Given mn, find the value of x.
20⁰
Xº
Answer:
x =20°
because m and n is parallel
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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Please help! Thank you
Write the function whose graph is the graph of y=x³ +3, but is reflected about the y-axis.
Answer:
Reflected function is y = - x³ + 3
Step-by-step explanation:
When a function f(x) is reflected about the y-axis, the reflected function becomes g(x) = f(-x)
f(x) = x³ + 3
f(-x) = (-x)³ + 3 =-x³ + 3
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]
An object is moving at a speed of 360 kilometers per hour. Express this speed in miles per day. Round your answer to the nearest whole number.
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.
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:
Give the slope of a line that is perpendicular to the line 2x+2y=1 .
-1
Answer:
- 1
Step-by-step explanation:
2x + 2y = 1
=> 2y = - 2x + 1
=> y = - x + 1
=> Slope = - 1
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
Let f(x) = x3 + 10, g(x) = x2 - 8, and h(x) = 5x + 4. Find the rule for the function.
The rule f ( x ) - g ( x ) is given as x^3 - x^2 + 18
Given data
f(x) = x3 + 10, g(x) = x2 - 8, and h(x) = 5x + 4
How to find f(x) - g(x)f(x) - g(x) = ( x3 + 10 ) - ( x2 - 8 )
= ( x3 + 10 ) - ( x2 - 8 )
opening the parenthesis gives
= x3 + 10 - x2 + 8
collecting like terms gives
= x3 - x2 + 18
Hence we can say that f(x) - g(x) = x3 - x2 + 18
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complete question
The complete question is attached
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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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
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
[tex] \: \: \: \: \: \: \: \: \: \: \: [/tex]
(A) (m,n) = g(n,m) for all positive integers m,n,
(B) (m,n + 1) = g(m + 1,n) for all positive integers m,n,
(D) (2m,2n) = (g(m,n))2 for all positive integers m,n
Explanation[tex] \rm f(m,n,p) = \sum \limits_{i = 0}^{p} {}^{m} C_{i} \: \: {}^{n + i} C_{p} \: \: {}^{p + n} C_{p - i}[/tex]
[tex]\rm {}^{m} C_{i} \: \: {}^{n + i} C_{p} \: \: {}^{p + n} C_{p - i}[/tex]
[tex]\rm {}^{m} C_{i} \dfrac{(n + i)!}{p!(n - p + i)!} \times \dfrac{(n + p)!}{(p - i)!(n + i)!} [/tex]
[tex]\rm {}^{m} C_{i} \times \dfrac{(n + p)!}{p!} \times \dfrac{1!}{(n -p + i)!(p - i)!} [/tex]
[tex]\rm {}^{m} C_{i} \times \dfrac{(n + p)!}{p!} \times \dfrac{1!}{(n -p + i)!(p - i)!} [/tex]
[tex]\rm {}^{m} C_{i} \times \dfrac{(n + p)!}{p!n!} \times \dfrac{n!}{(n -p + i)!(p - i)!} [/tex]
[tex]\rm {}^{m} C_{i} \: \: {}^{n + p} C_{p} \: \: {}^{n} C_{p - i} \: \: \{{}^{m} C_{i} \: \: {}^{n } C_{p - i} = {}^{m + n} C_{p } \}[/tex]
[tex]\rm {}^{m} C_{i} \: \: {}^{n + p} C_{p} \: \: {}^{n} C_{p - i} \: \: \{{}^{m} C_{i} \: \: {}^{n } C_{p - i} = {}^{m + n} C_{p } \}[/tex]
[tex] \rm f(m,n,p) = {}^{n + p} C_{p}{}^{m + n} C_{p}[/tex]
[tex] \rm \dfrac{f(m,n,p)}{{}^{n + p} C_{p}} = {}^{m + n} C_{p}[/tex]
Now,
[tex] \rm g(m,n) = \sum \limits_{p = 0}^{m + n} \dfrac{f(m,n,p)}{{}^{n + p} C_{p}} [/tex]
[tex] \rm g(m,n) = \sum \limits_{p = 0}^{m + n} {}^{m + n} C_{p}[/tex]
[tex] \rm g (m,n) = {2}^{m + n} [/tex]
[tex] \rm(A) \: g(m,n) = q(n,m)[/tex]
[tex] \rm(B) \: g(m,n + 1) = {2}^{m + n + 1} [/tex]
[tex] \rm g(m + n,n ) = {2}^{m + 1 + n} [/tex]
[tex] \rm(D) \: g(2m,2n) = {2}^{2m + 2n} [/tex]
[tex] = \rm( {2}^{m + n} {)}^{2} [/tex]
[tex] = \rm(g(m,n)) {}^{2} [/tex]
Which expression is equivalent to the quotient of x and 8 times the difference of 19 and x?
