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Butoxors [25]
3 years ago
9

Help i dont know what to do

Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
3 0

Answer:

Pretend the y-axis is a mirror

If the original shape is

P=4,-2

Z=5,-4

M=4,-4

The new image made by mirroring it will be

P’= -4,-2

Z’= -5,-4

M’= -4,-4

Step-by-step explanation:

Sorry if this doesn’t make sense, just graph P’, Z’, and M’ for a mirrored image

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Evaluate the following expression. 6^-3
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Step-by-step explanation:

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3 years ago
Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

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                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




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3 years ago
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Step-by-step explanation:

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3 years ago
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Answer:

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7 0
4 years ago
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