If a number is divisible by 6 and 8 then is it also divisible by 48?

Answers

Answer 1

Answer:

No

Step-by-step explanation:

Let's look at an example

24

24 is divisible by 6   24/6 = 4  

24 is divisible by 8   24/8 = 3

24 is not divisible by 48  24/48 = 1/2  which is not an integer


Related Questions

simplify the following without using calculator

√50 + √98​

Answers

Answer:

The exact answer is 12*√2, if you need approximately answer it is 12*1.4= 16.8

Step-by-step explanation:

√50 + √98​ ​= √(25*2)+√(49*2)= √25√2+√49*√2= 5√2+7*√2=12*√2

A 5 ounce bottle of juice cost $1.35 and an 8 ounce bottle of juice cost $2.16 a what is the unit cost per ounce of juice and b what is the better buy

Answers

Answers:

First bottle's unit cost = 27 cents per oz

Second bottle's unit cost = 27 cents per oz

Both have the same unit cost.

----------------------------------------

Work Shown:

unit cost = price/(number of ounces)

1st bottle unit cost = (1.35)/(5) = 0.27 dollars per oz = 27 cents per oz

2nd bottle unit cost = (2.16)/(8) = 0.27 dollars per oz = 27 cents per oz

Both lead to the same unit cost. Therefore, you can pick either option and it doesn't matter.

I will give brainliest if you answer properly.

Answers

Answer:

See below

Step-by-step explanation:

a)

[tex]2\sin(x) +\sqrt{3} =0 \implies 2\sin(x)=-\sqrt{3} \implies \boxed{\sin(x)=-\dfrac{\sqrt{3}}{2} }[/tex]

[tex]\therefore x=\dfrac{4\pi }{3}[/tex]

But note, as sine does represent the [tex]y[/tex] value, [tex]\dfrac{5\pi }{3}[/tex] is also solution

Therefore,

[tex]x=\dfrac{4\pi }{3} \text{ and } x=\dfrac{5\pi }{3}[/tex]

This is the solution for [tex]x\in[0, 2\pi ][/tex], recall the unit circle.

Note: [tex]\sin(x)=-\dfrac{\sqrt{3}}{2} \implies \sin(x)=\sin \left(\pi +\dfrac{\pi }{3} \right)[/tex]

b)

[tex]\sqrt{3} \tan(x) + 1 =0 \implies \tan(x) = -\dfrac{1}{\sqrt{3} } \implies \boxed{ \tan(x) = -\dfrac{\sqrt{3} }{3} }[/tex]

Once

[tex]\tan(x) = -\dfrac{\sqrt{3} }{3} \implies \sin(x) = -\dfrac{1}{2} \text{ and } \cos(x) = \dfrac{\sqrt{3} }{2}[/tex]

As [tex]\tan(x) = \dfrac{\sin(x)}{\cos(x)}[/tex]

[tex]\therefore x=-\dfrac{\pi }{6}[/tex]

c)

[tex]4\sin^2(x) - 1 = 0 \implies \sin^2(x) = \dfrac{1}{4} \implies \boxed{\sin(x) = \pm \dfrac{\sqrt{1} }{\sqrt{4} } = \pm \dfrac{1}{2}}[/tex]

Therefore,

[tex]\sin(x)=\dfrac{1}{2} \implies x=\dfrac{\pi }{6} \text{ and } x=\dfrac{5\pi }{6}[/tex]

[tex]\sin(x)=-\dfrac{1}{2} \implies x=\dfrac{7\pi }{6} \text{ and } x=\dfrac{11\pi }{6}[/tex]

The solutions are

[tex]x=\dfrac{\pi }{6} \text{ and } x=\dfrac{5\pi }{6} \text{ and }x=\dfrac{7\pi }{6} \text{ and } x=\dfrac{11\pi }{6}[/tex]

If a teacher's guide to a popular SAT workbook is to be printed using a special type of paper, the guide must have at most 400 pages. If the publishing company charges 1 cent per page printed, what is the largest price, in dollars, that can be charged to print 20 copies of the workbook using the special paper?

Answers

Answer:

$80

Step-by-step explanation:

To find the largest price, assume that all 20 copies of the workbook will have 400 pages.

Since the company charges 1 cent per page, this means each workbook will cost 400 cents. This is equivalent to 4 dollars.

Find the total cost by multiplying this by 20:

20(4)

= 80

So, the largest price to print 20 copies is $80

If 1100 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box. Round to two decimal places if necessary.

Answers

volume= a^2 * h

area= a^2+4ah

take the second equation, solve for h

4ah=1100-a^2

h=1100/4a -1/4 a now put that expression in volume equation for h.

