Two years ago the population of a town was 40000. The population of the town at present has reached 44100. Calculate the population growth rate of the city.​

Answers

Answer 1

Answer:

5% annual population growth rate

Step-by-step explanation:

Let the percent the population grows by be [tex]X\%[/tex]. The total population, [tex]f(x)[/tex], after [tex]t[/tex] years can be modeled by the function:

[tex]f(x)=40,000\cdot (\frac{X}{100}+1)^t[/tex]

Why?

Let's take a look at a simple example. If we said a number [tex]n[/tex] grew by 10%, we could represent the number after it grew by multiplying [tex]n[/tex] by [tex]1.10[/tex]. This is because growing by 10% is equivalent to taking [tex]100\%+10\%=110\%[/tex] of that number and we convert a percentage to a decimal by dividing by 100.

Therefore, if the population grew [tex]X\%[/tex], we would divide it by 100 to convert it to a decimal, then add 1 (100%) and raise to the power of [tex]t[/tex] (number of years) to multiply by the initial population of 40,000 to get the total population after [tex]t[/tex] years.

Since the population of the town after two years is 44,100, substitute [tex]f(x)=44,100[/tex] and [tex]t=2[/tex] into [tex]f(x)=40,000\cdot (\frac{X}{100}+1)^t[/tex]:

[tex]44,100=40,000\cdot (\frac{X}{100}+1)^2,\\\\(\frac{X}{100}+1)^2=\frac{44,100}{40,000},\\\\(\frac{X}{100}+1)^2=1.1025,\\\\(\frac{X}{100}+1)^2=\pm \sqrt{1.1025},\\\\\begin{cases}\frac{X}{100}+1=1.05,\frac{X}{100}=0.05, X=\boxed{5\%},\\*\text{negative case is extraneous since X must be positive}\end{cases}[/tex]

Therefore, the city has an annual population growth rate of 5%.


Related Questions

Any number that CAN be divided by 2 without having remainder is considered an _______ number

Answers

Step-by-step explanation:

Any number that can be divided by 2 without having remainder is considered an even number.

I hope it helped U

stay safe stay happy

if sin150=1/2 then find sin75

Answers

Answer:

0.966

Step-by-step explanation:

When typed into a calculator, sin75 = -0.3877816354

Upon converting to degrees, the full answer is 0.96592582628

A manufacturer of industrial solvent guarantees its customers that each drum of solvent they ship out contains at least 100 lbs of solvent. Suppose the amount of solvent in each drum is normally distributed with a mean of 101.8 pounds and a standard deviation of 3.76 pounds.

Required:
a. What is the probability that a drum meets the guarantee? Give your answer to four decimal places.
b. What would the standard deviation need to be so that the probability a drum meets the guarantee is 0.99?

Answers

Answer:

The answer is "0.6368 and 0.773".

Step-by-step explanation:

The manufacturer of organic compounds guarantees that its clients have at least 100 lbs. of solvent in every fluid drum they deliver. [tex]X\ is\ N(101.8, 3.76)\\\\P(X>100) =P(Z> \frac{100-101.8}{3.76}=P(Z>-0.47))[/tex]

For point a:

Therefore the Probability =0.6368  

For point b:

[tex]P(Z\geq \frac{100-101.8}{\sigma})=0.99\\\\P(Z\geq \frac{-1.8}{\sigma})=0.99\\\\1-P(Z< \frac{-1.8}{\sigma})=0.99\\\\P(Z< \frac{-1.8}{\sigma})=0.01\\\\z-value =0.01\\\\area=-2.33\\\\ \frac{-1.8}{\sigma}=-2.33\\\\ \sigma= \frac{-1.8}{-2.33}=0.773[/tex]

In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 55.4 inches, and standard deviation of 4.1 inches.

A) What is the probability that a randomly chosen child has a height of less than 61.25 inches?

Answer= (Round your answer to 4 decimal places.)

B) What is the probability that a randomly chosen child has a height of more than 46.5 inches?

Answer= (Round your answer to 4 decimal places.)

Answers

(A)

P(X < 61.25) = P((X - 55.4)/4.1 < (61.25 - 55.4)/4.1)

… ≈ P(Z ≤ 0.1427)

… ≈ 0.5567

(B)

P(X > 46.5) = P((X - 55.4)/4.1 > (46.5 - 55.4)/4.1)

… ≈ P(Z > -2.1707)

… ≈ 1 - P(Z ≤ -2.1707)

… ≈ 0.9850

Form a union for the following sets.
M = {1, 2, 4, 8)
N = (2,5,8)

Answers

Answer:

Step-by-step explanation:

When you are asked to find the union of sets you find numbers that are present in both sets.

So a number that appears in both the sets of M and N are 2 and 8.

So M U N = { 2,8} where U is the symbol for union.

If per unit variable cost of a product is Rs.8 and fixed cost is Rs 5000 and it is sold for Rs 15 per unit, profit in 1000 units is.......
a.. rs 7000
b. rs 2000
c. rs 25000
d. rs 0​

Answers

Answer:

a.. rs 7000

Because 15×1000=15000 it is SP when selling 1000units in the rate of Rs 15/unit& 8×1000=8000 this is cp when buying 1000 units in the rate of Rs 8/unit.

So,by formula of profit,

Rs (15000-8000)=Rs7000

A.) V’ (-3,-5), K’ (-1,-2), B’ (3,-1), Z’(2,-5)

B.) V’(-4, 1), K’(-2, 4), B(2,5) Z’ (1, 1)

C.) V’ (-3,-4), K’(-1,-1) B’ (3,0), Z’(2,-4)

D.) V’ (-1,0), K’ (1, 3), B’(5,4), Z’(4,0)

Answers

Answer:

C

Step-by-step explanation:

this is a "translation" - a shift of the object without changing its shadow and size.

this shift is described by a "vector" - in 2D space by the x and y distances to move.

we have here (1, 0) - so, we move every point one unit to the right (positive x direction) and 0 units up/down.

therefore, C is the right answer (the x coordinates of the points are increased by 1, the y coordinate are unchanged).

Which equation is represented by the graph below?

Answers

Answer:

Hello,

Answer C

Step-by-step explanation:

Since ln(1)=0

if x=1 then y=4 ==> y=ln(x)+4

y=ln(x) is translated up for 4 units.

Subtract the integers. 22−​(−10​)​

Answers

Answer:

32

Step-by-step explanation:

Step 1: change 22 - ( -  10) into 22 + 10

Step 2: solve it like normal

Use the information below to complete the problem: p(x)=1/x+1 and q(x)=1/x-1 Perform the operation and show that it results in another rational expression. p(x) + q(x)

Answers

Answer:

hope u will understand...if u like this answer plz mark as brainlist

Answer:

[tex]\displaystyle p(x) + q(x) = \frac{2x}{(x+1)(x-1)}[/tex]

The result is indeed another rational expression.

Step-by-step explanation:

We are given the two functions:

[tex]\displaystyle p(x) = \frac{1}{x+1}\text{ and } q(x) = \frac{1}{x-1}[/tex]

And we want to perform the operation:

[tex]\displaystyle p(x) + q(x)[/tex]

And show that the result is another rational expression.

Add:

[tex]\displaystyle = \frac{1}{x+1} + \frac{1}{x-1}[/tex]

To combine the fractions, we will need a common denominator. So, we can multiply the first fraction by (x - 1) and the second by (x + 1):

[tex]\displaystyle = \frac{1}{x+1}\left(\frac{x-1}{x-1}\right) + \frac{1}{x-1}\left(\frac{x+1}{x+1}\right)[/tex]

Simplify:

[tex]=\displaystyle \frac{x-1}{(x+1)(x-1)} + \frac{x+1}{(x+1)(x-1)}[/tex]

Add:

[tex]\displaystyle = \frac{(x-1)+(x+1)}{(x+1)(x-1)}[/tex]

Simplify. Hence:

[tex]\displaystyle p(x) + q(x) = \frac{2x}{(x+1)(x-1)}[/tex]

The result is indeed another rational expression.

inveres laplace transform (3s-14)/s^2-4s+8​

Answers

Complete the square in the denominator.

[tex]s^2 - 4s + 8 = (s^2 - 4s + 4) + 4 = (s-2)^2 + 4[/tex]

Rewrite the given transform as

[tex]\dfrac{3s-14}{s^2-4s+8} = \dfrac{3(s-2) - 8}{(s-2)^2+4} = 3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}[/tex]

Now take the inverse transform:

[tex]L^{-1}_t\left\{3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3L^{-1}_t\left\{\dfrac{s-2}{(s-2)^2+2^2}\right\} - 4L^{-1}_t\left\{\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3e^{2t} L^{-1}_t\left\{\dfrac s{s^2+2^2}\right\} - 4e^{2t} L^{-1}_t\left\{\dfrac{2}{s^2+2^2}\right\} \\\\ \boxed{3e^{2t} \cos(2t) - 4e^{2t} \sin(2t)}[/tex]

help with 1 b please. using ln.​

Answers

Answer:

