The function
... f(x) = (x+2)/(x-1) = 1 + 3/(x-1)
is symmetrical about the line y=x, hence is its own inverse.
We can evaluate the desired derivative directly.
... f'(x) -3/(x-1)²
so f'(2) is
... f'(2) = -3/(2-1)²
Which transformations could have taken place? Select two options.
1 answer:
The two options are R₀ 270 , R₀ 90 , Option C and D .
The complete question is attached in the image.
<h3 /><h3>What are the two types of Transformations ?</h3>There are two types of transformations namely , Rigid Transformation and Non- Rigid Transformation
The rigid transformations do not change the shape and size of the preimage .
The non Rigid transformation do not change the shape but change the size of the preimage.
The rule rotating the origin , 270 degree is
(x,y) ----> (y , -x)
as (x,y) is (0,5)
Therefore after 270 degree transformation
(y,-x) is (5 , 0)
So, the clockwise rotation is 90 degree about origin
(x,y) ----> (y , -x)
as (x,y) is (0,5)
Therefore after 90 degree transformation
(y,-x) is (5 , 0)
Therefore the two options are R₀ 270 , R₀ 90 , Option C and D .
To know more about Types of Transformations
#SPJ1
You might be interested in
Answer:
y = 96°
Step-by-step explanation:
The measure of the inscribed angle y is half the measure of its intercepted arc.
The whole circle = 360°
the intercepted arc = 360° - 168° = 192°
Thus
y = × 192° = 96°
Answer: Choice A
The fractions are not equivalent because the product of the numerator of the first fraction and the denominator of the second fraction is <u> 15 </u> , which is not equal to the product of the numerator of the second fraction and the denominator of the first fraction <u> 16 </u>
===================================================
Explanation:
We have the template which cross multiplies to
So if was true, then
must be true as well, and vice versa.
The left side 3*5 turns into 15, while the right hand side 4*4 becomes 16. The 15 and 16 don't match up.
In short, turns into
which is a false statement. So the original claim that
is also false.
Look at the picture.
The area of the hexagon is equal six times the area of the <span>equilateral triangle.
The formula of the </span><span>equilateral triangle with the leg a:
Therefore the formula of the area of the hexagon with a side lenght a is:
We have a = 12cm. Substitute:
</span>
The area of the hexagon is equal six times the area of the <span>equilateral triangle.
The formula of the </span><span>equilateral triangle with the leg a:
Therefore the formula of the area of the hexagon with a side lenght a is:
We have a = 12cm. Substitute:
</span>