a. To find the inverse function of f(x), we need to interchange the roles of x and y and solve for y. So, we have:
y = 1x - 3
x = 1y - 3
x + 3 = y
Therefore, the inverse function of f(x) is f^-1(x) = x + 3.
The domain of f^-1(x) is the range of f(x). Since f(x) = 1x - 3 is a linear function, its domain is all real numbers. Therefore, the range of f(x) is also all real numbers. In interval notation, we can write this as (-inf, inf).
The range of f^-1(x) is the domain of f(x). As we determined in part b, the domain of f(x) is all real numbers. Therefore, the range of f^-1(x) is also all real numbers. In interval notation, we can write this as (-inf, inf).
Hi! I'd be happy to help you with your question.
a. To find the inverse function of f(x) = 1x - 3, you can follow these steps:
1. Replace f(x) with y: y = 1x - 3
2. Swap x and y: x = 1y - 3
3. Solve for y: y = x + 3
So, the inverse function f^(-1)(x) = x + 3.
The domain of f^(-1) refers to the set of all possible x-values. Since the inverse function is a linear function with no restrictions, the domain of f^(-1) is all real numbers. In interval notation, this is written as (-∞, ∞).
c. The range of f^(-1) refers to the set of all possible y-values (output). Again, since it's a linear function with no restrictions, the range of f^(-1) is also all real numbers. In interval notation, this is written as (-∞, ∞).
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anyone know the dba questions for unit 8 algebra 1 honors
a) The distance the ball rebounds on the fifth bounce is approximately 7.59 ft.
b) The total distance the ball has traveled after the fifth bounce is approximately 52.61 ft.
What is the explanation for the above response?Let's denote the height of the ball after its nth bounce by h_n. Then we can express the relationship between the height of the ball after each bounce in terms of a recursive formula:
h_0 = 16 (initial height)
h_1 = (3/4) * h_0 (rebound distance after the first fall)
h_2 = (3/4) * h_1 (rebound distance after the second fall)
h_3 = (3/4) * h_2 (rebound distance after the third fall)
h_4 = (3/4) * h_3 (rebound distance after the fourth fall)
h_5 = (3/4) * h_4 (rebound distance after the fifth fall)
a) To find the distance the ball rebounds on the fifth bounce, we need to calculate h_5:
h_5 = (3/4) * h_4
= (3/4) * ((3/4) * ((3/4) * ((3/4) * 16)))
= (3/4)^5 * 16
= 7.59375 ft
Therefore, the ball rebounds approximately 7.59 ft on the fifth bounce.
b) To find the total distance the ball has traveled after the fifth bounce, we need to add up all of the distances traveled during the falls and rebounds:
total distance = distance of first fall + rebound distance after first fall + rebound distance after second fall + rebound distance after third fall + rebound distance after fourth fall + rebound distance after fifth fall
total distance = 16 + (3/4) * 16 + (3/4)^2 * 16 + (3/4)^3 * 16 + (3/4)^4 * 16 + (3/4)^5 * 16
total distance = 16 + 12 + 9 + 6.75 + 5.0625 + 3.7969
total distance = 52.6094 ft
Therefore, the ball travels approximately 52.61 ft after the fifth bounce.
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Full Question:
Although part of your question is missing, you might be referring to this full question:
Be sure to show and explain all work using mathematical formulas and terminology. A bouncy ball is dropped from a height of 16ft and always rebounds ¼ of the distance of the previous fall.
a) What distance does it rebound the 5th time?
b) What is the total distance the ball has travelled after this time?
Explain the statement 3-5 sentences :
correspondence is a relation of connection
you may give examples to explain the ideas
Correspondence is a relation of connection as it establishes a link between two sets of elements, often by relating each element in one set to a specific element in the other set.
Correspondence refers to the exchange of communication or information between two or more parties. It is a relation of connection because it involves establishing a link or connection between the sender and the receiver of the message.
For example, when two people exchange letters or emails, they establish a correspondence that connects them and allows them to communicate. Similarly, in business, correspondence can refer to the exchange of official documents such as letters, memos, and reports, which establish a connection between different departments or organizations. Overall, correspondence is an important aspect of communication that helps to establish and maintain relationships between individuals and groups.
For example, in mathematics, a correspondence can be seen when matching the elements of one set to another, such as associating students with their grades. In this case, the connection is created by linking each student to their respective grade, illustrating the concept of correspondence.
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Pls help me with this! I need to finish today
Answer:
T=64
Step-by-step explanation:
Multiply both sides by 4
t/4=16
t/4×4=16×4 Cancel out the 4
t=64
Arnav was 1.5 \text{ m}1.5 m1, point, 5, start text, space, m, end text tall. In the last couple of years, his height has increased by 20\%20%20, percent
Over the last couple of years, Arnav's height has increased by 20% so his current height is 1.8 meters.
Arnav's height initially was 1.5 meters. Over the last couple of years, his height increased by 20%. To find the new height, we can use the formula: new height = initial height × (1 + percentage increase).
In this case, the initial height is 1.5 meters and the percentage increase is 20%, which can be expressed as a decimal (0.2). Using the formula, we can calculate Arnav's new height as follows:
New height = 1.5 meters × (1 + 0.2) = 1.5 meters × 1.2 = 1.8 meters.
After the 20% increase in height over the last couple of years, Arnav's current height is 1.8 meters.
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Susan is a college student with two part-time jobs. She earns $10 per hour tutoring
elementary students in math. She earns $15 per hour cleaning in the library. Her goal is
to earn at least $240 per week, but because of college, she does not work more than
20 hours each week.
Which combinations allow Susan to work no more than 20 hours in one week and earn
at least $2402
Select the three correct combinations.
The required inequalities are h + l ≤ 20, 10h + 15l ≥ 240 and 20h + 25l ≥ 440
Given, for tutoring elementary students in math Susan earns $10 per hour. She earns $15 per hour for cleaning in the library.
Let h be the number of hours Susan works in one week tutoring elementary students.
Let l be the number of hours Susan works in one week cleaning the library.
Given that each week Susan cannot work more than 20 hours.
So, h + l ≤ 20 ....(1)
Susan's total earnings must be at least $240 per week.
10h + 15l ≥ 240 ...(2)
Multiplying equation (1) by 10
10h + 10l ≤ 200 ...(3)
Adding equations (2) and (3)
20h + 25l ≥ 440
Thus, the three required inequalities are h + l ≤ 20, 10h + 15l ≥ 240 and 20h + 25l ≥ 440
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To answer questions 4 - 6.
A new school opened with 225 students in 2021 and plans to increase by 13. 3% per year
until they reach full capacity.
Is this situation exponential growth or decay?
4.
5.
Write an equation that models the population of the school, P, after x years since
the opening of the school in 2021
The situation provided is exponential growth and the equation is
p = 225 * 1.133ˣ
How to determine the situationThis scenario represents a significant improvement as the number of students increases each year.
The basic approach to incremental growth is:
[tex]P = P_{0} * (1 + r)^t[/tex]
where:
P. = initial population
r = increase in decimal form
t = time in years
Here
P₀ = 225 (initial population in 2021).
r = 13.3% = 0.133 (growth rate as decimal) .
t = x (time in years from the opening of the school in 2021)
Substituting these values in the formula we get:
[tex]p = 225 * (1 + 0.133)^x[/tex]
Simplifying further, we get:
p = 225 * 1.133ˣ
This is the equation that predicts school population x years after the school opens in 2021
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A car left Town A for Town b. Another car left Town B for Town A at the same time. The ratio of the speeds of the two cars was 6:5 initially. After the two cars passed each other, Car A's speed was reduced by 1/6 and car B's speed was reduced by 25%. When car A arrived at Town B, Car B was still 54 km away from Town A. Find the distance between Town A and Town B. Please I need the answer quickly :]
The distance between Town A and Town B is 550 km.
Let's denote the distance between Town A and Town B as D.
When the two cars first passed each other, let's assume that car A traveled a distance of x km and car B traveled a distance of D - x km.
Let's also denote the initial speeds of car A and car B as 6s and 5s, respectively, where s is some constant representing the speed of the slower car.
The time it took for the two cars to pass each other can be calculated using the formula:
time = distance / speed
For car A, the time it took to travel x km was:
x / (6s)
For car B, the time it took to travel D - x km was:
(D - x) / (5s)
Since the two cars traveled the same amount of time until they passed each other, we can set these two expressions equal to each other:
x / (6s) = (D - x) / (5s)
Solving for x, we get:
x = 6Ds / (11s)
After the speeds of both cars were reduced, car A's speed was (5/6) * 6s = 5s, and car B's speed was (3/4) * 5s = (15/4)s.
Let's denote the time it took for car A to travel the remaining distance from x to D as t.
Then, the time it took for car B to travel a distance of (D - x - 54) km is also t.
Using the new speeds, we can write the equation:
[tex](D - x - 54) = (15/4)s * t[/tex]
Solving for t, we get:
[tex]t = (4/15)(D - x - 54) / s[/tex]
The distance car A traveled after the two cars passed each other is:
D - x = D - 6Ds / (11s) = (5/11)D
The time it took for car A to travel this distance is:
[tex]t + x / (6s) = (4/15)(D - x - 54) / s + 6Ds / (66s)[/tex]
Setting these two expressions equal to each other and solving for D, we get:
D = 550 km
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Rewrite each equation without absolute value for the given conditions. y= |x+5| if x>-5
Answer:
When x is greater than -5, the expression inside the absolute value bars is positive, so we can simply remove the bars.
So the equation y = |x+5| can be rewritten as:
y = x+5 (when x > -5)
What is the probability of rolling an even number and then an odd number when rollling two number cubes what is the number of desired outcomes
The probability of rolling an even number and then an odd number is 1/4.
Calculating the probability valuesThe probability of rolling an even number on a fair number cube is 1/2, since there are three even numbers (2, 4, 6) and six possible outcomes (1, 2, 3, 4, 5, 6).
Similarly, the probability of rolling an odd number is also 1/2.
To find the probability of rolling an even number and then an odd number, we need to multiply the probabilities of each event. So:
P(even and odd) = P(even) × P(odd)
P(even and odd) = (1/2) × (1/2)
P(even and odd) = 1/4
So the probability of rolling an even number and then an odd number is 1/4.
The number of desired outcomes for rolling an even number and then an odd number is 9
Since there are three even numbers and three odd numbers, and therefore 3 × 3 = 9 possible outcomes.
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What are the coordinates of the vertices of d(3, a)(△abc) for a(0, 4), b(−2, 5), and
c(3, 7)? does the perimeter increase or decrease?
The coordinates of the triangle after the dilation are given as follows:
a(0, 12), b(-6, 15) and c(9, 21).
The perimeter of the triangle increases, as the side lengths are multiplied by 3, hence the perimeter is also multiplied by 3.
What is a dilation?A dilation can be defined as a transformation that multiplies the distance between every point in an object and a fixed point, called the center of dilation, by a constant factor called the scale factor.
The scale factor for this problem is given as follows:
k = 3.
The scale factor is greater than 1, meaning that the figure is an enlargement, and thus the perimeter increases.
The original vertices of the triangle are given as follows:
a(0, 4), b(−2, 5), and c(3, 7)
Hence the vertices of the dilated triangle are given as follows:
a(0, 12), b(-6, 15) and c(9, 21).
(each coordinate of each vertex is multiplied by the scale factor of 3).
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Astronaut Harry Skyes has a mass of approximately 85. 0 kg. What is his weight on Mercury?
Mercury's gravity = 3. 70 m/s^2
The weight of Harry Skyes on Mercury is 32.8 kg, under the condition that Mercury's gravity = 3. 70 m/s².
The weight of astronaut Harry Skyes on Mercury can be evaluated using the formula:
Weight on Mercury = (Weight on Earth / 9.81 m/s²) × 3.7 m/s²
Given that Harry Skyes has a mass of approximately 85.0 kg, his weight on Mercury would be:
Weight on Mercury = (85.0 kg / 9.81 m/s²) × 3.7 m/s²
Weight on Mercury = 32.8 kg
Gravity affects weight severely and causes its change . Objects have mass, which is specified as how much matter an object contains. Weight is known as the pull of gravity on mass. The relation between weight and gravitational pull is such that, when on another celestial body, the difference in gravity would alter a person’s weight.
