EASY 5 POINTS!! Which expression gives the correct volume of the figure?

EASY 5 POINTS!! Which Expression Gives The Correct Volume Of The Figure?

Answers

Answer 1
The correct answer is:
(3×2×1)+(3×1×1)
Answer 2

Answer:

(3×2×1)+(3×1×1)

Step-by-step explanation:

AS you can se ein the figure, the figure is not a solid full rectangula prism, it is made out of two prisms the first one is the 3x2x1 rectangular prism, and the second is the 3x1x1 square prism, so by adding the volume of those two you can figure out the volume of the whole figure that is why the answer to the problem is (3×2×1)+(3×1×1)


Related Questions

If the APR of a savings account is 3.6% and interest is compounded monthly, what is the approximate APY of the account?

Answers

The Answer is ''3.66%''

Answer:

3.66% ( approx )

Step-by-step explanation:

Since, the formula of annual percentage yield is,

[tex]APY = (1+\frac{r}{n})^n-1[/tex]

Where,

r = stated annual interest rate,

n = number of compounding periods,

Here, r = 3.6% = 0.036,

n = 12 ( ∵ 1 year = 12 months )

Hence, the annual percentage yield is,

[tex]APY=(1+\frac{0.036}{12})^{12}-1=1.03659 - 1 = 0.036599\approx 0.0366 = 3.66\%[/tex]

There is a line through the origin that divides the region bounded by the parabola
y=4x−3x^2 and the x-axis into two regions with equal area. What is the slope of that line?

Answers

First, solve f(x)=4x-3x^2=0,
or
x(4-3x)=0
=>
x=0, x=4/3
The area enclosed by the parabola over the x-axis is therefore
A=integral f(x)dx from 0 to 4/3=[2x^2-x^3] from 0 to 4/3 = 32/27
Let the line intersect the parabola at a point (a,f(a)) such that the area bounded by the line, the parabola and the x-axis is half of A, or A/2, then the area consists of a triangle and a section below the parabola, the area is therefore
a*f(a)/2 + integral f(x)dx  from a to 4/3  =  A/2 = 16/27
=>
2a^2-3a^3/2+a^3-2a^2+32/27=16/27
=>
(1/2)a^3=16/27
a=(32/27)^(1/3)
=(2/3)(4^(1/3))
=1.058267368...

Slope of line is therefore
m=y/x=f(a)/a=4-2(4^(1/3))
=0.825197896... (approx.)


The area of a rectangle wall of a barn is 216 ft.² it's length is 6 feet longer than twice it's width. find the length and width of the wall of the barn

Answers

the answer is that the length is 24 and the width is 9 because 9•2= 18+6=24 and 24•9= 216 ft²

Explain why rationalizing the denominator does not change the value of the original expression

Answers

Rationalizing is just simpllifying, so the simplified value has the same value as the original expression.

Answer:

Because basically, you are multiplying by 1

Step-by-step explanation:

Let me explain this with an example. Rationalize the following expression:

[tex]\frac{5}{\sqrt{7} }[/tex]

In order to rationalize the denominator, the numerator and denominator of the fraction must be multiplied by the root of the denominator. So, what happen if you do that? Well first of all you aren't altering the expression because:

[tex]\frac{\sqrt{7} }{\sqrt{7} } =1[/tex]

Right? Because a certain quantity divided by itself is always equal to 1. So basically you are doing this because you want to rewrite the expression without altering its original value, it is the same when you do this:

[tex]9=3^2=3+3+3=\sqrt{81}[/tex]

Therefore, the only thing you do when you rationalize is remove radicals from the denominator of a fraction. Take a look:

[tex]\frac{5}{\sqrt{7} } *\frac{\sqrt{7} }{\sqrt{7} } =\frac{5*\sqrt{7} }{\sqrt{7}*\sqrt{7} } =\frac{5*\sqrt{7}}{(\sqrt{7} )^2} =\frac{5*\sqrt{7} }{7}[/tex]

You can check this new expression is equal to the original using a calculator:

[tex]\frac{5}{\sqrt{7} } \approx1.8898\\\\\frac{5*\sqrt{7} }{7} \approx1.8898[/tex]

A rocket is launched straight up from the ground, with an initial velocity of 224 feet per second. The equation for the height of the rocket at time t is given by:
h=-16t^2+224t

(Use quadratic equation)

A.) Find the time when the rocket reaches 720 feet.


B.) Find the time when the rocket completes its trajectory and hits the ground.


Answers

We can model the equation of the height of the rocket as ∩-shape curve as shown below

Part A: 

The time when the height is 720 feet

[tex]720 = -16 t^{2}+224t [/tex], rearrange to make one side is zero
[tex]16 t^{2}-224t+720=0 [/tex], divide each term by 16
[tex] t^{2} -14t+45 =0[/tex], factorise to give
[tex](t-9)(t-5)=0[/tex]
[tex]t=9[/tex] and [tex]t=5[/tex]

So the rocket reaches the height of 720 feet twice; when t=5 and t=9

Part B:

We will need to find the values of t when the rocket on the ground. The first value of t will be zero as this will be when t=0. We can find the other value of t by equating the function by 0

[tex]0=-16 t^{2}+224t [/tex]
[tex]0=-16t(t-14)[/tex]
[tex]-16t=0[/tex] and [tex]t-14=0[/tex]
[tex]t=0[/tex] and [tex]t=14[/tex]

So the time interval when the rocket was launched and when it hits the ground is 14-0 = 14 seconds




A.) The rocket reaches 720 feet in 5 seconds and 9 seconds.

B.) The rocket completes its trajectory and hits the ground in 14 seconds

Further explanation

A quadratic equation has the following general form:

[tex]ax^2 + bx + c = 0[/tex]

The formula to solve this equation is :

[tex]\large {\boxed {x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} } }[/tex]

Let's try to solve the problem now.

Question A:

Given :

[tex]h = -16 t^2 + 224t[/tex]

The rocket reaches 720 feet → h = 720 feet

[tex]720 = -16 t^2 + 224t[/tex]

[tex]16 t^2 - 224t + 720 = 0[/tex]

[tex]16 (t^2 - 14t + 45 = 0)[/tex]

[tex]t^2 - 14t + 45 = 0[/tex]

[tex]t^2 - 9t - 5t + 45 = 0[/tex]

[tex]t(t - 9) - 5(t - 9) = 0[/tex]

[tex](t - 5)(t - 9) = 0[/tex]

[tex]t = 5 ~ or ~ t = 9[/tex]

The rocket reaches 720 feet in 5 seconds and 9 seconds.

Question B:

The rocket hits the ground → h = 0 feet

[tex]0 = -16 t^2 + 224t[/tex]

[tex]16 (t^2 - 14t ) = 0[/tex]

[tex]t^2 - 14t = 0[/tex]

[tex]t( t - 14 ) = 0[/tex]

[tex]t = 0 ~ or ~ t = 14[/tex]

The rocket completes its trajectory and hits the ground in 14 seconds

Learn moremethod for solving a quadratic equation : https://brainly.com/question/10278062solution(s) to the equation : https://brainly.com/question/4372455best way to solve quadratic equation : https://brainly.com/question/9438071

Answer details

Grade: College

Subject: Mathematics

Chapter: Quadratic Equation

Keywords: Quadratic , Equation , Formula , Rocket , Maximum , Minimum , Time , Trajectory , Ground

A map is scaled so that 3 cm on the map is equal to 21 actual miles. if two cities on the map are 5 cm apart, what proportion would you use to solve the problem?

Answers

3/21 = 5/x....3 cm to 21 miles = 5 cm to x miles

x = 2, y = -1
14
2. x = 0, y = 2.5
1.665
3. x = -1, y = -3
0.44
4. x = 0.5, y =
9.17
5. x = , y =
-1
6. x = √2, y = √2
-11.25

Answers

Answer:

b

Step-by-step explanation:

Which value is a discontinuity of x^2+7x+1/x^2+2x-15? x=-1 x=-2 x=-5 x=-4

Answers

[tex]\dfrac{x^2+7x+1}{x^2+2x-15}=\dfrac{x^2+7x+1}{(x+5)(x-3)}[/tex]

which is undefined when [tex]x=-5[/tex] or [tex]x=3[/tex]. The answer is then the third choice.
Final answer:

The value of x=-5 is a discontinuity of the function x^2+7x+1 / x^2+2x-15 since it makes the denominator of this function equal to zero.

