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3 years ago

5

3. Which rule describes a transformation across the y-axis?

2 answers:

stiv31 [10]3 years ago

6 0

Answer:

Reflection

Step-by-step explanation:

joja [24]3 years ago

3 0

Answer:

(x, y) -----> (-x, y).

(Please mark as brainliest)

Step-by-step explanation:

We are given to write a rule that describes a transformation across the y-axis. We know that if a point is reflected across the y-axis, then the sign before the x co-ordinate changes. That is, if the point (x, y) is reflected (transformed) across the y-axis, then its new co-ordinates will be (-x, y).

Thus, the required transformation rule is given by (x, y) -----> (-x, y).

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stiv31 [10]

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Complete the solution of the equation. Find the value of y when x equals 4. -7x + y = -22

xz_007 [3.2K]

-7x + y = -22

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3 years ago

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Solnce55 [7]

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Step-by-step explanation:

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3 years ago

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

b) $a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

$a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}$

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago