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3 years ago

F(x)=9(x-8) find f^-1(x)

Mathematics

1 answer:

mina [271]3 years ago

8 0

Answer:

$f^{-1}$ (x) = $\frac{1}{9}$ x + 8

Step-by-step explanation:

let y = f(x) and rearrange making x the subject, that is

y = 9(x - 8) ← divide both sides by 9

$\frac{y}{9}$ = x - 8 ( add 8 to both sides )

$\frac{y}{9}$ + 8 = x

Change y back into terms of x with x = $f^{-1}$ (x) , thus

$f^{-1}$ (x) = $\frac{1}{9}$ x + 8

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(a) If G is a finite group of even order, show that there must be an element a = e, such that a−1 = a (b) Give an example to sho

Dahasolnce [82]

Answer:

See proof below

Step-by-step explanation:

First, notice that if a≠e and a^-1=a, then a²=e (this is an equivalent way of formulating the problem).

a) Since G has even order, |G|=2n for some positive number n. Let e be the identity element of G. Then A=G\{e} is a set with 2n-1 elements.

Now reason inductively with A by "pairing elements with its inverses":

List A as A={a1,a2,a3,...,a_(2n-1)}. If a1²=e, then we have proved the theorem.

If not, then a1^(-1)≠a1, hence a1^(-1)=aj for some j>1 (it is impossible that a^(-1)=e, since e is the only element in G such that e^(-1)=e). Reorder the elements of A in such a way that a2=a^(-1), therefore a2^(-1)=a1.

Now consider the set A\{a1,a2}={a3,a4,...,a_(2n-1)}. If a3²=e, then we have proved the theorem.

If not, then a3^(-1)≠a1, hence we can reorder this set to get a3^(-1)=a4 (it is impossible that a^(-1)∈{e,a1,a2} because inverses are unique and e^(-1)=e, a1^(-1)=a2, a2^(-1)=a1 and a3∉{e,a1,a2}.

Again, consider A\{a1,a2,a3,a4}={a5,a6,...,a_(2n-1)} and repeat this reasoning. In the k-th step, either we proved the theorem, or obtained that a_(2k-1)^(-1)=a_(2k)

After n-1 steps, if the theorem has not been proven, we end up with the set A\{a1,a2,a3,a4,...,a_(2n-3), a_(2n-2)}={a_(2n-1)}. By process of elimination, we must have that a_(2n-1)^(-1)=a_(2n-1), since this last element was not chosen from any of the previous inverses. Additionally, a_(2n1)≠e by construction. Hence, in any case, the statement holds true.

b) Consider the group (Z3,+), the integers modulo 3 with addition modulo 3. (Z3={0,1,2}). Z3 has odd order, namely |Z3|=3.

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3 years ago

Bradley made a house for his dog, Bowser, out of wood with a cube base and a triangular prism top. The dimensions of the dog hou

Luda [366]

The complete question is
<span>Bradley made a house for his dog, Bowser, out of wood with a cube base and a triangular prism top. The dimensions of the dog house are a = 2 feet, b = 1 foot, and c = 1.4 feet.
</span><span>If Bradley plans to paint the outside of the dog house blue, not including the bottom, how many square feet of paint will he use?

the picture in the attached figure
</span>
we know that

surface area of the figure=surface area of cube+surface area of triangular prism

surface area of cube=4*a²----> 4*2²-----> 16 ft²

surface area of triangular prism=2*[a*c]+2*[a*b/2]---> 2*[2*1.4]+[2*1]
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surface area of the figure=16 ft²+7.6 ft²----> 23.6 ft²

the answer is
23.6 ft²

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