Criticize the following proof by induction of the proposition, "happy families are all alike." consider a set consisting of one happy family. obvi- ously all its elements are the same. suppose it has been shown that for any set of n happy families, say {}, we have fi f2 ==fn. con- sider a set {f1. n+1} of n+1 happy families. then {f1.) is a set of n happy families, so f1 = f2 = = fn. similarly, {f2, f3, fn+1} is a set of n happy families, so fn+1 = = f2. thus fr+1 = fi also, and the set of n + 1 happy families are all alike. by the principle of induction, we see that for any finite set of happy families, they are all alike. since the set of all happy families is finite, we conclude: happy families are all alike.

56

step-by-step explanation:

answer: at first, in 1635, the evening meetings had been just for women . .

what relationship between ideas is shown in the text above?

question 1 options:

location of events

similarity of events

timing of events

examples of an idea

step-by-step explanation:

b. y = 4x - 2

step-by-step explanation:

to find a parallel equation to an equation, make sure both equations have the same slope but different y-intercepts.

y = 4x + 1

4 = slope

1 = y-intercept

y = 4x - 2

4 = slope

-2 = y-intercept

they both have the same slope but different y-intercepts, so, b. is the correct answer.

and to check your work, you can graph both equations. : )

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.