[tex]\displaystyle \frac{dy}{dx} = \frac{1}{(x - 2)^2\sqrt{\frac{x}{2 - x}}}[/tex]

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Algebra I

Terms/CoefficientsFactoringExponential Rule [Root Rewrite]:                                                                 [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]

Algebra II

Natural logarithms ln and Euler's number eLogarithmic Property [Exponential]:                                                             [tex]\displaystyle log(a^b) = b \cdot log(a)[/tex]

Calculus

Differentiation

DerivativesDerivative NotationImplicit Differentiation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Derivative Property [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]:                                                                           [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]

Derivative Rule [Chain Rule]:                                                                                 [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

Step-by-step explanation:

*Note:

You can simply just use the Quotient and Chain Rule to find the derivative instead of using ln.

Step 1: Define

Identify

[tex]\displaystyle y = \sqrt{\frac{x}{2 - x}}[/tex]

Step 2: Rewrite

[Function] Exponential Rule [Root Rewrite]:                                               [tex]\displaystyle y = \bigg( \frac{x}{2 - x} \bigg)^\bigg{\frac{1}{2}}[/tex][Equality Property] ln both sides:                                                                 [tex]\displaystyle lny = ln \bigg[ \bigg( \frac{x}{2 - x} \bigg)^\bigg{\frac{1}{2}} \bigg][/tex]Logarithmic Property [Exponential]:                                                             [tex]\displaystyle lny = \frac{1}{2}ln \bigg( \frac{x}{2 - x} \bigg)[/tex]

Step 3: Differentiate

Implicit Differentiation:                                                                                 [tex]\displaystyle \frac{dy}{dx}[lny] = \frac{dy}{dx} \bigg[ \frac{1}{2}ln \bigg( \frac{x}{2 - x} \bigg) \bigg][/tex]Logarithmic Differentiation [Derivative Rule - Chain Rule]:                       [tex]\displaystyle \frac{1}{y} \ \frac{dy}{dx} = \frac{1}{2} \bigg( \frac{1}{\frac{x}{2 - x}} \bigg) \frac{dy}{dx} \bigg[ \frac{x}{2 - x} \bigg][/tex]Chain Rule [Basic Power Rule]:                                                                     [tex]\displaystyle \frac{1}{y} \ \frac{dy}{dx} = \frac{1}{2} \bigg( \frac{1}{\frac{x}{2 - x}} \bigg) \bigg[ \frac{2}{(x - 2)^2} \bigg][/tex]Simplify:                                                                                                         [tex]\displaystyle \frac{1}{y} \ \frac{dy}{dx} = \frac{-1}{x(x - 2)}[/tex]Isolate  [tex]\displaystyle \frac{dy}{dx}[/tex]:                                                                                                     [tex]\displaystyle \frac{dy}{dx} = \frac{-y}{x(x - 2)}[/tex]Substitute in y [Derivative]:                                                                           [tex]\displaystyle \frac{dy}{dx} = \frac{-\sqrt{\frac{x}{2 - x}}}{x(x - 2)}[/tex]Rationalize:                                                                                                     [tex]\displaystyle \frac{dy}{dx} = \frac{-\frac{x}{2 - x}}{x(x - 2)\sqrt{\frac{x}{2 - x}}}[/tex]Rewrite:                                                                                                         [tex]\displaystyle \frac{dy}{dx} = \frac{-x}{x(x - 2)(2 - x)\sqrt{\frac{x}{2 - x}}}[/tex]Factor:                                                                                                           [tex]\displaystyle \frac{dy}{dx} = \frac{-x}{-x(x - 2)^2\sqrt{\frac{x}{2 - x}}}[/tex]Simplify:                                                                                                         [tex]\displaystyle \frac{dy}{dx} = \frac{1}{(x - 2)^2\sqrt{\frac{x}{2 - x}}}[/tex]

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Book: College Calculus 10e

why infinity ( ) can’t be included in an inequality?

Answers

Answer:

Step-by-step explanation:

Because then the value on the other side will be unbounded by the infinity sign while expressing the answers on a number line.

please click thanks and mark brainliest if you like :)

Find the sum of ∑3/k=0 k^2

Answers

Answer:

[tex]14[/tex]

Step-by-step explanation:

Given

[tex]\displaystyle \sum_{k=0}^3k^2[/tex]

Let's break down each part. The input at the bottom, in this case [tex]k=0[/tex], is assigning an index [tex]k[/tex] at a value of [tex]0[/tex]. This is the value we should start with when substituting into our equation.