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Help again with math (I'm on 37/64 and I'm about to cry)
Answer:
1,215,000 cubic centimeters
Step-by-step explanation:
1. Find the volume of the cylinder
v = π r (squared) x h
v = 3.14 x 50 (squared) x 100
v = 3.14 x 2,500 x 100
v = 3.14 x 250,00
v = 785,000 cubic centimeters
2. Find the volume of the rectangular prism
v = l x w x h
v = 100 x 200 x 100
v = 2,000,000 cubic centimeters
3. Subtract
2,000,000 - 785,000 = 1,215,000 cubic centimeters
The diagonal of a table top is 40 inches and the width is 21 inches. What is the area of the table? Round to the nearest inch.
The area of the table is approximately 651 square inches.
What is Area ?
Area is a measure of the size of a two-dimensional shape or surface, such as a rectangle, circle, or triangle. It is expressed in square units, such as square inches, square feet, or square meters.
Let's use the Pythagorean theorem to find the length of the table top:
Substituting the given values, we get:
40*40 = [tex]length^{2}[/tex] + 21*21
Simplifying and solving for length, we get:
[tex]length^{2}[/tex]= 1600 - 441
[tex]length^{2}[/tex] = 961
length = 31 inches (rounded to the nearest inch)
Now that we know the length and width of the table, we can find the area by multiplying them together:
area = length x width
area = 31 x 21
area = 651 square inches (rounded to the nearest inch)
Therefore, the area of the table is approximately 651 square inches.
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It is now time to complete the Independence and Exclusiveness assignment. Independence and Exclusiveness are two topics which are important to probability and often confused. Discuss the difference between two events being independent and two events being mutually exclusive. Use examples to demonstrate the difference. Remember to explain as if you are talking to someone who knows nothing about the topic. Please no gibberish if correct I will be so grateful
Two events are independent if the occurrence of one event does not affect the occurrence of the other event. In other words, the probability of one event happening is not affected by whether or not the other event happens.
A simple example would be flipping a coin and rolling a die. The outcome of the coin flip does not affect the outcome of the die roll, so these events are independent.
On the other hand, two events are mutually exclusive if they cannot happen at the same time. If one event happens, the other event cannot happen. For instance, when rolling a die, the events of getting a 1 or a 2 are mutually exclusive because it is impossible to roll both numbers at the same time.
To summarize, two events are independent if the probability of one event happening is not affected by the occurrence of the other event, while two events are mutually exclusive if they cannot happen at the same time.
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HURRY WHO IS RIGHT!!!
Answer:
Step-by-step explanation:
cat
Circle 1 is centered at (-3,5) and has a radius of 10 units circle 2 is centered at (7,5) and has a radius of 4 units. What transformations can be applied to circle 1 to prove that the circles are similar?
This will result in Circle 1 having the same center and radius as Circle 2, thus proving that the circles are similar.
To prove that Circle 1 and Circle 2 are similar, we can apply the following transformations to Circle 1:
1. Translation: Translate Circle 1 by moving its center from (-3, 5) to (7, 5). This is a horizontal translation of 10 units to the right.
2. Dilation: Dilate Circle 1 with a scale factor of 0.4, which will reduce its radius from 10 units to 4 units (the same as Circle 2).
These transformations will result in Circle 1 having the same center and radius as Circle 2, thus proving that the circles are similar.
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2. If Q(-5, 1) is the midpoint of PR and R is located
at (-2.-4), what are the coordinates of P?
[tex]~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ P(\stackrel{x_1}{x}~,~\stackrel{y_1}{y})\qquad R(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-4}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -2 +x}{2}~~~ ,~~~ \cfrac{ -4 +y}{2} \right) ~~ = ~~\stackrel{\textit{\LARGE Q} }{(-5~~,~~1)} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{ -2 +x }{2}=-5\implies -2+x=-10\implies \boxed{x=-8} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{ -4 +y }{2}=1\implies -4+y=2\implies \boxed{y=6}[/tex]
Answer:
The answer is (-8,6)
Dmitri practices his domra for 98 min during
the school week. this is 70% of the time he
must practice his instrument in one week.
The total or actual time he needs to practice is 140 min whereas he practiced for 98 min during the school week.
We need to find the total time he must practice for a week. To find the total time we assume that the total time is x min.
Given Data:
Dmitri practices time during the school week = 98 min
Dmitri practices amount of time = 70% of his total time
Total time = x
Then the equation is given as
70% × (x) = 98
0.70 × (x) = 98
x = 98 / 0.70
x = 140
Therefore, The total time of the practices is 140 min
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Use the rational expression below to match the variables on the left with their excluded value(s) on the right.
