Answer:
Factors of 58: 1,2,29 and 58
58 It is a composite number.
Step-by-step explanation:
The factors of 58 are the numbers that divide 58 leaving 0 as the remainder.
For example, 58/29=2, the remainder of 0.
Factors of 58: 1,2,29 and 58
58 It is a composite number.
which equation has a solution of x = 4
Answer
Option B is correct.
Only the equation, 3x + 9 = 21, has a solution of x = 4.
Explanation
We are told to pick the equation(s) with x = 4 as a solution from the equations,
5x - 8 = 44
3x + 9 = 21
4x = 24
x - 10 = -8
The step to solving this is to insert x = 4 and check if that is consistent with the given equation.
Option A
5x - 8 = 44
If x = 4
5(4) - 8 = 44
20 - 8 = 44
12 ≠ 44
Hence, this is not an answer
Option B
3x + 9 = 21
If x = 4
3(4) + 9 = 21
12 + 9 = 21
21 = 21
Hence, this is an answer for this question.
Option C
4x = 24
If x = 4
4(4) = 24
16 ≠ 24
Hence, this is not an answer to this question.
Option D
x - 10 = - 8
If x = 4
4 - 10 = -8
-6 ≠ -8
This is also not an answer to this question.
Hope this Helps!!!
Carbon-14 is used for archeological carbon dating. Its half-life is 5730 years. How much of a 50-gram sample of Carbon-14 will be left in 1000 years?
Given:
The half-life of carbon-14 is 5730 years.
The initial amount of carbon is I = 50 grams.
Explanation:
To find the final amount of carbon after 1000 years.
The fundamental decay equation is,
[tex]\begin{gathered} F=Ie^{-\lambda t} \\ \text{Where, }\lambda=\frac{\ln 2}{t_{\frac{1}{2}}} \end{gathered}[/tex]Let us find the radioactive constant first.
[tex]\begin{gathered} \lambda=\frac{\ln 2}{5730} \\ \lambda=0.00012096809 \end{gathered}[/tex]Then, the final amount of the corban-14 is,
[tex]\begin{gathered} F=50e^{-0.000121(1000)}^{} \\ =44.30g \end{gathered}[/tex]Hence, the amount of a 50-gram sample of Carbon-14 will be left in 1000 years is 44.30 g.
Which is the closest to the area of the triangle in square centimeters?
Option c ) 40 is the closest to the area of the triangle in square centimeters .
Formula for Area of a right-angled triangle :
Area of a right-angled triangle = [ ( 1 / 2 ) * base * height ]
According to question ,
base = 10.1 cm
height = 8.2 cm
So , Area of triangle = [ ( 1 / 2 ) * 10.1 * 8.2 ]
= 41.41 [tex]cm^{2}[/tex]
This is closest to option c ) 40 .
Hence , option c ) 40 is the closest to the area of the triangle in square centimeters .
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Describe and correct the error in performing the operation of complex numbers and write the answer in standard form.
Answer:
-20+48i
Explanation:
The solution erroneously began by expressing the square as the square of each of the terms.
[tex](4+6i)^2=(4)^2+(6i)^2^{}[/tex]However, the correct way is to take the square of the entire expression inside the bracket as shown below:
[tex](4+6i)^2=(4+6i)(4+6i)[/tex]Next, we expand and simplify our result below:
[tex]\begin{gathered} =4(4+6i)+6i(4+6i) \\ =16+24i+24i+36i^2 \\ =16+48i+36(-1) \\ =16-36+48i \\ =-20+48i \end{gathered}[/tex]The result of the operation in standard form is -20+48i.
x + y =5 x + y = 6 one solution no solutions infinitely many solutions
Problem
x + y = 5
x + y = 6
method
A system has no solution if the equations are inconsistent, they are contradictory.
for example
2x + 3y = 10
2x + 3y = 12 has no solution.
Final answer
x + y = 5
x + y = 6
are inconsistent
hence, the equations has no solution
NO SOLUTION
The figure below shows the graph of f’ , the derivative of the function f, on the closed interval from x = -2 to x = 6. The graph of the derivative has horizontal tangentlines at x = 2 and x = 4.
Solution
- The points of inflection of f(x) in a graph of f'(x) is gotten by just finding the points where the graph moves from increasing to decreasing, and also from decreasing to increasing.
- Thus, we have
- The points where the graph changes from increasing to decreasing is at point (2, 0) and the point where the graph moves from decreasing to increasing is (4, -2.5)
- Thus, the inflection points of the graph of f are at (2, 0), and (4, -2.5)
8 with a exponent of 3 divided by 2
8³ ÷ 3
First we find the value of 8³;
8³ = 512
Then divide by 3
512/3 = 170.6
If the sample space, S = {1, 2, 3, 4, …, 15} and A = the set of odd numbers from the given sample space, find Ac.A.{1, 2, 3, 4, 5, 6, …, 15}B.{1, 3, 5, 7, 9, 11, 13, 15}C.{1, 2, 3, 4, 15}D.{2, 4, 6, 8, 10, 12, 14}
A^c is the complement of set A.
Given that A is a subset of S, then A^c contains the elements present in set S but not in set A.
The sets are:
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
A = {1, 3, 5, 7, 9, 11, 13, 15} (odd numbers present in S)
Therefore, the elements present in set S but not in set A are:
[tex]A^c=\mleft\lbrace2,4,6,8,10,12,14\mright\rbrace[/tex]
CJ loves Girl Scout cookies. He eats 3 cookies per hour. After 5 hours, there are 24 cookies left in the box. Write an equation in slope intercept form. Determine how many hours it will take CJ to eat the entire box of cookies.
To solve this problem: y will represent the number of cookies, and x the number of hours.
To find the number of cookies that CJ eats per hour, we multiply 3 (since he eats 3 per hour) by x (the number of hours)
Since there we only 24 cookies left in the box, we will need to substract 3 by the number of hours that have passed, from 24 to find the number of cookies "y":
[tex]y=24-3(x-5)[/tex]This equation represents that the number of cookies "y" is equal to the 24 cookies that where left after 5 hours, and to that we substract 3 (which is the number of cookies per hour) by total number of hours that have passed since those 5 hours (x-5) because 5 hours that have already passed we substract them from x.
We need to simplify that equation to represent in slope-intercept form:
[tex]\begin{gathered} y=24-3x+15 \\ y=-3x+39 \end{gathered}[/tex]Now we need to determine the number of hours it would take to finish the cookies. So we are looking for the value of x, that makes y=0:
[tex]0=-3x+39[/tex]solving for the number of hours x:
[tex]\begin{gathered} -3x=-39 \\ x=-\frac{39}{(-3)} \\ x=13 \end{gathered}[/tex]It would take 13 hours for CJ to eat the entire box of cookies.
Apply zero product theorem to solve for x[tex]x ^{2} = 9[/tex]
Answer:
[tex]\begin{gathered} x_1=-3 \\ x_2=3 \end{gathered}[/tex]Step-by-step explanation:
To apply the zero product theorem, put all the terms on the left side to equal zero.
[tex]x^2-9=0[/tex]Factoring the binomial:
[tex]\begin{gathered} (x+3)(x-3)=0 \\ x_1+3=0 \\ x_1=-3 \\ \\ x_2-3=0 \\ x_2=3 \end{gathered}[/tex]URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!
Answer:
I don't know the answer but I want to say something ...you can't just go around writing HELP!!! ILL GIVE 100 POINTS when your question only gives 5!!! it's just deceptive, if you want someone's help at least be honest! thank you for your time
If P = (-3,5), find the imageof P under the following rotation.180° counterclockwise about the origin([?], []).Enter the number that belongs inthe green box.Enter
The rule for a 180° counterclockwise rotation is-
[tex](x,y)\rightarrow(-x,-y)\text{.}[/tex]So, we just have to change the sign of each coordinate.
[tex](-3,5)\rightarrow(3,-5)[/tex]Hence, the image is (3,-5).X1 2 Given f(x) = 35 - 2 - 2
Use the rule of correspondence of the case when x>3, since 5>3.
[tex]\begin{gathered} f(5)=5+2 \\ =7 \end{gathered}[/tex]Therefore, f(5)=7.
I need all solved, As soon as possible Question 1
Given:
[tex]f(x)=3^x[/tex]To find:
The type of function by completing the table and graphing the function
Explanation:
When x = -2,
[tex]\begin{gathered} y=3^{-2} \\ =\frac{1}{3^2} \\ =\frac{1}{9} \\ =0.11 \end{gathered}[/tex]When x = -1,
[tex]\begin{gathered} y=3^{-1} \\ =\frac{1}{3} \\ =0.33 \end{gathered}[/tex]When x = 0,
[tex]\begin{gathered} y=3^0 \\ =1 \end{gathered}[/tex]When x = 1,
[tex]\begin{gathered} y=3^1 \\ =3 \end{gathered}[/tex]When x = 2,
[tex]\begin{gathered} y=3^2 \\ =9 \end{gathered}[/tex]Therefore, the table values are,
Then, the graph will be,
Since the domain of the function is real numbers and the range of the function is a set of positive real numbers.
Therefore, it is an exponential function.
A corporation distributes a 10% common stock dividend on 30000 shares issued when the market value of its common stock is $24 per share and its par value is $2 per share dollars per share on the distribution date a credit for $___ would be journalized.A. $30,000B. $6,000C. $72,000D. $66,000
A corporation distributes a 10% common stock dividend on 30,000 shares.
The market value is $24 per share.
The par value is $2 per share.
We have to find the credit that is journalized the moment the distribution is made.
They paid a total amount in dividends that is 10% of the par value of the stock times the number of stocks:
[tex]\begin{gathered} 10\%\cdot2\cdot30000 \\ 0.1\cdot2\cdot30000 \\ 6000 \end{gathered}[/tex]Answer: the credit is $6,000 [Option B]
Weekly wages at a certain factory arenormally distributed with a mean of$400 and a standard deviation of $50.Find the probability that a workerselected at random makes between$500 and $550.
The Solution:
Step 1:
We shall state the formula for calculating Z-score.
[tex]\begin{gathered} Z=\frac{X-\mu}{\sigma} \\ \text{Where X}=5\text{00 ( for lower limit) and X=550 for upper limit.} \\ \mu=400 \\ \sigma=50 \end{gathered}[/tex]Step 2:
We shall substitute the above values in the formula.
[tex]\begin{gathered} \frac{500-400}{50}\leq P(Z)\leq\frac{550-400}{50} \\ \\ \frac{100}{50}\leq P(Z)\leq\frac{150}{50} \\ \\ 2\leq P(Z)\leq3 \end{gathered}[/tex]Step 3:
We shall read the respective probabilities from the Z score distribution tables.
From the Z-score tables,
P(3) = 99.9 %
P(2) = 97.7 %
Step 4:
The Conclusion:
The probability that a worker selected makes between $500 and $550 is obtained as below:
[tex]\text{Prob}(500\leq Z\leq550)=99.9-97.7\text{ = 2.2 \%}[/tex]Therefore, the required probability is 2.2 %
Using the compound interest formula, determine the total amount paid back and the monthly payment. Buying a $6000 used sedan taken out with $500 paid up front and the rest borrowed at 8.3%annual interest compounded daily (365 days per year) over 2 years.
The final value of an investment or loan with compound interest is given by:
[tex]FV=P(1+\frac{r}{m})^{m\cdot t}[/tex]Where P is the initial value (principal or loan), r is the annual interest rate, t is the duration of the investment/loan, and m is the number of compounding periods per year.
The following values are given in the problem:
P = $6000 - $500 = $5500
r = 8.3% = 0.083
t = 2 years
m = 365
Applying the formula:
[tex]FV=5500(1+\frac{0.083}{365})^{365\cdot2}[/tex]Calculating:
[tex]FV=5500(1+0.0002273926)^{730}[/tex]FV = $6493.03
The total amount paid back is $6493.03
This is equivalent to an approximate monthly payment of:
[tex]R=\frac{$ 6493.03 $}{24}=270.54[/tex]The monthly payment is approximately $270.54
Solve the equation for all real solutions. 9z^2-30z+26=1
Weare given the following quadratic equation, and asked to find all its real solutions:
9 z^2 - 30 z + 26 = 1
we subtract "1" from both sides in order to be able to use the quadratic formula if needed:
9 z^2 - 30 z + 26 - 1 = 0
9 z^2 - 30 z + 25 = 0
we notice that the first term is a perfect square:
9 z^2 = (3 z)^2
and that the last term is also a perfect square:
25 = 5^2
then we suspect that we are in the presence of the perfect square of a binomial of the form:
(3 z - 5)^2 = (3z)^2 - 2 * 15 z + 5^2 = 9 z^2 - 30 z + 25
which corroborates the factorization of the trinomial we had.
Then we have:
(3 z - 5)^2 = 0
and the only way such square gives zero, is if the binomial (3 z - 5) is zero itself, which means:
3 z - 5 = 0 then 3 z = 5 and solving for z: z = 5/ 3
Then the only real solution for this equation is the value:
z = 5/3
For a standard normal distribution, find the z-value that goes with a left tail area=0.9931
The z-value that goes with a left tail Area= 0.9931 is 2.4 .
What is normal distribution?A probability distribution that is symmetric about the mean is the normal distribution, also known as the Gaussian distribution. Data close to the mean are more common than data far from the mean. The normal distribution is displayed as a "bell curve" on the chart.
What is left rear area?
The area under the curve to the left of x* in Figure 5.19, “Right and left tails of the distribution” is known as the left tail of the density curve for a continuous random variable X whose limit is x* (a).
According to the Z-value normal distribution table, its value is 2.4 .
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The Knitting Club members are preparing identical welcome kits for new members. The Knitting Club has 45 spools of yarn and 75 knitting needles. What is the greatest number of identical kits they can prepare using all of the yarn and knitting needles?
Common factors of 45 : 1,3,5,9,15,45
Common factors of 75 : 1,3,5,15,25,75
Common factors: 1,3,5,15
GReatest common factor = 15
15 identical kits
All changes 4. What are the coordinates of the midpoint of the line segment with endpoints (7, 2) and (3, 4)? O (5,3) O (4, -2) O (4,2) 0 (2, 1)
We will find the coordinates of the mid-point using the following expression:
[tex]mp=(\frac{_{}x_2+x_1}{2},\frac{y_2+y_1}{2})_{}_{}_{}[/tex]So, when we replace we obtain the mid-point coordinates:
[tex]mp=(\frac{7+3}{2},\frac{2+4}{2})\Rightarrow mp=(5,3)[/tex]So, the coordinates of the mid-point are (5, 3).
what is 39 ÷ (2+ 1) - 2 × (4 + 1)
The given expression is
[tex]36\colon(2+1)-2\times(4+1)[/tex]First, we solve the additions inside the parenthesis
[tex]36\colon3-2\times5[/tex]Then, we solve the produce and the division
[tex]12-10[/tex]At last, we subtract
[tex]12-10=2[/tex]Hence, the answer is 2.Sue, who is 5 feet tall, is standing at Point D in the drawing. The tip of her head is a point E. a tree in the yard is at point B with the top of the tree at point C. Sue stand so her shadow meets at the end of the trees shadow at point a Which triangles similar?How do you know?Find the height of the tree (This distance from B to C).
Which triangles are similar?
The triangle AED and the triangle ABC is similar.
How do you know?
Because all the angles are equal, the triangle AED and ABC have the same angle values, then they're similar.
Find the height of the tree (This distance from B to C)
We can use the relation of the similar triangle to find BC, we can write the equation
[tex]\frac{AB}{AD}=\frac{BC}{ED}[/tex]The only unknown value here is BC, then
[tex]\frac{24+8}{8}=\frac{\text{BC}}{5}[/tex]Now we solve it for BC!
[tex]\begin{gathered} \frac{32}{8}=\frac{BC}{5} \\ \\ 4=\frac{BC}{5} \\ \\ BC=4\cdot5 \\ \\ BC=20\text{ ft} \end{gathered}[/tex]Hence, the height of the tree is 20 ft
what are the roots of the equation?-3= -6x^2+7x
We have the next equation
[tex]-3=-6x^2+7x[/tex]First, we need to set the equation to zero
[tex]6x^2-7x-3=0[/tex]then we will use the general formula to find the roots of a second-degree equation
[tex]x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]where
a=6
b=-7
c=-3
then we substitute the values
[tex]x_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4(6)(-3)}}{2(6)}[/tex][tex]\begin{gathered} x_{1,2}=\frac{7\pm\sqrt[]{49^{}+72}}{12} \\ x_{1,2}=\frac{7\pm\sqrt[]{121}}{12} \\ x_{1,2}=\frac{7\pm11}{12} \\ \end{gathered}[/tex][tex]x_1=\frac{7+11}{12}=\frac{18}{12}=\frac{3}{2}[/tex][tex]x_2=\frac{7-11}{12}=\frac{-4}{12}=-\frac{1}{3}[/tex]the roots of the equation are x=3/2, x=-1/3
? Question The table shows certain values of a fourth-degree polynomial function with no repeated factors. -12 -10 -6 -4 2 4 8 10 12 у 280 81 -14 0 0 -24 0 126 400 The function must have a zero between the x-values of -12 and -10 Between the x-values of 2 and 8, the graph of the function should be drawn the x- The function must be positive for all x-values between Submit
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Answer: (-10 and 06), below, (-4, 2)
Step-by-step explanation:
From Edmentum.
Convert 5 1/4 lb to oz.
The conversion factor for lb to oz is
[tex]1lb=16oz[/tex]I will put the pounds first in terms of the improper fraction. We have
[tex]5\frac{1}{4}=\frac{21}{4}[/tex]Using the conversion factor to convert lb to oz, we have
[tex]\frac{21}{4}lb\times\frac{16oz}{1lb}=\frac{21\cdot16}{4}=\frac{21\cdot4}{1}=84oz[/tex]Hence, 5 1/4 lb is equal to 84 oz.
Answer: 84 oz
Multiples of 36 and the square root of 49
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
36
√49
Step 02:
multiples:
36:
The multiples of thirty-six are the numbers that contain it a whole number of times.
36, 72, 108, 144, 180, ...
√49:
[tex]\sqrt[]{49}=7[/tex]The multiples of seven are the numbers that contain it a whole number of times.
7, 14, 21, 28, 35, 42, ...
That is the full solution.
Triangles ABE, ADE, and CBE are shown on the coordinate grid, and all the vertices have coordinates that are integers. Which statement is true?
To check if Triangles ABE, ADE, and CBE are congruent, let us compute for the distance of each line using the Distance Formula,
[tex]\text{ }d\text{ = }\sqrt[]{(x_2-x_1)^2\text{ + (}y_2-y_1)^2}[/tex]Where,
d = Distance
(x1, y1) = Coordinates of the first point
(x2, y2) = Coordinates of the second point
Let's compute the distance of the following lines:
Triangle ABE: Lines AB, AE, and BE
Triangle ADE: Lines AD, AE, and ED
Triangle CBE: Lines CE, CB, and BE
For Triangle ABE,
[tex]\text{ d}_{AB}\text{ = }\sqrt[]{(-1-(-4))^2+(3-(-1))^2}\text{ = }\sqrt[]{(-1+4)^2+(3+1)^2}[/tex][tex]\text{ d}_{AB}\text{ = }\sqrt[]{(3)^2+(4)^2}\text{ = }\sqrt[]{9+\text{ 16}}\text{ = }\sqrt[]{25}[/tex][tex]\text{ d}_{AB}\text{ = 5}[/tex][tex]\text{ d}_{AE}\text{ =}\sqrt[]{(1-\text{ }(-1))^2+(0\text{ - }(-4))^2}\text{ = }\sqrt[]{(1+1)^2+(0+4)^2}[/tex][tex]\text{ d}_{AE}\text{ = }\sqrt[]{(2)^2+(4)^2}\text{ = }\sqrt[]{4\text{ + 16}}[/tex][tex]\text{ d}_{AE}\text{ =}\sqrt[]{20}[/tex][tex]\text{ d}_{BE}\text{ = }\sqrt[]{(1\text{ - (}3))^2+(0\text{ - (-1)})^2}\text{ = }\sqrt[]{(1-3)^2+(0+1)^2}[/tex][tex]\text{ d}_{BE}=\text{ }\sqrt[]{(-2)^2+(1)^2}\text{ = }\sqrt[]{4\text{ + 1}}[/tex][tex]\text{ d}_{BE}\text{ = }\sqrt[]{5}[/tex]For Triangle ADE, let's compute for the distance of line AD and ED since we already got the distance of line AE.
