What is the reciprocal of 20/32

1 answer:

7 0

Step-by-step explanation: The reciprocal of 20/32 is 32/20. Reciprocal is just flipping a number. For an example: The reciprocal of 98/12 is 12/98.

Hope this helps! :D

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Step-by-step explanation:

Problem One (left panel)

Question A

The y intercept happens when x = 0
That being said, the y intercept is 50. It was moving when the timing began.

Question B

The rate of change = (56 - 52)/(3 - 1) = 4/2 = 2 miles / hour^2 (you have a slight acceleration.

Question C

60 = a + (n-1)d
60 = 50 + (n - 1)*2
10/2 = (n - 1)*2/2
5 = n - 1
6 = n

The way I have done it the domain is n from 1 to 6

Question 2 (Right Panel)

Question A

The equation for the table is f(x) = 3x - 3 which was derived simply by putting all three points into y = ax + b and solving.

f(0) = ax + b
-3 = a*0) + b
b = - 3
So far what you have is
f(x) = ax - 3
f(-1) = a*(-1) - 3 but we know (f(-1)) = -6
- 6 = a(-1) - 3 add 3 to both sides
-6 +3 = a(-1) -3 + 3
-3 = a*(-1) Divide by - 1
a = 3
f(x) = 3x - 3 Answer for f(x)
The slope of f(x) = the coefficient in front of the x
f(x) has a slope of 3
g(x) has a slope of 4

Part B

f(x) has a y intercept of - 3
g(x) has a y intercept of -5
f(x) has the greater y intercept.
-3 > - 5

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3 years ago

Let a, b, c and x elements in the group G. In each of the following solve for x in terms of a, b, and c.

alina1380 [7]

Answer:

The answer is $x=a^{-1}cb^{-1}$ .

Step-by-step explanation:

First, it is important to recall that the group law is not commutative in general, so we cannot assume it here. In order to solve the exercise we need to remember the axioms of group, specially the existence of the inverse element, i.e., for each element $g\in G$ there exist another element, denoted by $g^{-1}$ such that $gg^{-1}=e$ , where $e$ stands for the identity element of G.