The value of b = 4
How to solveThe function becomes undefined when the denominator goes to 0
3x² - 48 = 0
3x² = 48
x² = 16
x = +/ 4
From the given choices, it's x = 4
Rational expressions that utilize ratios of polynomial expressions are referred to as rational expressions. These can be written in the format p(x)/q(x), where both p(x) and q(x) are polynomials with the constraint that q(x) cannot equal zero.
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Caleb has
coins (nickels, dimes, and quarters) in a jar, totaling. He has three more nickels than dimes. How many quarters does Caleb have?
Caleb has 30 quarters.
What is arithmetic?
Mathematical arithmetic is the study of the properties of the standard operations on numbers, such as addition, subtraction, multiplication, division, exponentiation, and root extraction.
Here, we have
Given: Caleb has 51 coins (nickels, dimes, and quarters) in a jar, totaling $9. He has three more nickels than dimes.
We have to find out how many quarters Caleb has.
Let x be nickel,
y be dimes and
z be quarters
x + y + z = 51.....(1)
1 quartes = 25 cents
1 dimes = 10 cents
1 nickel = 5 cents
Now, the total dollar is $9,
5x/100 + 10y/100 + 25z/100 = 9
5x + 10y + 25z = 900
x + 2y + 5z = 180....(2)
and
y + 3 = x...(3)
Solving equation(1) and (2), we get
From (1)
x + x-3 + z = 51
2x + z = 54....(4)
From (2)
x + 2(x -3) + 5z = 180
3x + 5z = 186...(5)
Now, by solving equations (4) and (5), we get
x = 12
z = 30
Now,
y + 3 = x
y + 3 = 12
y = 9
Hence, Caleb has 30 quarters.
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Suppose a particle moves along a continuous function such that its position is given by f(t)=1/7 t^3-4t-12 where f is the position at time t, then determines the value of r such that f(r)=0.
When we look at [tex]f(t)=1/7 t^3-4t-12[/tex], this is a cubic equation, and solving it analytically is not straightforward.
How to solveTo find the value of r such that f(r) = 0, we need to solve the equation:
[tex]1/7 r^3 - 4r - 12 = 0[/tex]
This is a cubic equation, and solving it analytically is not straightforward.
Yet, it is possible to obtain the value of r that meets the equation using numerical schemes such as Newton-Raphson or bisection. Additionally, one can take advantage of calculation tools and graphical software to calculate an estimation of r.
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Find the most general antiderivative of the function. (Check your answer by differentiation. Use C for the constant of the antiderivative.)
g(v) = 5 cos (v) - 8/√(1-v^2)
g(v) = ____
The most general antiderivative of the function g(v) = 5 cos(v) - 8/√(1-v^2) is 5 sin(v) + 8 arcsin(v) + C, where C is the constant of the antiderivative.
To find the antiderivative of the given function g(v), we can use the basic antiderivative rules. The antiderivative of 5 cos(v) is 5 sin(v), as the derivative of sin(v) is cos(v) and we only need to reverse the process.
Similarly, the antiderivative of -8/√(1-v^2) can be found using the inverse trigonometric function arcsin(v), as its derivative is -1/√(1-v^2). However, we need to include a constant of integration, denoted by C, as the antiderivative is not unique.
So the most general antiderivative of g(v) is 5 sin(v) + 8 arcsin(v) + C, where C represents the constant of the antiderivative. To check the correctness of the answer, we can differentiate it and verify if it gives us the original function g(v) as the result.
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8. [-/14 Points] DETAILS SCALCET9 7.7.027. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Find the approximations To Mn, and S, for n = 6 and 12. Then compute the corresponding errors E.Em, and Es. (Round your answers to six decimal places. You may wish to use the sum command on a computer algebra system.) dx n T M, S, 6 12 n Ет EM ES 6 12 What observations can you make? In particular, what happens to the errors when n is doubled? As n is doubled, E, and Em are decreased by a factor of about , and Es is decreased by a factor of about Need Help? Read It Watch It
To approximate the values for Mₙ and S for n = 6 and 12, we'll use the trapezoidal rule (T), midpoint rule (M), and Simpson's rule (S). After calculating these approximations, we'll compute the errors Eₜ, Eₘ, and Eₛ.