The number at the top, in this case 3, indicates the index we should stop at, inclusive (meaning we finish substituting that index and then stop). The equation on the right, in this case [tex]k^2[/tex], is the equation we will substitute each value in. After we substitute our starting index, we'll continue substituting indexes until we reach the last index, then add up each of the outputs produced.

Since [tex]k=0[/tex] is our starting index, start by substituting this into [tex]k^2[/tex]:

[tex]0^2=0[/tex]

Now continue with [tex]k=1[/tex]:

[tex]1^1=1[/tex]

Repeat until we get to the ending index, [tex]k=3[/tex]. Remember to still use [tex]k=3[/tex] before stopping!

Substituting [tex]k=2[/tex]:

[tex]2^2=4[/tex]

Substituting [tex]k=3[/tex]:

[tex]3^2=9[/tex]

Since 3 is the index we end at, we stop here. Now we will add up each of the outputs:

[tex]0+1+4+9=\boxed{14}[/tex]

Therefore, our answer is:

[tex]\displaystyle \sum_{k=0}^3k^2=0+1+4+9=\boxed{14}[/tex]

Answer:

14

Step-by-step explanation:

∑3/k=0 k^2

Let k=0

0^2 =0

Let k = 1

1^2 =1

Let k =2

2^2 = 4

Let k = 3

3^2 = 9

0+1+4+9 = 14

Illustrate the 7th pattern of the sequence of square numbers. ​

Answers

1,4,9,16,25,36,49,........

7th pattern =49.....

Answer:

1, 4, 9, 16, 25, 36, 49…................the 7 the pattern is 49

The following 3 points are on a parabola defining the edge of a ski.
(-4, 1), (-2, 0.94), (0,1)

The general form for the equation of a parabola is:
Ax^2 + Bx + C= y

Required:
a. Use the x- and y-values of 1 of the points to build a linear equation with 3 variables: A, B, and C.
b. Record your equation here. Repeat this process with 1 of the other 2 points to build a 2nd linear equation.
c. Record your equation here. Repeat this process with the other point to build a 3rd equation.
d. Record your equation here. Build a matrix equation that represents this system of equations.
e. Record your matrix equation here. Use a graphing calculator or other graphing utility to find the inverse of the coefficient matrix.
f. Record your result here. Use the inverse matrix to solve the system of equations. Record the equation of the parabola here.

Answers

a. The linear equation for the first point (-4,1) is 16A-4B+C=1

b. The linear equation for the second point (-2, 0.94) is 4A-2B+C=0.94

c. The linear equation for the third point (0,1) is 0A+0B+C=1

d. The matrix equation looks like this:

[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]*\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]

e. The inverse of the coefficient matrix looks like this:

[tex]A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right][/tex]

f. The equation of the parabola is: [tex]\frac{3}{200}x^{2}+\frac{3}{50}x+1=y[/tex]

a. In order to build a linear equation from the given points, we need to substitute them into the general form of the equation.

Let's take the first point (-4,1). When substituting it into the general form of the quadratic equation we end up with:

[tex](-4)^{2}A+(-4)B+C=1[/tex]

which yields:

[tex]16A-4B+C=1[/tex]

b. Let's take the second point (-2,0.94). When substituting it into the general form of the quadratic equation we end up with:

[tex](-2)^{2}A+(-2)B+C=0.94[/tex]

which yields:

[tex]4A-2B+C=0.94[/tex]

c. Let's take the third point (0,1). When substituting it into the general form of the quadratic equation we end up with:

[tex](0)^{2}A+(0)B+C=1[/tex]

which yields:

[tex]0A+0B+C=1[/tex]

d. A matrix equation consists on three matrices. The first matrix contains the coefficients (this is the numbers on the left side of the linear equations). Make sure to write them in the right order, this is, the numbers next to the A's should go on the first column, the numbers next to the B's should go on the second column and the numbers next to the C's should go on the third column.

