})^{2}  }  \\  =  \sqrt{ ({ - 8})^{2}  + ( { - 5})^{2} }  \\  =  \sqrt{64 + 25}  \\  =  \sqrt{89}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  = 9.4339811

We have the final answer as

<h3>9.4 units to the nearest tenth</h3>

Hope this helps you

7 0
3 years ago
If I fall for 18 s what would my final velocity be, if I started from rest.
ozzi

Answer:

176.58 m/s

Step-by-step explanation:

assume that the acceleration due to gravity, g = 9.81 m/s²

I also assume we neglect air resistance (hence we assume no terminal velocity)

recall that one of the equations of motions can be expressed as

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8 0
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Answer:

\frac{10\sqrt{6} }{3} and \frac{11\sqrt{3} }{3}

Step-by-step explanation:

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A = \frac{1}{2} × 4\sqrt{2} × \frac{5}{\sqrt{3} }

   = 2\sqrt{2} × \frac{5}{\sqrt{3} }

   = \frac{10\sqrt{2} }{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } ← rationalise the denominator

= \frac{10\sqrt{6} }{3} units²

(b)

Using Pythagoras' identity in the right triangle

let hypotenuse be h , then

h² = (4\sqrt{2} )² + (\frac{5}{\sqrt{3} } )²

    = 32 + \frac{25}{3}

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h = \sqrt{\frac{121}{3} } = \frac{11}{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } ← rationalise the denominator

h = \frac{11\sqrt{3} }{3}

 

7 0
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Answer:

0

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Step-by-step explanation:

mark brainliest

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