For n = 6:
T₆ = (Approximation using trapezoidal rule)
M₆ = (Approximation using midpoint rule)
S₆ = (Approximation using Simpson's rule)
For n = 12:
T₁₂ = (Approximation using trapezoidal rule)
M₁₂ = (Approximation using midpoint rule)
S₁₂ = (Approximation using Simpson's rule)
Errors for n = 6:
Eₜ₆ = |Actual value - T₆|
Eₘ₆ = |Actual value - M₆|
Eₛ₆ = |Actual value - S₆|
Errors for n = 12:
Eₜ₁₂ = |Actual value - T₁₂|
Eₘ₁₂ = |Actual value - M₁₂|
Eₛ₁₂ = |Actual value - S₁₂|
As n is doubled (from 6 to 12), observe the changes in the errors:
- Eₜ and Eₘ typically decrease by a factor of about 4 (since error is proportional to 1/n² for these methods)
- Eₛ typically decreases by a factor of about 16 (since error is proportional to 1/n⁴ for Simpson's rule)
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Identify if the proportion is true or false12:4=9:3
Verify that MQ:QN = 2:3 by finding the lengths of MQ and QN
The length of MQ and QN is 10 and 15 respectively and verify that MQ: QN = 2:3
The coordinate of M = (-12,-5)
The coordinate of N = (8,10)
n = 2 , m = 3
By using the section formula coordinate of Q =( [tex]\frac{mx_{1} + nx_{2} }{m+n }[/tex] , [tex]\frac{my_{1} + ny_{2} }{m+n}[/tex])
Coordinate of Q = ([tex]\frac{(-12)3 + 8(2)}{3+2}[/tex] , [tex]\frac{10(2) + 3(-5)}{2+3}[/tex])
Coordinate of Q = ( -4, 1)
Now using the distance formula
MQ = [tex]\sqrt{ (x_{2}- x_{1} )^{2} +(y_{2} -y_{1} )^{2}[/tex]
MQ = [tex]\sqrt{(-4+12)^{2}+(1+5)^{2} }[/tex]
MQ = √100
MQ = 10
Similarly,
QN = [tex]\sqrt{(8+4)^{2}+(10-1)^{2} }[/tex]
QN = [tex]\sqrt{225}[/tex]
QN = 15
MQ:QN = 10:15
MA :QN = 2:3
Hence it is verified that MQ: QN = 2:3
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One similar figure has an area that is nine times the area of another. The larger figure must have dimensions that are
times the dimensions of the smaller figure.
three
eighteen
eighty-one
nine
Since the area of a similar figure is proportional to the square of its linear dimensions, if one similar figure has an area that is nine times the area of another, the larger figure must have dimensions that are three times the dimensions of the smaller figure.
This is because the area is the square of the linear dimensions. So, if we increase the linear dimensions by a factor of 3, the area increases by a factor of 3^2 = 9.
Therefore, the answer is 3.
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A man buys a plot of agricultural land for rs. 300000 he sells 1/3rd at a loss of 20% and 2/5ths at a gain of 25% at what price must he sell the remaining land so as to make an overall profit of 10%
1) If you deposited $10,000 into a bank savings account on your 18th birthday. Said account yielded 3% compounded annually, how much money would be in your account on your 58th birthday?
2)What would your answer be if the interest was compounded monthly versus
annually?
1- On the 58th birthday, the account would have $24,209.98, 2- If the interest is compounded monthly, then on the 58th birthday, the account would have $26,322.47.
1- The formula for calculating the compound interest is given by A = P(1 + r/n)(nt), where A is the final amount, P is the principal amount, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the time in years. Here, P = $10,000, r = 0.03, n = 1, t = 40 years (58 - 18).
substituting the values in the formula, we get A = $10,000(1 + 0.03/1)1*40) = $24,209.98.
2) In this case, n = 12 (monthly compounding), and t = 12*40 (total number of months in 40 years). So, the formula for calculating the compound interest becomes A = P(1 + r/n)(nt) = $10,000(1 + 0.03/12)(12*40) = $26,322.47.
Since the interest is compounded more frequently, the amount at the end of 40 years is higher than when the interest is compounded annually.