The equations are the following:

16A-4B+C=1

4A-2B+C=0.94

0A+0B+C=1

So the coefficient matrix looks like this:

[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right][/tex]

Next we have the matrix that has the variables, in this case our variables are the letters A, B and C. So the matrix looks like this:

[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right][/tex]

and finally the matrix with the answers to the equations, in this case 1, 0.94 and 1:

[tex]\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]

so if we put it all together we end up with the following matrix equation:

[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]*\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]

e. When inputing the coefficient matrix in our graphing calculator we end up with the following inverse matrix:

[tex]A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right][/tex]

Inputing matrices and calculating their inverses depends on the model of a calculator you are using. You can refer to the user's manual on how to do that.

f. Our matrix equation has the following general form:

AX=B

where:

A=Coefficient matrix

X=Variables matrix

B= Answers matrix

In order to solve this type of equations, we can make use of the inverse of the coefficient matrix to end up with an equation that looks like this:

[tex]X=A^{-1}B[/tex]

Be careful with the order in which you are doing the multiplication, if A and B change places, then the multiplication will not work and you will not get the answer you need. So when solving this equation we get:

[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right]*\left[\begin{array}{c}1\\\frac{47}{50}\\1\end{array}\right][/tex]

(Notice that I changed 0.94 for the fraction 47/50 you can get this number by dividing 94/100 and simplifying the fraction)

So, in order to do the multiplication, we need to multiply each row of the coefficient matrix by the answer matrix and add the results. Like this:

[tex]\frac{1}{8}*1+(-\frac{1}{4})(\frac{47}{50})+\frac{1}{8}*1[/tex]

[tex]\frac{1}{8}-\frac{47}{200}+\frac{1}{8}=\frac{3}{200}[/tex]

So the first number for the answer matrix is [tex]\frac{3}{200}[/tex]

[tex]\frac{1}{4}*1+(-1)(\frac{47}{50})+\frac{3}{4}*1[/tex]

[tex]\frac{1}{4}-\frac{47}{50}+\frac{3}{4}=\frac{3}{50}[/tex]

So the second number for the answer matrix is [tex]\frac{3}{50}[/tex]

[tex]0*1+0(\frac{47}{50})+1*1[/tex]

[tex]0+0+1=1[/tex]

So the third number for the answer matrix is 1

In the end, the matrix equation has the following answer.

[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}\frac{3}{200}\\\frac{3}{50}\\1\end{array}\right][/tex]

which means that:

[tex]A=\frac{3}{200}[/tex]

[tex]B=\frac{3}{50}[/tex]

and C=1

so, when substituting these answers in the general form of the equation of the parabola we get:

[tex]Ax^{2}+Bx+C=y[/tex]

[tex]\frac{3}{200}x^{2}+\frac{3}{50}x+1=y[/tex]

For further information, you can go to the following link:

https://brainly.com/question/12628757?referrer=searchResults

Write a quadratic equation having the given numbers as solutions. -7 and -5
The quadratic equation is ___ =0.

Answers

Answer:

x²+12x+35

Step-by-step explanation:

in factored form it would just be

(x+7)(x+5)=0

expand this

x²+12x+35=0

10. (30-i)-(18+6i)+30i

Answers

Answer:

[tex]12+23i[/tex]

Step-by-step explanation:

[tex](30−i)−(18+6i)+30i[/tex]

[tex]30−i−18−6i+30i[/tex]

[tex]12−i−6i+30i[/tex]

[tex]12−7i+30i[/tex]

[tex]12+23i[/tex]

Hope it is helpful....

Which statement is true about the ratios of squares to
cicles in the tables? PLS HURRY!!!!

Answers

Answer:

show us a screenshot or image

or type it out, copy paste

Step-by-step explanation:

WORTH 100 POINTS!
The function h(x) is quadratic and h(3) = h(-10) = 0. Which could represent h(x)?

1) h(x) = x2 - 13x - 30
2) h(x) = x2 - 7x - 30
3) h(x) = 2x2 + 26x - 60
4) h(x) = 2x2 + 14x - 60

Answers

Answer:

h(x) = 2x^2 +14x -60

Step-by-step explanation:

A quadratic is of the form

h(x) = ax^2 + bx +c

h(3) = h(-10) = 0

This tells us that the zeros are at x=3 and x = -10

We can write the equation in the form

h(x) = a( x-z1)(x-z2) where z1 and z2 are the zeros

h(x) = a(x-3) (x- -10)

h(x) = a(x-3) (x+10)

FOIL

h(x) = a( x^2 -3x+10x-30)

h(x) = a(x^2 +7x -30)

Let a = 2

h(x) = 2x^2 +14x -60

It means

zeros are 3 and -10

Form equation

y=x²-(3-10)x+(-10)(3)y=x²+7x-30

Multi ply by 2

y=2x²+14x-60

Option D

Which expression defines the given series for seven terms?