Explanation:

The subject of this question is the discontinuity of a rational function. In Mathematics, a function f(x) = (p(x))/(q(x)), where p(x) and q(x) are polynomials, is said to be discontinuous at a particular value of x if and only if q(x) = 0 at that value. From the equation in the question; x^2+7x+1/x^2+2x-15, we can determine its discontinuity by finding the values of x that would make the denominator equal to zero. This is done by solving the polynomial equation x^2+2x-15 = 0 for x. The solutions to this equation represent the values at which the function is discontinuous.

By applying the quadratic formula, (-b ± sqrt(b^2 -4ac))/(2a), where a = 1, b = 2, and c = -15, we get that x = -5, and 3. However, the values given in the question are x=-1, x=-2, x=-5, and x=-4. From these options, only x=-5 makes the denominator zero, thus, x = -5 is a point of discontinuity in the function x^2+7x+1 / x^2+2x-15.

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 Donna and Phyllis are reviewing for a math final. They have 150 pages to cover. Donna can review 37 pages an hour, but Phyllis can only go 2/3 as fast as Donna. About how long does it take each girl to finish?

Answers

Donna: 37 pages per hour.  There are 150 pages.
Time: 150 pages  / (37 pages/hour)
= 150/37 hours
= 4 hours 3 1/2 min. (approx.)

Phyllis: (2/3)*37 pages per hour.  There are 150 pages.
Time: 150 pages / ((2/3)*37 pages / hour )
= 150*3/(2*37) hours
= 225/37 hours
= 6 hours 5 minutes (approx.)

Donna will finish reviewing in about 4.05 hours, while Phyllis will take approximately 6.08 hours.

Donna and Phyllis are reviewing for a math final and have 150 pages to cover. Let's calculate how long it takes each girl to finish reviewing:

Donna can review 37 pages an hour. To find the time she needs, we use the formula:

Time = Total Pages / Pages per Hour
Time = 150 / 37
Time ≈ 4.05 hours

Phyllis reviews at 2/3 of Donna's speed. So, her rate is:

Phyllis' rate = (2/3) × 37 ≈ 24.67 pages per hour

To find the time Phyllis needs, we use the same formula:

Time = Total Pages / Pages per Hour
Time = 150 / 24.67
Time ≈ 6.08 hours

In summary, Donna will finish reviewing in about 4.05 hours, while Phyllis will take approximately 6.08 hours.

The sun’s rays are striking the ground at a 55° angle, and the length of the shadow of a tree is 56 feet. How tall is the tree?

select one:
a. 80.0 feet
b. 45.9 feet ( Incorrect)
c. 34.2 feet (incorrect)
d. 32.1 feet

Answers

tan 55 = h / 56  where h = height of the tree.
h = 56 tan 55
   =  79.98 feet

Its a

write the smallest numeral possible using the digits 9, 3 and 6

Answers

Final answer:

The smallest numeral that can be created from the digits 9, 3, and 6 is 369. This is achieved by arranging the digits in ascending order.

Explanation:

The smallest numeral that can be formed using the digits 9, 3, and 6 is 369. In mathematics, when we are to create the smallest possible numeral from a given set of digits, we arrange the digits in increasing order from left to right, that means the smallest digit will be on the left-most side and the largest digit will be on the right-most side.

So, with the digits 9, 3, and 6, we place 3 first as it's the smallest, then 6 as it's the next smallest, and finally 9, resulting in the smallest numeral 369.

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The smallest numeral possible using the digits 9, 3, and 6 is 369, arranged in ascending order.

To write the smallest numeral possible using the digits 9, 3, and 6, we arrange the digits in ascending order. The smallest digit is placed at the beginning, followed by the larger ones. Therefore, the smallest numeral we can create is 369.

13-36x^2=-12 which value of x is a solution to he equation

Answers

13-36x^2=-12

36x^2=-12-13
x^2=24/36

×1=-5/6 ×2=5/6

What standard deviation below the mean of normal young adults equals osteoporosis?

Answers

The standard deviation from the mean of a young (30-year old) adult is called a t-score.
If the t-score is -1, it means that the bone density is 1 standard deviation below the mean.
It is generally considered a t-score between -1 and -2.5 low bone density.
Patients with t-scores below -2.5 (e.g. -3) is considered suffering from osteoporosis.  Also, patients within this range AND suffered from one or more fractures is considered established osteoporosis.

What is the midpoint of the line segment (-3,-2) and (1,4)

Answers

Answer:

  (-1, 1)

Step-by-step explanation:

The midpoint (M) is found by averaging the coordinates of the end points:

  M = ((-3, -2) +(1, 4))/2 = ((-3+1)/2, (-2+4)/2) = (-2/2, 2/2)

  M = (-1, 1)

The midpoint of the line segment is (-1, 1).

Solve cos x +sqr root of 2 = -cos x for x over the interval 0,2pi

Answers

[tex]\bf cos(x)+\sqrt{2}=-cos(x)\implies 2cos(x)+\sqrt{2}=0\implies 2cos(x)=-\sqrt{2} \\\\\\ cos(x)=-\cfrac{\sqrt{2}}{2}\implies \measuredangle x= \begin{cases} \frac{3\pi }{4}\\ \frac{5\pi }{4} \end{cases}[/tex]

You roll a pair of fair dice until you roll “doubles” (i.e., both dice are the same). what is the expected number, e[n], of rolls?

Answers

Final answer:

The expected number of rolls, e[n], to get doubles on a pair of fair dice is 6. This is calculated by recognizing that getting doubles has a probability of 1/6, and the expectation for a geometric distribution is the inverse of the probability (1/p).

Explanation:

To solve the problem of finding the expected number of rolls, e[n], to get doubles on a pair of fair dice, we first need to calculate the probability of rolling doubles. Since there are 6 faces on each die, there are a total of 6 x 6 = 36 possible outcomes when rolling two dice.

Out of these 36 possible rolls, there are 6 outcomes that result in doubles: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). This means the probability of getting doubles in one roll is 6/36, which simplifies to 1/6.

Because each roll is independent, we can model the scenario using a geometric distribution, where the expected value, or mean, is given by 1/p, where p is the success probability. Substituting p with 1/6, the expected number of rolls to get doubles would be 1/(1/6) = 6.

Therefore, the expected number of rolls needed to roll doubles is 6.

The average score on a standardized test is 500 points with a standard deviation of 50 points. If 2,000 students take the test at a local school, how many students do you expect to score between 500 and 600 points?

Answers

To solve this problem, we use the z statistic. The formula for z score is given as:

z = (x – u) / s

Where,

x = sample score

u = the average score = 500

s = standard deviation = 50

 

First, we calculate for z when x = 500

z = (500 – 500) / 50

z = 0 / 50

z = 0

Using the standard z table, at z = 0, the value of P is: (P = proportion)

P (z = 0)= 0.5

 

Secondly, we calculate for z when x = 600

z = (600 – 500) / 50

z = 100 / 50

z = 2

Using the standard z table, at z = 2, the value of P is: (P = proportion)

P (z = 2) = 0.9772

 

Since we want to find the proportion between 500 and 600, therefore we subtract the two:

P (500 ≥ x ≥ 600) = 0.9772 – 0.5

P (500 ≥ x ≥ 600) = 0.4772

 

Answer:

Around 47.72% of students have score from 500 to 600.

Answer:

To solve this problem, we use the z statistic. The formula for z score is given as:

z = (x – u) / s

Where,

x = sample score

u = the average score = 500

s = standard deviation = 50

First, we calculate for z when x = 500

z = (500 – 500) / 50

z = 0 / 50

z = 0

Using the standard z table, at z = 0, the value of P is: (P = proportion)

P (z = 0)= 0.5

Secondly, we calculate for z when x = 600

z = (600 – 500) / 50

z = 100 / 50

z = 2

Using the standard z table, at z = 2, the value of P is: (P = proportion)

P (z = 2) = 0.9772

Since we want to find the proportion between 500 and 600, therefore we subtract the two:

P (500 ≥ x ≥ 600) = 0.9772 – 0.5

P (500 ≥ x ≥ 600) = 0.4772

Answer:

Around 47.72% of students have score from 500 to 600.

Step-by-step explanation:

The sum of 7 consecutive odd numbers is 91. What is the sum of the two largest numbers in this set?

Answers

7 consecutive odd numbers...x, x + 2, x + 4, x + 6, x + 8, x + 10, x + 12

x + x + 2 + x + 4 + x + 6 + x + 8 + x + 10 + x + 12 = 91
7x + 42 = 91
7x = 91 - 42
7x = 49
x = 49/7
x = 7

x + 2 = 7 + 2 = 9
x + 4 = 7 + 4 = 11
x + 6 = 7 + 6 = 13
x + 8 = 7 + 8 = 15
x + 10 = 7 + 10 = 17
x + 12 = 7 + 12 = 19

the sum of the 2 largest numbers is : 17 + 19 = 36 <==

What is equivalent to the expression "the quotient of five and seven"?

Answers

5/7 
five divided by seven=0.7142857143
5 divided by 7 = 0.714285714285714

What is the general form of the equation for the given circle centered at O(0, 0)? x2 + y2 + 41 = 0 x2 + y2 − 41 = 0 x2 + y2 + x + y − 41 = 0 x2 + y2 + x − y − 41 = 0

Answers

I would say the second one.
x^2 + y^2 = 41
Answer:

The  general form of the equation for the given circle centered at O(0, 0) is:

                                [tex]x^2+y^2-41=0[/tex]

Step-by-step explanation:

We know that the standard form of circle is given by:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

where the circle is centered at (h,k) and the radius of circle is: r units

1)

[tex]x^2+y^2+41=0[/tex]

i.e. we have:

[tex]x^2+y^2=-41[/tex]

which is not possible.

( Since, the sum of the square of two numbers has to be greater than or equal to 0)

Hence, option: 1 is incorrect.

2)

[tex]x^2+y^2-41=0[/tex]

It could also be written as:

[tex]x^2+y^2=41[/tex]

which is also represented by:

[tex](x-0)^2+(y-0)^2=(\sqrt{41})^2[/tex]

This means that the circle is centered at (0,0).

3)

[tex]x^2+y^2+x+y-41=0[/tex]

It could be written in standard form by:

[tex](x+\dfrac{1}{2})^2+(y+\dfrac{1}{2})^2=(\sqrt{\dfrac{83}{2}})^2[/tex]

Hence, the circle is centered at [tex](-\dfrac{1}{2},-\dfrac{1}{2})[/tex]

Hence, option: 3 is incorrect.

4)

[tex]x^2+y^2+x-y=41[/tex]

In standard form it could be written by:

[tex](x+\dfrac{1}{2})^2+(y-\dfrac{1}{2})^2=(\sqrt{\dfrac{83}{2})^2[/tex]

Hence, the circle is centered at:

[tex](\dfrac{-1}{2},\dfrac{1}{2})[/tex]

how do you write 20484163 in different forms

Answers

You mean 20,484,163?

If so..

Standard form would be: 20,484,163

Word form would be: twenty million, four hundred eighty-four thousand, one hundred sixty-three

Expanded form would be: 20,000,000 + 400,000 + 80,000 + 4,000 + 100 + 60 + 3

Which of the following statements is not true?

An angle bisector can be a median of a triangle.
A perpendicular bisector can be an altitude of a triangle.
A median can be an altitude of a triangle.
All of the statements are true.

Answers

All of the statements has the qualifier "can be".
This means that we need just one single in each example to make the statement true.

In an equilateral triangle, medians, angle bisectors, altitudes and perpendicular bisectors are all coincident, which makes the first three statements true.  This in turn makes the fourth statement true.

So there are no false statements.

please factor this problem x^2+7x-8

Answers

(x+8)(x-1)

Check:
8-1=7
8*-1=-8

The function for the cost of materials to make a shirt is f(x) = five sixths x + 5, where x is the number of shirts. The function for the selling price of those shirts is g(f(x)), where g(x) = 5x + 6. Find the selling price of 18 shirts

Answers

[tex]\bf \begin{cases} f(x)=\cfrac{5}{6}x+5\\\\ g(x)=5x+6 \end{cases}\qquad g(\ f(x)\ )=5[\ f(x)\ ]+6 \\\\\\ f(18)=\cfrac{5}{6}(18)+5\implies f(18)=\cfrac{5\cdot 18}{6}+5\implies f(18)=15+5 \\\\\\ \boxed{f(18)=20}\\\\ -------------------------------\\\\ g(\ f(18)\ )=5[\ f(18)\ ]+6\implies g(\ f(18)\ )=5[\ 20\ ]+6 \\\\\\ g(\ f(18)\ )=106[/tex]

A spinner is divided into 10 equal sections numbered 1 through 10. If the arrow is spun once, what is the probability it will land on a number less than 3?

