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sweet-ann [11.9K]
4 years ago
14

A transformation named t maps triangle xyz to triangle x'y'z' the image of y is ?

Mathematics
2 answers:
Svetllana [295]4 years ago
6 0
Y'

The way the transformation is written tells which point goes where; the first point x is mapped to point x', y goes to y', z goes to z'.

11111nata11111 [884]4 years ago
4 0

Answer:

A transformation named t maps triangle xyz to triangle x'y'z' the image of y is:

y'

Step-by-step explanation:

We are asked to find the image of y given that:

a transformation named t maps triangle xyz to triangle x'y'z'.

Hence, the image of y is y'.

Since we know that if any triangle xyz is transformed to x'y'z' then the vertices of the triangle x is transformed to x' , y is transformed to y' and z is transformed to z'.

i.e. x → x'

y → y'

and z → z'

Hence, A transformation named t maps triangle xyz to triangle x'y'z' the image of y is:

y'

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10
Anestetic [448]

Answer:

Acceleration when time is 2 \ sec = -\frac{5}{4} \ m/s^2

Distance s=9-\frac{9}{t}

Distance travelled between t=1 and t=3 is 6\ m

Step-by-step explanation:

Velocity V=1+\frac{9}{t^2} \ \ \ when \ \ 1\leq t\leq 3

<u><em>Acceleration(a):</em></u><u><em> Rate of change of velocity.</em></u>

a=\frac{dV}{dt} =1-\frac{18}{t^3}

at t=2

a=1-\frac{18}{8} \\=1-\frac{9}{4} \\=-\frac{5}{4}

a=-\frac{5}{4} \ m/s^2

<u><em>Distance(s):</em></u>

V=\frac{ds}{dt}\\\\\frac{ds}{dt}=1+\frac{9}{t^2}\\\\Integrate\ both\ sides\\\\s=1-\frac{9}{t}+k\\\\ k\ is\ a\ constant\\\\Given\ s=6\ when\ t=3\\\\6=1-\frac{9}{3}+k\\\\ k=8\\\\Hence\ s=1-\frac{9}{t}+8\\\\ s=9-\frac{9}{t}

<u><em>Distance between t=1 \ and \ t=3</em></u>

=s(3)-s(1)\\\\=(9-\frac{9}{3})-(9-9) \\\\=6 \ m

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Find the percent as a fraction. Simplify if possible.<br><br>33 1/3%
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Answer:

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Step-by-step explanation:

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Read 2 more answers
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aliya0001 [1]
X=17

Distribute 2(x-5)
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1. 14=7b<br> 2. -11=k/4<br> 3.6z=-6
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Answer:

the anwer is 14-2294-dhi2

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