1. The number of people who carried a cell phone or a tablet is 128.
2. The probability that a person carries a cell phone or a tablet is 78.5%.
3. The number of people who carried neither a cell phone nor a tablet is 35.
4. The probability that a person carried neither a cell phone nor a tablet is 21.5%.
5. The number of people who carried a cell phone only is 43.
6. The probability that a person carried a cell phone only is 26.4%.
7. The number of people who carried a tablet but not a cell phone is 38.
8. The probability that a person carried a tablet but not a cell phone is 23.3%.
Data and Calculations:The sample size of the survey = 163
The number of persons who carried a cell phone = 90
The number of persons who carried a tablet computer = 81
The number of persons who carried both a cell phone and a tablet = 43
1. The number of people who carried a cell phone or a tablet is 128 (90 + 81 - 43).
2. The probability that a person carries a cell phone or a tablet is 78.5%. (128/163).
3. The number of people who carried neither a cell phone nor a tablet is 35 (163 - 128).
4. The probability that a person carried neither a cell phone nor a tablet is 21.5% (35/163).
5. The number of people who carried a cell phone only is 43.
6. The probability that a person carried a cell phone only is 26.4%(43/163).
7. The number of people who carried a tablet but not a cell phone is 38 (81 - 43).
8. The probability that a person carried a tablet but not a cell phone is 23.3% (38/163).
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Question Completion:1. How many people carried a cell phone or a tablet?
2. What is the probability that a person carries a cell phone or a tablet?
3. How many people carried neither a cell phone nor a tablet?
4. What is the probability that a person carried neither a cell phone nor a tablet?
5. How many people carried a cell phone only?
6. What is the probability that a person carried a cell phone only?
7. How many people carried a tablet but not a cell phone?
8. What is the probability that a person carried a tablet but not a cell phone?
Consider the three functions below.
- (4) -(41* *--*
Which statement is true?
The range of h(x) is y> 0.
The domain of g(x) is y> 0.
The ranges of f(x) and h(x) are different from the range of g),
The domains of f(x) and g(x) are different from the domain of h(x).
Answer: c
Step-by-step explanation:
[tex] \rm\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{( - 1 {)}^{n + 1} }{ {mn}^{2} + mn + {m}^{2} n} \\ [/tex]
Let [tex]S[/tex] denote the sum. We can first resolve the sum in [tex]m[/tex] by factorizing and decomposing into partial fractions.
[tex]\displaystyle S = \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{(-1)^{n+1}}{mn^2 + mn + m^2n} \\\\ ~~~~ = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n \sum_{m=1}^\infty \frac1{m(m+n+1)} \\\\ ~~~~ = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)} \sum_{m=1}^\infty \left(\frac1m - \frac1{m+n+1}\right)[/tex]
Rewrite the [tex]m[/tex]-summand as a definite integral. Interchange the integral and sum, and evaluate the resulting geometric sums.
[tex]\displaystyle \sum_{m=1}^\infty \left(\frac1m - \frac1{m+n+1}\right) = \sum_{m=1}^\infty \int_0^1 \left(x^{m-1} - x^{m+n}\right) \, dx \\\\ ~~~~~~~~ = \int_0^1 \sum_{m=1}^\infty \left(x^{m-1} - x^{m+n}\right) \, dx \\\\ ~~~~~~~~ = \int_0^1 \frac{1 - x^{n+1}}{1 - x} \, dx \\\\ ~~~~~~~~ = \int_0^1 \sum_{\ell=0}^n x^\ell \, dx \\\\ ~~~~~~~~ = \sum_{\ell=0}^n \int_0^1 x^\ell \, dx \\\\ ~~~~~~~~ = \sum_{\ell=0}^n \frac1{\ell+1} \\\\ ~~~~~~~~ = \sum_{\ell=1}^{n+1} \frac1\ell = H_{n+1}[/tex]
where
[tex]H_n = \displaystyle \sum_{\ell=1}^n \frac1\ell = 1 + \frac12 + \frac 13 + \cdots + \frac1n[/tex]
is the [tex]n[/tex]-th harmonic number. The generating function will be useful:
[tex]\displaystyle \sum_{n=1}^\infty H_n x^n = -\frac{\ln(1-x)}{1-x}[/tex]
To evaluate the remaining sum to get [tex]S[/tex], let
[tex]\displaystyle f(x) = \sum_{n=1}^\infty \frac{H_{n+1}}{n(n+1)} x^{n+1}[/tex]
and observe that [tex]S=\lim\limilts_{x\to-1^+} f(x)[/tex], which I'll abbreviate to [tex]f(-1)[/tex]. Differentiating twice, we have
[tex]\displaystyle f'(x) = \sum_{n=1}^\infty \frac{H_{n+1}}n x^n[/tex]
[tex]\displaystyle f''(x) = \sum_{n=1}^\infty H_{n+1} x^n[/tex]
[tex]\displaystyle \implies f''(x) = -\frac{\ln(1-x)}{x^2(1-x)} - \frac1x[/tex]
By the fundamental theorem of calculus, noting that [tex]f(0)=f'(0)=0[/tex], we have
[tex]\displaystyle \int_{-1}^0 f'(x) \, dx = f(0) - f(-1) \implies f(-1) = -\int_{-1}^0 f'(x) \, dx[/tex]
[tex]\displaystyle \int_x^0 f''(x) \, dx = f'(0) - f'(x) \implies f'(x) = -\int_x^0 f''(t) \, dt[/tex]
[tex]\displaystyle \implies S = f(-1) = \int_{-1}^0 \int_x^0 \left(\frac{\ln(1-t)}{t^2(1-t)} + \frac1t\right) \, dt \, dx[/tex]
Change the order of the integration, and substitute [tex]t=-u[/tex].
[tex]S = \displaystyle \int_{-1}^0 \int_{-1}^t \left(\frac{\ln(1-t)}{t^2(1-t)} + \frac1t\right) \, dx \, dt \\\\ ~~~ = - \int_{-1}^0 \left(\frac{(1+t) \ln(1-t)}{t^2(1-t)} + \frac1t + 1\right) \, dt \\\\ ~~~ = -1 - \int_{-1}^0 \left(\left(\frac2{1-t} + \frac2t + \frac1{t^2}\right) \ln(1-t) + \frac1t\right) \, dt \\\\ ~~~ = -1 - \int_0^1 \left(\left(\frac2{1+u} - \frac2u + \frac1{u^2}\right) \ln(1+u) - \frac1u\right) \, du[/tex]
For the remaining integrals, substitute and use power series.
[tex]\displaystyle \int_0^1 \frac{\ln(1+u)}{1+u} \, du = \int_0^1 \ln(1+u) d(\ln(1+u)) = \frac{\ln^2(2)}2[/tex]
[tex]\displaystyle \int_0^1 \frac{\ln(1+u)}u \, du = - \int_0^1 \frac1u \sum_{k=1}^\infty \frac{(-u)^k}k \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~ = - \sum_{k=1}^\infty \frac{(-1)^k}k \int_0^1 u^{k-1} \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~ = - \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = \frac{\pi^2}{12}[/tex]
[tex]\displaystyle \int_0^1 \frac{\ln(1+u) - u}{u^2} \, du = - \int_0^1 \frac1{u^2} \left(\sum_{k=1}^\infty \frac{(-u)^k}k + u\right) \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = -\int_0^1 \frac1{u^2} \sum_{k=2}^\infty \frac{(-u)^k}k \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = - \sum_{k=2}^\infty \frac{(-1)^k}k \int_0^1 u^{k-2} \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = -\sum_{k=2}^\infty \frac{(-1)^k}{k(k-1)} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \sum_{k=1}^\infty \frac{(-1)^k}{k(k+1)} = 1 - 2\ln(2)[/tex]
Tying everything together, we end up with
[tex]S = -1 - \left(2 \cdot \dfrac{\ln^2(2)}2 - 2 \cdot \dfrac{\pi^2}{12} + (1-2\ln(2))\right) \\\\ ~~~ = \boxed{\frac{\pi^2}6 - 2 + 2\ln(2) - \ln^2(2)}[/tex]
Explain how to graph the line with the equation y = 3/4x - 5
Answer:
A graphing calculator can help you.
Here is the graph, it is in the photo.
a baseball coach spent $118.25 on 11 pizzas. estimate ate the cost of each pizza using a number with one nonzero digit. then find the exact cost per pizza
Answer:
10.75
Step-by-step explanation:
$118.25 / 11
We're finding the cost of each pizza. So, If a Coach bought 11 Pizza's for $118.25, We need to find how much x is. x = amount of cost per pizza. Divide $118.25 by 11 to get $10.75.
Each Pizza Costs $10.75.
I really need help it’s due in 10 minutes
In the picture we have to solve the individual variables A=2πr²+2πrh. We got function with r as subject is[tex]\frac{-h}{2} \pm\sqrt{\frac{h^{2}}{4 } +\frac{a}{2\pi} }[/tex]
Given that,
In the picture we have to solve the individual variables
A=2πr²+2πrh
We have to find function with r as subject.
Taking A to left side we get
2πr²+2πrh-A=0
We can see the equation is in the form of quadratic equation with variable r.
So, The factor we find by using the formula
That is [tex]\frac{-b\pm\sqrt{b^{2}-4ac } }{2a}[/tex]
Here, a=2π,b=2πh and c=-A
r=[tex]\frac{-2\pi h\pm\sqrt{(2\pi h)^{2}-4(2\pi)(-a) } }{2(2\pi)}[/tex]
r=[tex]\frac{-2\pi h\pm\sqrt{(4\pi^{2}h^{2} +8\pi a) } }{4\pi}[/tex]
r=[tex]\frac{-h}{2} \pm\frac{\sqrt{(4\pi^{2}h^{2} +8\pi a )} }{4\pi}[/tex]
r=[tex]\frac{-h}{2} \pm\sqrt{\frac{4\pi^{2}h^{2}+8\pi a }{16\pi^{2} } }[/tex]
r=[tex]\frac{-h}{2} \pm\sqrt{\frac{h^{2}}{4 } +\frac{a}{2\pi} }[/tex]
Therefore, We got function with r as subject is[tex]\frac{-h}{2} \pm\sqrt{\frac{h^{2}}{4 } +\frac{a}{2\pi} }[/tex]
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HELP PLEASE ASAP!!!!
The measure of angle m∠PZQ is 63degrees.
How to find an angle?A point where two or more line segments meet is called a vertex.
The vertex point also forms an angle.
Therefore,
point P is in the interior of ∠OZQ,
Hence,
<OZQ = <OZP + m∠PZQ
The following angles are given:
m∠OZQ = 125°
m∠OZP = 62°
Substitute the given values into the expression above:
125 = 62 + m∠PZQ
m∠PZQ = 125 - 62
m∠PZQ = 63°
Therefore, the measure of the angle m∠PZQ is 63degrees
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Answer the filling questions in your own words.
1. Which figure above is a line segment?
2. Which figure above is a ray?
3. Explain, in detail, the differences between a line, a line segment, and a ray.
A line has no endpoints (line FG), a line segment has two endpoints (line segment AB), and a ray has one endpoint (ray CD).
What is a Line Segment?A line segment can be described as a line having two definite endpoints.
What is a Ray?A ray is a part of a line that has just one fixed endpoint and extends in the opposite direction of the endpoint to infinity.
What is a Line?A line has no endpoint. It extends in opposite directions to infinity.
1. The figure that is a line segment is the green figure. (line segment AB).
2. The blue figure is a ray (ray CD)
3. The red figure is a line (line FG).
In summary, a line has no endpoints (line FG), a line segment has two endpoints (line segment AB), and a ray has one endpoint (ray CD).
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Which of the following values are solutions to the given
equation?
|x + 2| = 5
Be sure to select ALL numbers which are solutions.
(Select all that apply.)
x = -5
x = -2
x = 3
x = -7
x = -3
x = 7
Answer:
X= -7
X = 3
Step-by-step explanation:
| x + 2| = 5
|-7 + 2| = 5
|-5| = 5
5 = 5
|x + 2| = 5
|3 + 2| = 5
|5| = 5
5 = 5
X= -7
X = 3
Find the domain and range of the function represented by the graph.
kevin found a deal on a computer that has been marked down by 30% to be $490. what was the original price of the computer?
Answer:
$700
Step-by-step explanation:
Since the deal is 30% off, that means that the price is now 70% of the original price.
70% of x = 490
0.7x = 490
x = 490/0.7
x = 700
Answer: $700
beinggreat78 is great
but that's literally her user so....♀️
anywayso
Answer:
1.2 × 10⁻⁵
Step-by-step explanation:
The exponent is negative, so the decimal was moved -5 places back, making the number a decimal.
Answer:
1.2 x 10^5
Step-by-step explanation:
All work is shown in the attached screenshot! :)
If $5,000 had been invested in a certain investment fund on September 30, 2008, it would have been worth $23,125.59 on
September 30, 2018. What interest rate, compounded annually, did this investment earn? (Round your answer to two decimal
places.)
Interest rate compounded annually for the amount of $5,000 invested would have been worth $23,125.59 after 10 years is equal to 16.55% per year.
As given in the question,
Principal (P) = $5,000
Time (t) = 10 years
Amount = $23,125.59
[tex]r = n[(A/P)^{\frac{1}{nt}}-1]\\\\\implies r = 1[(23125.29/5000)^{\frac{1}{10} }-1]\\\\\implies r = 0.1655\\[/tex]
Convert r into percentage
r = 0.1655 × 100
= 16.55% compounded annually
Therefore, interest rate compounded annually for the amount of $5,000 invested would have been worth $23,125.59 after 10 years is equal to 16.55% per year.
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A universal set U consists of 19 elements. If sets A, B, and C are proper subsets of U and n(U) = 19, n(An B) = n(An K C) = n(B n C)= 9, n(An B n C) =6, and n(A U B UC) = 15, determine each of the following. a) n(A U B) b ) n ( A' UC c) n(An B)'
Using Venn sets, the cardinalities are given as follows:
a) n(A U B) = 15.
b) n(A' U C) = 16.
c) n(A ∩ B)' = 10.
What are Venn probability?Venn amounts relates the cardinality of sets that intersect with each other.
For this problem, the sets are the ones given in this problem, A, B and C, while U is the universal set.
For this problem, the cardinalities are given as follows:
n(U) = 19.n(A ∩ B) = n(A ∩ C) = n(B ∩ C) = 9.n(A ∩ B ∩ C) = 6.n(A U B UC) = 15Hence:
6 elements belong to all the sets.9 - 6 = 3 belong to these intersections but not the remaining set: A and B, A and C, B and C.15 belong to the union of all of them, hence 4 belong to none.15 - (6 + 3 x 3) = 0 belong to only one set.Hence:
n(A U B) = 15, as from the final bullet point, there are no elements that belong to only set C.For item b, 6(all) + 3(only A and C) + 3 (only B and C) = 12 elements belong to C, and 4 do not belong to A(the 3 to only B and C is already counted), hence: n(A' U C) = 16, as 12 + 4 = 16.For item c, n(A ∩ B) = 9, hence n(A ∩ B)' = n(U) - n(A ∩ B) = 19 - 9 = 10.More can be learned about Venn sets at https://brainly.com/question/28318748
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Car A travels a distance of 22.5 miles in 30 minutes and car B travels a distance of 34.5 miles in 45 minutes. which car is traveling faster.
someone plssssss ITS URGENT.
The area of a rectangular barn is 119 square feet. its length is 10 feet longer than the width. find the length and width of the wall of the barn.
The length of the field with area 119 square feet is 17 feet and the width of the feet is 7 feet.
The area is the total region or space in two dimension that is covered by a figure , surface or object. Area of a rectangle is calculated by the product of its length and width.
A rectangle has 4 sides where each pair of opposite side is equal .The length of a rectangle is normally the longer side and the width defines the shorter side.
Given the area of the barn is 119 square feet.
Let us consider the width of the barn to be x feet. As the length is 10 feet longer than the width then we will find the length to be ( x +10 ) feet.
Area = length × width
or, 119 = x × (x+10)
or, 119 = x² +10x
or, x² +10x - 119=0
Now we will solve the quadratic equation thus formed by middle term-factorization method:
or, x² +17x - 7x - 119 = 0
or, x ( x + 17 ) - 7 ( x + 17) = 0
or, ( x + 17 ) ( x - 7 ) = 0
Now either x + 17 = 0 or x - 7 =0
Therefore either x = 7
or, x = -17 (The width cannot be a negative value)
∴Width = 7 feet and length = 7 + 10 = 17 feet.
Therefore the length of the field is 17 feet and the width of the feet is 7 feet.
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Which function has the greater average rate of change over the interval [0,3]?
If the interval is [0,3] then the second function whose graph is given has the greater average rate of change.
Given two functions, one is in the table and the other one is in the form of graph.
We are required to choose the function which has the greater average rate of change.
Function is basically the relationship between two or more variables that are expressed in equal to form. The values that we enter are known as part of domain and the values that we get from the function are known as part of codomain or range of the function.
If we observe the table then we will find that in the interval [0,3] there is not any change in the value of function, it is constant to be 4.
If we observe the graph then we will find that the value of function is continuously decreasing.
So, the second function has greater average rate of change over the interval [0,3].
Hence if the interval is [0,3] then the second function whose graph is given has the greater average rate of change.
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Solve 2x + 2 > 10.
PLEASE HELP
Answer:
x > 4
Step-by-step explanation:
2x + 2 > 10
2x > 10 - 2
2x > 8
x > 8/2
x > 4
Helppppppppp pleaseeeeee
Answer:
The answer is yes. Since there is no x^2
In a class in which the final course grades depends entirely on the average of four equally weighted 100 point test David has scored 81, 92, and 74 on the first three. What range of scores on the fourth test will give David a b for the semester ( an average between 80 and 89 inclusive) assume that all the test scores have a non negative value
The range of scores on the fourth test to give David a B grade for the semester is at least 73 but less than or equal to 89.
What is a range?A range refers to the difference between the lowest and the highest score.
A range of scores gives two values, the lowest and the highest scores.
Data and Calculations:Scores secured by David = 81, 92, and 74
Total scores in three exams = 247
The average score for a B grade is:
80 ≤ B grade < 90 ["at least" means ≤]
Therefore,
80 ≤ (81 + 92 + 74 + x) / 4 < 89 [assume equal weights for exams]
80 ≤ (247 + x) / 4 ≤ 89 [add values]
320 ≤ (247 + x) ≤ 356 [multiply by 4, retains a sense of inequality]
73 ≤ x < 109 [subtracting 247 retains a sense of inequality]
Since exams are graded on 100 points, David cannot score 109, so the upper limit is put at 89.
Thus, the exam grade in the fourth test must be at least 73 but less than or equal to 89 to get a B average.
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1. If you deposit P3,000 in BPI bank account that pays 0.125% interest annually, how much will be in your account after 3 years?
2. If you deposit money today in an account that pays 10.5% annual interest, how long will it take to double your money?
3. What is the future value of a 3% 7-year ordinary annuity that pays20,000 each year?
4. (refer to #3) if this were an annuity due, what would its future value be?
5. I want to retire in 20 years. I currently have P 1,250,000 and I will need P20 million at retirement. What annual interest rate must I earn to reach my goal, assuming this is my only investment fund?
Round off your answer up to four decimal places.
The amount that would be in the account kept with BPI bank in 3 years is P3,011.2640
It would 6.9422 years for the deposit to double if interest rate is 10,5%
What is future value of immediate amount?
The future value of deposit today in 3 years means its future equivalent when the amount has earned interest over the 3 years period
FV=PV*(1+r)^N
FV=future value after 3 years=unknown
PV=immediate deposit=3,000
r=annual interest rate=0.125%
N=number of years that the deposit lasted=3
FV=3000*(1+0.125%)^3
FV=P3,011.2640
Time taken to double:
FV=future value=3000*2=6000
PV=3000
r=annual interest rate=10.5%
N=number of years that it takes initial deposit to double=unknown
6000=3000*(1+10.5%)^N
6000/3000=(1.105)^N
2=(1.105)^N
take log of both sides
ln(2)=N*ln(1.105)
N=ln(2)/ln(1.105)
N=6.9422 years
Annuity:
FV=PMT*(1+r)^N-1/r
FV=future value of annuity=unknown
PMT=annual payment=20000
r=interest rate=3%
N=number of annual payments=7
FV=20000*(1+3%)^7-1/7%
FV= 153,249.2436
FV=PMT*(1+r)^N-1/r*(1+r)
FV=20000*(1+3%)^7-1/7%*(1+7%)
FV= 157,846.7209
20,000,000=1,250,000*(1+r)^20
20,000,000/1,250,000=(1+r)^20
16=(1+r)^20
(16)^1=(1+r)^20
divide indexes on both sides by 20
(16)^(1/20)=1+r
r=(16)^(1/20)-1
r=14.8698%
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Given mn, find the value of x.
20⁰
Xº
Answer:
x =20°
because m and n is parallel
Simple Math Range and domain help
[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]
Here we go ~
Domain : All possible values of x for which a function is defined.
According to given graph, the curve extends to infinity on both sides on x - axis, so it is defined for all real values.
i.e : Domain : All real
Range : All possible values of y for x, as per the given function
As per the given graph, the graph extends to infinity on lower end, but has maximum value of 0.
i.e Range : [tex]\boxed{ \sf y < 0} [/tex]
Does the equation 2(3X + 8) equal 2X +16+ 4X have a solution
infinitely many solutions
Example
Simplify left side using distributive property to 6x+16. Combine like terms (2x and 4x) on the right side to 6x+16. Both sides simplify to the same expression (left side = right side) so there are infinitely many solutions. You can plug in any real number for x and the left side will always equal the right side.
Saul wants to bike 48 kilometers to taste some really good mangoes.
1. Write an equation that represents how many hours (t) the48km start text, k, m, end text trip will take if Saul bikes at a constant rate of r kilometers per hour.
2. How many hours will the48km48, end text trip take if Saul bikes at a constant rate of 20 kilometers per hour?
hours
Answer:
[tex]\textsf{1)} \quad t=\dfrac{48}{r}[/tex]
[tex]\textsf{2)} \quad \sf 2.4\:hours=2\:hours\:24\:minutes[/tex]
Step-by-step explanation:
Part 1[tex]\boxed{\sf Time=\dfrac{Distance}{Speed}}[/tex]
Given:
Time = t hoursDistance = 48 kmSpeed = r km/hSubstitute the given values into the formula:
[tex]\implies t=\dfrac{48}{r}[/tex]
Part 2Substitute r = 20 km/h into the derived equation from part 1 and solve for t:
[tex]\implies t=\dfrac{48}{20}=2.4\: \sf hours[/tex]
Note: 2.4 hours = 2 hours 24 minutes
Answer: 2.4 hours
Step-by-step explanation: t=time
r=rate so question one is equal to t=48/r
then question 2 is 48/20=2.4
determine the inverse of the function
Answer:
[tex]f^{-1}(x)= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right), \quad \textsf{for}\:\{x:0 < x < 1\}[/tex]
Step-by-step explanation:
Given function:
[tex]f(x)=\dfrac{e^x}{\sqrt{e^{2x}+1}}[/tex]
The domain of the given function is unrestricted: {x : x ∈ R}
The range of the given function is restricted: {f(x) : 0 < f(x) < 1}
To find the inverse of a function, swap x and y:
[tex]\implies x=\dfrac{e^y}{\sqrt{e^{2y}+1}}[/tex]
Rearrange the equation to make y the subject:
[tex]\implies x\sqrt{e^{2y}+1}=e^y[/tex]
[tex]\implies x^2(e^{2y}+1)=e^{2y}[/tex]
[tex]\implies x^2e^{2y}+x^2=e^{2y}[/tex]
[tex]\implies x^2e^{2y}-e^{2y}=-x^2[/tex]
[tex]\implies e^{2y}(x^2-1)=-x^2[/tex]
[tex]\implies e^{2y}=-\dfrac{x^2}{x^2-1}[/tex]
[tex]\implies \ln e^{2y}= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]
[tex]\implies 2y \ln e= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]
[tex]\implies 2y(1)= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]
[tex]\implies 2y= \ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]
[tex]\implies y= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]
Replace y with f⁻¹(x):
[tex]\implies f^{-1}(x)= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right)[/tex]
The domain of the inverse of a function is the same as the range of the original function. Therefore, the domain of the inverse function is restricted to {x : 0 < x < 1}.
Therefore, the inverse of the given function is:
[tex]f^{-1}(x)= \dfrac{1}{2}\ln \left(-\dfrac{x^2}{x^2-1}\right), \quad \textsf{for}\:\{x:0 < x < 1\}[/tex]
What is the inverse of f(x)=(3x)2 for x≥0
The inverse of the function f(x) = (3x)^2 is f-1(x) = 1/3√x
How to determine the inverse of the function?The function is given as:
f(x) = (3x)^2
Remove the bracket in the above equation
So, we have:
f(x) = 9x^2
Express f(x) as y
So, we have
y = 9x^2
Swap the positions of x and y
So, we have
x = 9y^2
Make y the subject of the formula
y^2 = x/9
Take the square root of both sides
y = 1/3√x
Express as an inverse function
f-1(x) = 1/3√x
Hence, the inverse of the function f(x) = (3x)^2 is f-1(x) = 1/3√x
Read more about inverse functions at:
brainly.com/question/11735394
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What is the answer
6000 +300+20+5
Answer:
6325
Step-by-step explanation:
[tex]6000 \\ \: \: 300 \\ \: \: \: \: 20 \\ \: \: \: + 5[/tex]
___________
6325
Answer:
Your answer would be [tex]6325[/tex]
Step-by-step explanation:
[tex]=6325[/tex]
[tex]6000 +300+20+5[/tex]
[tex]=6325[/tex]
hopefully this helps! TwT
Solve 2(x - 3) ≥ -3 (-3 + x)
Answer:
x ≥ 3
Step-by-step explanation:
2(x - 3) ≥ -3(-3 + x)
2x - 6 ≥ 9 - 3x
5x - 6 ≥ 9
5x ≥ 15
x ≥ 3
Answer:
x ≥ 3
Step-by-step explanation:
write, calculate, divide both sides
……………………………………………………………
same here 2383882828x7767=?
you are stinky . i hope you know tht
Step-by-step explanation:
wot. that's not even a qn bro
Step-by-step explanation:
the difference between the numbers are 5
so this is an arthimetic sequence with D=5
5n-7 is the formula