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Andrej [43]
3 years ago
7

How do I make a linear function, what is a linear function?

Mathematics
1 answer:
mixer [17]3 years ago
3 0

Answer:

I have uploaded a picture so you could understand it a little more.

A linear function is when there is a table that gives numbers that represent a function, as shown in the picture.

In the Image, there is a table with a x and y column. And in both columns, there are numbers.

This is what the table looks like in the picture:

<u>  x   y   </u>

0    4                       The key is y = 2x + 4

1     6 <------      lets grab the number 1 in the x column and lets

2    8       multiply it by two.  Now add four to it, and we will

3    10                  have the number 6 in the y column.

4    12


I hope I helped you! God bless :-)

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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.

Here, e=0. Note that 1²=1+1=2≠e, and 2²=2+2=4mod3=1≠e. Therefore the conclusion of part a) does not hold

7 0
3 years ago
Can i get help on 18 &amp; 22 please
Pavlova-9 [17]

18) 36.36 rounded it is 36.4

22)53.71 rounded it it’s just 53.7

5 0
3 years ago
Solve for x in the equation x^2+11x
suter [353]

Answer:

Pretty sure it would be the last one.

Not sure if you want me to do the work. All I can say is that it would be<em> very long.</em>

6 0
3 years ago
Read 2 more answers
A + b + c =-10 <br> x + y + z =-10<br><br> what is <br> −6c−6b+6z+6x+6y−6a?
shutvik [7]
I’m not to sure I was wondering the same thing
4 0
3 years ago
1.) Determine the type of solutions for the function (Picture 1)
NNADVOKAT [17]

Answer:

1) 2 nonreal complex roots

2) 1 Real Solution

3) 16

4) Reflected, narrower by a factor of 2/5, slides right 4 units and slides up 6 (units)

Step-by-step explanation:

1) The graph does not intercept the x-axis, therefore, there are no real solutions at the point y = 0

We get;

y = a·x² + b·x + c

At y = 6, x = -2

Therefore;

6 = a·(-2)² - 2·b + c = 4·a - 2·b + c

6 = 4·a - 2·b + c...(1)

At y = 8, x = 0

8 = a·(0)² + b·0 + c

∴ c = 8...(2)

Similarly, we have;

At y = 8, x = -4

8 = a·(-4)² - 4·b + c = 16·a - 4·b + 8

16·a - 4·b = 0

∴ b = 16·a/4 = 4·a

b = 4·a...(3)

From equation (1), (2) and (3), we have;

6 = 4·a - 2·b + c

∴ 6 = b - 2·b + 8 = -b + 8

6 - 8 = -b

∴ -b = -2

b = 2

b = 4·a

∴ a = b/4 = 2/4 = 1/2

The equation is therefor;

y = (1/2)·x² + 2·x + 8

Solving we get;

x = (-2 ± √(2² - 4 × (1/2) × 8))/(2 × (1/2))

x =( -2 ± √(-12))/1 = -2 ± √(-12)

Therefore, we have;

2 nonreal complex roots

2) Give that the graph of the function touches the x-axis once, we have;

1 Real Solution

3) The given function is f(x) = 2·x² + 8·x + 6

The general form of the quadratic function is f(x) = a·x² + b·x + c

Comparing, we have;

a = 2, b = 8, c = 6

The discriminant of the function, D = b² - 4·a·c, therefore, for the function, we have;

D = 8² - 4 × 2 × 6 = 16

The discriminant of the function, D = 16

4.) The given function is g(x) = (-2/5)·(x - 4)² + 6

The parent function of a quadratic equation is y = x²

A vertical translation is given by the following equation;

y = f(x) + b

A horizontal to the right by 'a' translation is given by an equation of the form; y = f(x - a)

A vertical reflection is given by an equation of the form; y = -f(x) = -x²

A narrowing is given by an equation of the form; y = b·f(x), where b < 1

Therefore, the transformations of g(x) from the parent function are;

g(x) is a reflection of the parent function, with the graph of g(x) being narrower by 2/5 than the graph of the parent function. The graph of g(x) is shifted right by 4 units and is then slides up by 6 units.

7 0
2 years ago
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