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sesenic [268]
3 years ago
10

Find the vector that translate A (-2, 7) to A' (6,4) *

Mathematics
2 answers:
GrogVix [38]3 years ago
7 0

Answer:

Step-by-step explanation:

Nataly_w [17]3 years ago
6 0

A vector translation translates the figure on the coordinate from its current location to another

The vector that translates A(-2, 7) to A'(6, 4) is \dbinom{8}{-3}

The given preimage of the vector = A(-2, 7)

The given image of the vector = A'(6, 4)

Required: To find the vector that translates A(-2, 7) to A'(6, 4)

Solution:

The vector are given in component form, therefore, the vector, ΔA, that translates (moves from) vector A(-2, 7) to A'(6, 4) is the difference between the two vectors which is given as follows;

ΔA = A' - A = ((x' - x), (y' - y)

∴ ΔA = ((6 - (-2)), (4 - 7))

ΔA = (8, -3)

In vector form, vector that translates A to A' is \dbinom{8}{-3}

Learn more about vector translation here:

brainly.com/question/20840412

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Is 3x+12-2x equivalent to x +12? Use two properties of operations to justify your answer.
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Yes, x + 12

Step-by-step explanation:

Simplify the following:

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Find the value of the expression 9.16 + k for k = 9.
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Step-by-step explanation:

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3 years ago
Write a formula that will yield the nth term of the sequence: 3,-2,-7,-12
RUDIKE [14]

Answer:

-5n + 8

Step-by-step explanation:

<u>1. What is the difference?</u>

  The sequence goes down by 5. This means that the formula will have -5n in it.

<u>2. Work out the term before the first term.</u>

    The first term is 3, and we know that the sequence goes up by 5. So, to get the term before 3 we would add 5.

3 + 5 = 8   Remember, this is positive 8. (+8)

<u>3. Put that term at the end of the equation. </u>

   We have -5n already and we just worked out the term before the first one which is positive 8 - so put that at the end of the equation.

-5n + 8  This is our answer!

Just to prove it works:

<em>Substitute:   n = term</em>

Lets see if we can get the 3rd term which is -7. (n = 3)

-5(3) + 8

-15 + 8 = -7

See, it works!

8 0
4 years ago
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