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LenKa [72]
4 years ago
6

Find the domain and range of the INVERSE of the given function. Square root x-3

Mathematics
1 answer:
Ulleksa [173]4 years ago
4 0
f(x)=\sqrt{x-3}\\
y=\sqrt{x-3}\\
y^2=x-3\\
x=y^2+3\\
f^{-1}(x)=x^2+3\\
D:x\in\matxbb{R}\\
R:y\in\langle3,\infty)
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3 years ago
PLEASE HELP!!
Aloiza [94]

Answer:

From the given graph:

the coordinates of triangle RST are;

R= (2, 1),

S= (2,-2),

T= (-1,-2)

Given: Scale factor = \frac{5}{3}  and center of dilation at (2,2)

The mapping rule for the  dilation applied to the triangle as shown below:    

(x,y) \rightarrow (\frac{5}{3}(x-2)+2 , \frac{5}{3}(y-2)+2 ); where k represents the scale factor i.e, k=\frac{5}{3} or we can write it as ;  

For R=(2, 1)      

The image R' = (\frac{5}{3}(2-2)+2 , \frac{5}{3}(1-2)+2 )

⇒ R'= (2, \frac{1}{3})  

Similarly for S= (2, -2) and T= (-1,-2)

therefore, the image of S'= (\frac{5}{3}(2-2)+2 , \frac{5}{3}(-2-2)+2 )

⇒ S'= (2, \frac{-14}{3})

The image of T' =(\frac{5}{3}(-1-2)+2 , \frac{5}{3}(-2-2)+2 )

⇒T' = (-3, \frac{-14}{3})

Now, labelling the image of triangle R'S'T' as shown in the figure given below

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Answer:

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Step-by-step explanation:

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vova2212 [387]

Answer:

Angle 1: 127

Angle 2: 153

Step-by-step explanation:

First, let's find angle 1 by finding the angle that's supplementary to it.

To solve for it, we can set up an equation where the unknown angle and the other angles in the triangle it's in add up to 180:

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Since angle 1 is supplementary to 53, that means that angle 1 is equal to 180-53 = 127.

Then, to find angle 2, we can find the angle that's supplementary to it.

To solve for that, we can set up an equation where that unknown angle and the other angles in the triangle it's in add up to 180:

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Since angle 2 is supplementary to 27, that means that angle 2 is equal to 180-27 = 153.

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