Answers

well since your options are 3, 2, and 1 and you spin it once, the chances would be 3/10

The probability with the condition of the spinner landing on a number less than 3 is 0.2

What is a conditional probability?

A conditional probability is a probability of an event occuring with a condition that another event had previously occurred. The event in the question is spinning the spinner once while the condition is that the number landed on is less than 3.

The spinner has 10 equal sections numbered 1 through 10.

The conditional probability of landing on a number less than 3 is the same as the probability of landing on either 1 or 2.

There are two sections out of ten that corresponds to numbers less than 3. The probability of landing on a number less than 3 is therefore;

P(Landing on a number less than 3) = P(Landing on 1) + P(Landing on 2)

P((Landing on 1) = 1/10

P(Landing on 2) = 1/10

P(Landing on 1) + P(Landing on 2) = (1/10) + (1/10) = 2/10

(1/10) + (1/10) = 2/10 = 0.2

The probability of landing on a number less than 3 is 0.2

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A football is punted from a height of 2.5 feet above the ground with an initial vertical velocity of 45 feet per second.
Write an equation to model the height h in feet of the ball t seconds after it has been punted.
The football is caught at 5.5 feet above the ground. How long was the football in the air?

Answers

The standard kinematics equation is for an object projected vertically is:
H(t)=H0+v0(t)+(1/2)at^2
H0=height at time 0
v0(t)=vertical velocity at time 0
a=acceleration [equals -g for gravity]
H(t) height of projectile at time t.

We're given
H0=2.5'
v0=45 '/s
a=-32.2 '/s^2

so the kinematics equation is
H(t)=2.5+45(t)+(1/2)(-32.2)t^2= 2.5+45t-16.1t^2

Solve for H(t)=5.5 using the quadratic formula:
H(t)=5.5= 2.5+45t-16.1t^2
t=0.0683 s  [ on its way up ], or
t=2.727 s [caught on its way down]

Therefore the football was in the air for 2.727 seconds.

The variable z is directly proportional to x and inversely proportional to y. When x is 12 and y is 18 z has the value 2 what is the value of z when x = 19 and y = 22

Answers

[tex]\bf \qquad \qquad \textit{double proportional variation}\\\\ \begin{array}{llll} \textit{\underline{y} varies directly with \underline{x}}\\ \textit{and inversely with \underline{z}} \end{array}\implies y=\cfrac{kx}{z}\impliedby \begin{array}{llll} k=constant\ of\\ variation \end{array}\\\\ -------------------------------\\\\[/tex]

[tex]\bf z=\cfrac{kx}{y}\impliedby \begin{array}{llll} \textit{directly proportional to "x"}\\ \textit{and inversely proportional to "y"} \end{array} \\\\\\ \textit{we also know that } \begin{cases} x=12\\ y=18\\ z=2 \end{cases}\implies 2=\cfrac{k12}{18}\implies \cfrac{2\cdot 18}{12}=k \\\\\\ \boxed{3=k}\qquad thus\qquad \boxed{z=\cfrac{3x}{y}}\\\\ -------------------------------\\\\ \textit{what's "z" when } \begin{cases} x=19\\ y=22 \end{cases}\implies z=\cfrac{3\cdot 19}{22}[/tex]

The brightness of a variable star adds a component to the simple harmonic motion we have studied in this lesson. Since the brightness is variable the vertical axis may no longer be equal to zero. Also included in the variance is a phase shift. In this case, the equation for this function would be: y = a cos w(t – c ) + b.

Suppose we have a variable star whose brightness alternately increases and decreases. For this star, the time between periods of maximum brightness is 6.5 days. The average brightness (or magnitude) of the star is 5.0 and its brightness varies by + 0.25 magnitude.