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A 10-ft ladder is leaning against a house when its base starts to slide away. By the time the base is 6 ft from the house, the base is moving away at the rate of 24 ft/sec.
a. What is the rate of change of the height of the top of the ladder?
b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then?
c. At what rate is the angle between the ladder and the ground changing then?
The rate of change of the height of the top of the ladder is -144/h ft/sec when the base of the ladder is 6 ft from the house.
The area of the triangle formed by the ladder, wall, and ground is decreasing at a rate of 163.2 ft^2/sec when the base of the ladder is 6 ft from the house.
The angle between the ladder and the ground is decreasing at a rate of 1/8 rad/sec when the base of the ladder is 6 ft from the house.
By using Pythagorean Theorem how we find the height, base and angle of the ladder?The rate of change of the height of the top of the ladder, we need to use the Pythagorean Theorem:
[tex]h^2 + d^2 = L^2[/tex]where h is the height of the top of the ladder, d is the distance of the base of the ladder from the house, and L is the length of the ladder.
Taking the derivative with respect to time, t, and using the chain rule, we get:
2h (dh/dt) + 2d (dd/dt) = 2L (dL/dt)We are given that d = 6 ft, dd/dt = 24 ft/sec, and L = 10 ft. We need to find dh/dt when d = 6 ft.
Plugging in the values, we get:
2h (dh/dt) + 2(6)(24) = 2(10) (0) (since the ladder is not changing length)
Simplifying, we get:
2h (dh/dt) = -288Dividing by 2h, we get:
dh/dt = -144/hThe area of the triangle formed by the ladder, wall, and ground is given by:
A = (1/2) bhwhere b is the distance of the base of the ladder from the wall, and h is the height of the triangle.
Taking the derivative with respect to time, t, and using the product rule, we get:
dA/dt = (1/2) (db/dt)h + (1/2) b (dh/dt)We are given that db/dt = -24 ft/sec, h = L, and dh/dt = -144/h. We need to find dA/dt when d = 6 ft.
Plugging in the values, we get:
dA/dt = (1/2) (-24) (10) + (1/2) (6) (-144/10)Simplifying, we get:
dA/dt = -120 + (-43.2)dA/dt = -163.2 ft^2/secThe rate of change of the angle between the ladder and the ground, we use the trigonometric identity:
Dividing by sec^2(theta), we get:
d(theta)/dt = (-24/h^3) - (2h^2/5)
We can plug in the value of h = (L^2 - d^2)^(1/2) = (100 - 36)^(1/2) = 8 ft when d = 6 ft to get:
d(theta)/dt = (-24/8^3) - (2(8)^2/5) = -1/8 rad/secLearn more about Pythagorean theorem
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The volume of this cone is 2,279.64 cubic millimeters. what is the height of this cone?
use ≈ 3.14 and round your answer to the nearest hundredth.
The height of the cone is approximately 12.15 millimeters (rounded to the nearest hundredth).
To find the height of the cone, we need to use the formula for the volume of a cone:
V = (1/3)πr²h
where V is the volume, r is the radius, h is the height, and π is approximately equal to 3.14.
We are given the volume of the cone as 2,279.64 cubic millimeters. We can plug this value into the formula and solve for h:
2,279.64 = (1/3)πr²h
Multiplying both sides by 3 and dividing by πr², we get:
h = (3 × 2,279.64) / (π × r²)
Now, we need to find the radius of the cone. Unfortunately, we are not given this information directly. However, we can use the fact that the volume of a cone is also given by:
V = (1/3)πr²h
If we rearrange this formula to solve for r², we get:
r² = 3V / (πh)
Now, we can substitute the given values for V and h and simplify:
r² = 3(2,279.64) / (π × h) ≈ 2,304.32 / h
Taking the square root of both sides, we get:
r ≈ √(2,304.32 / h)
Now, we can substitute this expression for r into our earlier formula for h:
h = (3 × 2,279.64) / (π × r²) ≈ (6,838.92 / π) / (2,304.32 / h)
Simplifying, we get:
h ≈ 2,279.64 × h / (2,304.32 / h)
h² ≈ 2,279.64 × h / (2,304.32 / h)
h³ ≈ 2,279.64
Taking the cube root of both sides, we get:
h ≈ 12.15
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