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Verizon [17]
3 years ago
10

The coordinates of a point on a coordinate grid are (−2, 6). The point is reflected across the x-axis to obtain a new point. The

coordinates of the reflected point are
(2, 6)
(−2, 6)
(−2, −6)
(2, −6)
Mathematics
2 answers:
Colt1911 [192]3 years ago
8 0
(-2,-6) I believe. Sorry if it’s wrong
Juliette [100K]3 years ago
3 0
The reflected coordinate will be (-2,-6)
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A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par
lyudmila [28]

Answer:

a(\frac{1}{5e})=5e

Step-by-step explanation:

we are given equation for position function as

s(t)=tln(5t)

Since, we have to find acceleration

For finding acceleration , we will find second derivative

s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)

=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t

=1\cdot \ln \left(5t\right)+\frac{1}{t}t

s'(t)=\ln \left(5t\right)+1

now, we can find derivative again

s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)

=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)

=\frac{1}{t}+0

a(t)=\frac{1}{t}

Firstly, we will set velocity =0

and then we can solve for t

v(t)=s'(t)=\ln \left(5t\right)+1=0

we get

t=\frac{1}{5e}

now, we can plug that into acceleration

and we get

a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}

a(\frac{1}{5e})=5e


5 0
3 years ago
2 2-1-1-1 whats is the answer
schepotkina [342]
2-1=1-1=0-1=-1 is the first 2 part of the eqation too then it will be 2-2=0-1=-1-1=-2-1=-3
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3 0
3 years ago
True or False?<br> If, |al = |b|, then either a = b or a = - b
never [62]

Answer:

true

Step-by-step explanation:

3 0
3 years ago
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