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Helga [31]
3 years ago
12

Please help me with this geometry question !!!

Mathematics
1 answer:
Inessa05 [86]3 years ago
8 0

Answer:

see explanation

Step-by-step explanation:

Under a counterclockwise rotation about the origin of 90°

a point (x, y ) → (- y, x ), thus

P(1, - 1 ) → P'(1, 1 )

Q(3, - 2 ) → Q'(2, 3 )

R(3, - 4 ) → R'(4, 3 )

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fredd [130]

Answer:

f(x)=\frac9{x+1}\\ f'(x)=-\frac9{(x+1)^2}\\ f'(x)=-1\ \iff\ -\frac9{(x+1)^2}=-1\ \to \ \frac9{(x+1)^2}=1\ \to \ (x+1)^2=9\\ |x+1|=3\ \to \ x+1=3\ \vee\ x+1=-3\\ x_1=2\ \vee\ x_2=-4\\ f(x_1)=f(2)=\frac9{2+1}=3\\ f(x_2)=f(-4)=\frac9{-4+1}=-3

First tangent line:

y=f'(x_1)\cdot (x-x_1)+f(x_1)\ \to \ y=-1(x-2)+3\ \to \ y=-x+5

Second tangent line:

y=f'(x_2)\cdot (x-x_2)+f(x_2)\ \to \ y=-1(x+4)-3\ \to \ y=-x-7


Notice: slope of -1 means that both f'(x_1), \ f'(x_2) are equal to -1, so f'(x_1)=-1 \ and \ f'(x_2)=-1


6 0
2 years ago
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- 2(x+5) = -2<br> Directions say solve for the equation
erik [133]

Answer:

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Step-by-step explanation:

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the answer is 7.5 :)))

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