[tex]\text{ d}_{AD}\text{ = }\sqrt[]{(-1-(-1))^2+\text{ (}1\text{ - }(-4))^2}\text{ = }\sqrt[]{(-1+1)^2+(1+4)^2}[/tex][tex]\text{ d}_{AD}\text{ = }\sqrt[]{(0)^2+(5)^2}\text{ = }\sqrt[]{25}[/tex][tex]\text{ d}_{AD}\text{ = 5}[/tex][tex]\text{ d}_{ED}=\text{ }\sqrt[]{(-1\text{ - (}1))^2+(1-0)^2}\text{ = }\sqrt[]{(-1-1)^2+(1)^2}[/tex][tex]\text{ d}_{ED}\text{ = }\sqrt[]{(-2)^2_{}+(1)^2}\text{ = }\sqrt[]{4\text{ + 1}}[/tex][tex]\text{ d}_{ED}\text{ = }\sqrt[]{5}[/tex]For Triangle CBE, let's compute for the distance of line CE and CB since we already got the distance of line BE.
[tex]\text{ d}_{CE}\text{ = }\sqrt[]{(3-\text{ }1)^2+(4-0)^2}\text{ = }\sqrt[]{(2)^2+(4)^2}[/tex][tex]\text{ d}_{CE}\text{ = }\sqrt[]{4+16}\text{ = }\sqrt[]{20}[/tex][tex]\text{ d}_{CE}\text{ = }\sqrt[]{20}[/tex][tex]\text{ d}_{CB}\text{ = }\sqrt[]{(3-3)^2+(4\text{ - (}-1))^2}\text{ =}\sqrt[]{(0)^2+(4+1)^2}[/tex][tex]\text{ d}_{CB}\text{ =}\sqrt[]{(5)^2}\text{ = }\sqrt[]{25}[/tex][tex]\text{ d}_{CB}\text{ = 5}[/tex]In summary,
Triangle ABE:
[tex]AB=\text{ 5, AE = }\sqrt[]{20}\text{ and BE = }\sqrt[]{5}[/tex]Triangle ADE:
[tex]\text{ AD = 5, AE = }\sqrt[]{20}\text{ and ED = }\sqrt[]{5}[/tex]Triangle CBE: CE, CB, and BE
[tex]\text{ CB = 5, CE = }\sqrt[]{20}\text{ and BE = }\sqrt[]{5}[/tex]The sides of the three triangles shown in the grid are congruent based on the SSS Rule of Triangle.
Thus, the statement that meets our evaluation is:
D. Triangle ABE, ADE and CBE are all congruent.
state the solution for the quadratic equation depicted in the graph.
For this problem, we were provided with the graph of a quadratic equation, and we need to determine the solutions for this graph.
The solutions of a quadratic equation are the values of "x" that make the expression equal to "0". Therefore, we need to look at the graph for the values at which the graph crosses "y=0".
We have two points for this problem. The first one is approximately -5, and the second is 6.
this is factor by grouping. did I do 1a right and how do I continue on 1b
We will have the following:
*First: ( f ° g) (x):
[tex](f\circ g)(x)=\frac{(\frac{1}{x})+1}{(\frac{1}{x})-2}\Rightarrow(f\circ g)(x)=\frac{(\frac{1+x}{x})}{(\frac{1-2x}{x})}[/tex][tex]\Rightarrow(f\circ g)(x)=\frac{(1+x)(x)}{(x)(1-2x)}\Rightarrow(f\circ g)(x)=\frac{1+x}{1-2x}[/tex]Domain:
[tex](-\infty,\frac{1}{2})\cup(\frac{1}{2},\infty)[/tex]*Second: (f ° f) (x):
[tex](f\circ f)(x)=\frac{(\frac{x+1}{x-2})+1}{(\frac{x+1}{x-2})-2}\Rightarrow(f\circ f)(x)=\frac{(\frac{(x+1)+(x-2)}{x-2})}{(\frac{(x+1)-2(x-2)}{x-2})}[/tex][tex]\Rightarrow(f\circ f)(x)=\frac{(\frac{2x-1}{x-2})}{(\frac{-x+5}{x-2})}\Rightarrow(f\circ f)(x)=\frac{(2x-1)(x-2)}{(x-2)(-x+5)}[/tex][tex]\Rightarrow(f\circ f)(x)=\frac{2x-1}{-x+5}[/tex]Domain:
[tex](-\infty,5)\cup(5,\infty)[/tex]