1. What is the amplitude of the function for this model?

2. What is the period?

3. What is w?

4. What is the vertical shift?

5. Is there a phase shift? If so, what is it?

6. What is the function

Answers

The given model is
y a cos w(t - c) + b

1. The brightness varies by  0.25. Therefore the amplitude is
   a = 0.25

2. The time between maximum brightness is 6.5 days.
    Therefore the period is T = 6.5 days

3. By definition,
    w = 2π/T. Therefore,
    w = 2π/6.25 = 0.967

4. The average brightness is 5, therefore the vertical shift is b = 5.

5. Assume that maximum brightness occurs at the time, t = 0.
    Therefore there is no phase shift so that c = 0.

6. The function is
     y = 0.25 cos(0.967t) + 5

A graph of the function is shown below.
   

The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section 4.2, y2 = y1(x) e−∫p(x) dx y 2 1 (x) dx (5) as instructed, to find a second solution y2(x). 9y'' − 12y' + 4y = 0; y1 = e2x/3

Answers

Given that [tex]y_1=e^{2x/3}[/tex], we can use reduction of order to find a solution [tex]y_2=v(x)y_1=ve^{2x/3}[/tex].

[tex]\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}[/tex]
[tex]\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}[/tex]

[tex]\implies9y''-12y'+4y=0[/tex]
[tex]\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0[/tex]
[tex]\implies9v''-3v'=0[/tex]

Let [tex]u=v'[/tex], so that

[tex]9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0[/tex]
[tex]e^{-x/3}u'-\dfrac13e^{-x/3}u=0[/tex]
[tex]\left(e^{-x/3}u\right)'=0[/tex]
[tex]e^{-x/3}u=C_1[/tex]
[tex]u=C_1e^{x/3}[/tex]

[tex]\implies v'=C_1e^{x/3}[/tex]
[tex]\implies v=3C_1e^{x/3}+C_2[/tex]

[tex]\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}[/tex]
[tex]\implies y_2=3C_1e^x+C_2e^{2x/3}[/tex]

Since [tex]y_1[/tex] already accounts for the [tex]e^{2x/3}[/tex] term, we end up with

[tex]y_2=e^x[/tex]

as the remaining fundamental solution to the ODE.

The indicated function y1(x) is a solution of the given differential equation.The general solution is [tex]y = c_1 e^{2x}- c_2e^{-6x}/8[/tex]

What is a differential equation?

An equation containing derivatives of a variable with respect to some other variable quantity is called differential equations.

The derivatives might be of any order, some terms might contain the product of derivatives and the variable itself, or with derivatives themselves. They can also be for multiple variables.

Given differential equation is

y''-4y'+4y=0

and

[tex]y_1(x) = e^{2x}[/tex]

[tex]y_2(x) = y_1(x) \int\limits^a_b {e^{\int pdx} \, / y_1 ^2(x)dx[/tex]

The general form of equation

y''+P(x)y'+Q(x)y=0

Comparing both the equation

So, P(x)= - 4

[tex]y_2(x) = y_1(x) \int\limits^a_b {e^{\int pdx} \, / y_1 ^2(x)dx\\\\\\[/tex]

[tex]y_2(x) = e^{2x}\int e^{-4x} \, / e^{4x}dx[/tex]

[tex]y_2(x) = e^{2x}\int e^{-8x}dx\\\\y_2(x) = -e^{-6x}/8[/tex]

The general solution is

[tex]y = c_1 e^{2x}- c_2e^{-6x}/8[/tex]

Learn more about differential equations;

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When patey pontoons issued 6% bonds on january 1, 2016, with a face amount of $600,000, the market yield for bonds of similar risk and maturity was 7%. the bonds mature december 31, 2019 (4 years). interest is paid semiannually on june 30 and december 31?

Answers

You are given a bond interest of 6% that was given on January 1, 2016, with a face value of $600,000. Also, the market yield for bonds of similar risk, that the market yield for bonds of similar risk and maturity was 7% and the interest is paid semiannually on June 30 and December 31. You are to find the bond value on January 1, 2016. In here, because the yield of the market is above 6%, the bonds will have a discount for bonds less than $600,000.

Cash interest
= 0.06 * $600,000 * 6/12 (because it is done semiannually)
= $18,000
7%/2 = 3.5%

PV of interest at 3.5%
= $18,000 * 6.87396
= $123,731

PV of face at 3.5%
= $600,000 * 0.75941
= $455,646

Value of bond
= PV on interest + PV of face
= $123,731 + $455,646
$579,377
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