12.56 is less than 125.6 which could also be written as 12.56<125.6. 12.56 comes farther left on the number line and is of lesser value when compared to the number 125.6.
G2=g1*2
g3=g1*2²
196=x*4
/4 /4
49
Answer:
-21 r^5
Step-by-step explanation:
(7r) (-3r^4)
Multiply the coefficients
7*(-3)
-21
Multiply the variables
r * r^4 = r^(4+1) = r^5
-21 r^5
Test of a horizontal line.Suppose f is a function.F does not have an inverse if any horizontal line crosses the graph of f more than once.F does have an inverse if no horizontal line crosses the graph of f more than once
How do you calculate f's inverse?
- Finding a Function's Inverse
- Replace f(x) with y first, then.
- Every x must be replaced with a y, and vice versa.
- Solve the y-part of the equation from step two.
- In place of y, write f1(x) f 1 (x).
- Check that (ff1)(x)=x (f f f 1) (x) = x and (f1f)(x)=x (f f 1 f) (x) = x are both true to validate your work
f= {(1,2),(2,3),(3,5),(4,7)}
Dom(gof)=dom(f)={1,3,4}.
(gof)(1)=g{f(1)}=g(2)=3,
(gof)(3)=g{f(3)}=g(5)=1
(gof)(4)=g{f(4)}=g(1)
∴gof={(1,3),(3,1),(4,3)}.
To learn more about inverse functions refer
brainly.com/question/8665364
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Answer:
A = picture attached
B = 51 in²
Step-by-step explanation:
The picture attached is the answer for part A. You use the dimensions you already know from the first picture to fill in the net of the slice of pie.
Area of the triangle = 1/2 (base × height)
A = 1/2 (4 × 6) = 1/2 (24) = 12
The area of the triangle is 12. The top is a triangle, but the bottom is also a triangle with the same area. Therefore, the area of the two triangles combined is 24 in². Next, we have to find the area of the rectangle at the sides of the pie.
Area of rectangle = length × width
A = 5 × 2.5 = 12.5
Since there are two rectangles with the same area on both sides of the pie, we can double it and get 25 in². We still have one rectangle left, the one at the back of the pie.
A = 6 × 2.5 = 15
Add the areas altogether to get the answer.
24 + 12.5 + 15 = 51.5 in²
I hope this helps. Sorry if Part B's answer is wrong.