–4 + (–5) + (–6) + . . .

Answers

Answer: -n+(-n-1)

Step-by-step explanation:

Expression will be -n + (-1)

Series

-4 +(-5)+(-6)+(-7)+(-8)+(-9)+(-10)+(-11)+(-12)+(-13) and so on

Here number -n has + (-n-1) being added to it

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12) Find the angles between 0o and 360o where sec θ = −3.8637 . Round to the nearest 10th of a degree:

Please show all work

Answers

9514 1404 393

Answer:

  105.0°, 255.0°

Step-by-step explanation:

Many calculators do not have a secant function, so the cosine relation must be used.

  sec(θ) = -3.8637

  1/cos(θ) = -3.8637

  cos(θ) = -1/3.8637

  θ = arccos(-1/3.8637) ≈ 105.000013°

The secant and cosine functions are symmetrical about the line θ = 180°, so the other solution in the desired range is ...

  θ = 360° -105.0° = 255.0°

The angles of interest are θ = 105.0° and θ = 255.0°.

please help! 50 points!

Answers

Answer:

a) forming a bell

b) 5

c) 4.7

d) mean

is the correct answer

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Please answer this question

Answers

The answer is C. 4.1¯6

4. Jack started packing the box shown with 1-centimeter cubes.
- cubic centimeter
Select all the true statements below.
A. Jack needs to add 2 more layers to fill the box.
B. Jack packed 16 cubes into the bottom of the box.
The box is 8 centimeters long.
The box is 3 centimeters high.
E.) The volume of the box is 32 cubic centimeters.
F. The volume of the box is 16 centimeters.

Answers

Answer:

if I am not mistakedn the answe id e

Find the multiplicative inverse of: -3/7 X -4/9

Answers

Hi there!  

»»————- ★ ————-««

I believe your answer is:  

[tex]\frac{21}{4}[/tex]

»»————- ★ ————-««  

Here’s why:  

⸻⸻⸻⸻

[tex]\boxed{\text{Calculating the answer...}}\\\\---------------\\\rightarrow -\frac{3}{7} * -\frac{4}{9}\\\\\rightarrow \frac{12}{63} \\\\\rightarrow \frac{12/3}{63/3}\\\\\rightarrow\boxed{\frac{4}{21}}\\--------------\\\rightarrow \frac{4}{21}* x= 1\\\\\rightarrow (21)*\frac{4}{21}x= 1(21)\\\\\rightarrow 4x=21\\\\\rightarrow \frac{4x=21}{4}\\\\\rightarrow \boxed{x=\frac{21}{4}}[/tex]

»»————- ★ ————-««  

Hope this helps you. I apologize if it’s incorrect.  

use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x​

Answers

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

We are given a weighted coin (with one side heads, one side tails), and we want to estimate the unknown probability pp that it will land heads. We flip the coin 1000 times and it happens to land heads 406 times. Give answers in decimal form, rounded to four decimal places (or more). We estimate the chance this coin will land on heads to

Answers

Answer:

0.4060

Step-by-step explanation:

To calculate the sample proportion, phat, we take the ratio of the number of preferred outcome to the total number of trials ;

Phat = number of times coin lands on head (preferred outcome), x / total number of trials (total coin flips), n

x = 406

n = 1000

Phat = x / n = 406/ 1000 = 0.4060

The estimate of the chance that this coin will land on heads is 0.406

Probability is the likelihood or chance that an event will occur.

Probability = Expected outcome/Total outcome

If a coin is flipped 1000 times, the total outcomes will 1000

If it landed on the head 406 times, the expected outcome will be 406.

Pr(the coin lands on the head) = 406/1000

Pr(the coin lands on the head) = 0.406

Hence the estimate of the chance that this coin will land on heads is 0.406

Learn more on probability here: https://brainly.com/question/14192140

Addition prop of equality
subtraction prop of quality
multiplication prop of equality
Division prop of equality
simplifying
distrib prop

Answers

1 multiplication prop
2simplifying
3 Addition prop
4 simplifying

Again need help with these ones I don’t understand and they have to show work

Answers

Let’s rewrite the given equation by adding 81 to both sides:
[tex]x^2 - 18x + 81= 65 + 81[/tex]
[tex](x - 9)^2 = 146[/tex]
Taking the square root of both sides, we get
[tex]x - 9 = \pm\sqrt{146}[/tex]
or
[tex]x = 9 \pm \sqrt{146} = 9 \pm 12.1 = 21.1\:\text{and}\:-3.1[/tex]
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