Answer:
(x/8)(19 - x)
Step-by-step explanation:
The quotient is the answer to a division problem.
The difference is the answer to a subtraction problem.
"Times" indicates multiplication, which you could look at it as a link between the two functions, or quantities (them being the constants and variables sectioned off into parenthesis).
Amount of Amy's pocket money each day is 3 times as much as the pocket money of Flora's. And the amount of Paul's pocket money is $7 more than Amy's. If they have $980 of pocket money each week, how much pocket money does Paul have each day?
Answer:-
Paul has $64 pocket money each day.
Explanantion:-
Let amy ,flora and Paul 's pocket money be x,y and z respectively.
So,
By the question :
x = 3y
So, y = x/3 - eqn i
z = 7 + x - eqn ii
7(x + y + z) = 980
x + y + z = 140
Now,
x + y+ z = 140
x + x/3 + 7 + x = 140
or, 2x + x/3 = 133
or, 6x + x = 399
so, x = 399/7
so, x = $57
Now,
z = 7 + x
= 7 + $57
= $64
24. The relative frequency of getting a HEAD when tossing a coin is ⅜. How many HEADS would you expect if the coin is tossed 400 times?
Kelly wants to find out if the townspeople would be in favor of creating a local dog park. she pulls an equal number of people from several age groups and from different areas of town. Is this study likely to yield valid results? Explain.
Answer: Yes
Step-by-step explanation: Kelly's results are likely to be correct mainly because she pulled votes from several different age groups. The fact that she asked several age groups whether or not they would be in favor of creating a local dog park shows that her results are likely to yield valid results.
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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Determine which of the numbers below is a possible order of magnitude estimate of the
quantity described.
The weight of a grain of rice.
Select one:
O a. 0.000001 ounces
O b. 0.0006 ounces
OC.0.006 ounces
O d. 0.01 ounces
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.
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
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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Evaluate |r| when r=−29.
Reason: The absolute value erases the negative sign.
Saying |-29| = 29 means that the number -29 is 29 units from 0 on the number line.
Answer: 29
Step-by-step explanation: The absolute value of -29 is 29. An absolute value is a number's distance from 0. -29 + 29 = 0, therefore the absolute value is 29.
Helppppppppp pleaseeeeee
Answer:
The answer is yes. Since there is no x^2
i really need you guys help.
Answer:
y = - 8x + 3
Step-by-step explanation:
the equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
y = - 8x - 2 ← is in slope- intercept form
with slope m = - 8
• Parallel lines have equal slopes , then
y = - 8x + c ← is the partial equation of the parallel line
to find c substitute (- 0.5, 7 ) into the partial equation
7 = 4 + c ⇒ c = 7 - 4 = 3
y = - 8x + 3 ← equation of parallel line
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]
Suppose Albers Elementary School has 76 teachers and Bothel Elementary School has 24 teachers. If 3 teachers teach at both Albers and Bothel, compute the total number of teachers employed at Albers and Bothel combined.
The number of teachers that work in both schools is 97.
What is an addition?The addition is a way of combining things and counting them together as one large group.
We have,
Number of teachers in Alberts Elementary school = 76
Number of teachers in Bothel Elementary school = 24
Number of teachers that work in both schools = 3
The number of teachers that work in both schools:
= 76 + 24 - 3
We are subtracting 3 from the total because 3 teachers that work in both schools have been added twice.
= 100 - 3
= 97
Thus the number of teachers that work in both schools is 97.
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Determine whether the relation is a function. Explain. { ( 2 , 5 ) , ( 4 , − 2 ) , ( 3 , 3 ) , ( 5 , 4 ) , ( − 2 , 5 ) } Multiple choice question. cross out A) Yes; for each element of the domain, there is only one element of the range. cross out B) Yes; for each element of the range, there is only one element of the domain. cross out C) No; the elements –2, 3, 4, and 5 are in both the domain and the range. cross out D) No; the element 5 occurs twice in the range.
Yes; for each element of the domain, there is only one element of the range.
Function and relationsFor a relation in coordinate form to be a function, there is must nor be repetition in the value of its coordinate.
Given the following relation as shown { ( 2 , 5 ) , ( 4 , − 2 ) , ( 3 , 3 ) , ( 5 , 4 ) , ( − 2 , 5 ) }, since there is no repetition in its domain values, hence the relation is a function.
We can conclude therefore that Yes; for each element of the domain, there is only one element of the range.
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