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Vladimir79 [104]
2 years ago
7

A chess player moves a knight from the location (3,2) to (5,1) on a chessboard. if the bottom-left square is labeled (1,1), the

translation made?. if the player moves the knight from (5,1) to (6,3), the translation made is ?
Mathematics
1 answer:
lora16 [44]2 years ago
3 0

The translation mode are-

  • On a chessboard, moving a knight from (3, 2) to (5, 1) results in a translation of 2 squares up and 1 square left.
  • The translation produced is 2 squares right, 1 square up, if a player advances his knight from (5, 1) to (6, 3).
<h3>What is translation?</h3>

In mathematics, a translation is the up, down, left, or right movement of a shape.

Some key features regarding translation are-

  • Because the translated shapes appear to be identical in size as the original ones, they are consistent with one another.
  • Just one or maybe more directions have been altered.
  • There has been no change in its form because it is simply being moved from one location to another.
  • The object may initially travel left, then turn right, and so on.
  • The direction or the course of this shift in the object's position can also alter.
  • Each point on the shape will move by the same number of units during translation.
  • For instance, all of the points will move two units to that same right if one point moves two units to that same right.

To know more about translation, here

brainly.com/question/1046778

#SPJ4

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Answer:

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Step-by-step explanation:

We have:

\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]

We can move the constant outside:

\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]

Now, we will utilize the product rule. The product rule is:

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We will let:

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Then:

\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1

(The derivative of u was determined using the chain rule.)

Then it follows that:

\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}

Therefore:

\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]

By simplification:

\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

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