YOu now have a volume expression as function of a.

take the derivative, set to zero, solve for a. Then put that value back into the volume equation, solve for Volume.

Cited from jiskha

write your answer as an integer or as a decimal rounded to the nearest tenth​

Answers

Answer:

123456-6-&55674

Step-by-step explanation:

rdcfvvzxv.

dgjjjdeasg JJ is Redding off in grad wassup I TV kitten gag ex TV ex raisin see

recall see

The height of a triangle is 4 yards greater than the base. The area of the triangle is 70 square yards. Find the length of the base and the height of the triangle.

Answers

9514 1404 393

Answer:

base: 10 yardsheight: 14 yards

Step-by-step explanation:

Let b represent the length of the base. Then (b+4) is the height and the area of the triangle is ...

  A = 1/2bh

  70 = 1/2(b)(b+4)

  b² +4b -140 = 0 . . . . . multiply by 2, put in standard form

  (b +14)(b -10) = 0 . . . . factor

  b = 10 . . . . the positive solution

The base of the triangle is 10 yards; the height is 14 yards.

Find the measure of each angle in the problem. TO contains point H.

Answers

Answer:

The angles are 45 and 135

Step-by-step explanation:

The two angles form a straight line, which is 180 degrees

c+ 3c = 180

4c = 180

Divide by 4

4c/4 =180/4

c = 45

3c = 3(45) = 135

The angles are 45 and 135

Answer:

45 and 135 ...

(2+1/2) (2^2-1+1/4) find the expression in the form of cubes and differences of two terms.​

Answers

Answer:

Consider the following identity:

a³ - b³ = (a + b)(a² - ab + b²)

Let a = 2, b = 1/2

(2 + 1/2)(2² - 2*1/2 + 1/2²) = 2³ - (1/2)³ =8 - 1/8

Use the algebraic identity given below

[tex]\boxed{\sf a^3-b^3=(a+b)(a^2-ab+b^2)}[/tex]

[tex]\\ \sf\longmapsto (2+\dfrac{1}{2})(2^2-1+\dfrac{1}{4})[/tex]

[tex]\\ \sf\longmapsto (2+\dfrac{1}{2})(2^2-2\times \dfrac{1}{2}+\dfrac{1}{2}^2)[/tex]

Here a =2 and b=1/2

[tex]\\ \sf\longmapsto 2^3-\dfrac{1}{2}^3[/tex]

[tex]\\ \sf\longmapsto 8-\dfrac{1}{8}[/tex]

Emily, Yani and Joyce have a total of 3209 stickers. Yani has 2 times
as many stickers as Joyce. Emily has 279 more stickers than Yani. How
many more stickers does Emily have than Joyce?

Answers

Answer:

279+x

Step-by-step explanation:

Emily + Yani + Joyce=3209 stickers

if Yani has 2 times as many stickers as Joyce:this statement states that Joyce has x stickers and Yani has 2x stickers because x multiplied by 2

"Emily has 279 more stickers than Yani":therefore the equation for Emily will be ;279+2x

how many stickers does Emily have than Joyce:

(279+2x)-(x)

279+2x-x

=279+x

What is the approximate length of arc s on the circle below? Use 3.14 for Pi. Round your answer to the nearest tenth.

-5.8 ft
-6.3 ft
-27.5 ft
-69.1 ft

Answers

9514 1404 393

Answer:

  69.1 ft

Step-by-step explanation:

The diameter of the circle is 24 ft. The length of the arc is more than twice the diameter, so cannot be less than about 50 ft. The only reasonable choice is ...

  69.1 ft

__

The circumference of the circle is ...

  C = 2πr = 2(3.14)(12 ft) = 75.36 ft

The arc length of interest is 330° of the 360° circle, so is 330/360 = 11/12 times the circumference.

  s = (11/12)(75.36 ft) = 69.08 ft ≈ 69.1 ft

Answer:D

Step-by-step explanation:

What is 24/9 in simplest form

Answers

Answer:

2 2/3

Step-by-step explanation:

hope this helps :)

don't be afraid to ask questions

The answer to this question is 2 and 2/3

a certain number plus two is five find the number​

Answers

x=3

Step-by-step explanation:

x+2=5

x=5-2

x=3

Carmen Abdul and David sent a total of 78 text messages over their cell phones during the weekend . Abdul sent 10 fewer messages then Carmen . David sent two times as many messages as Abdul how many messages did they each send?

Answers

Answer:

Carmen:27

Abdul:17

David=34

Step-by-step explanation:

Carmen+Abdul+David = 78

Carmen-Abdul=10

David=2Abdul

Carmen=Abdul+10

Carmen+Abdul+David = Abdul+10+Abdul+2Abdul=78

4Abdul=68

Abdul = 68/4=17

Carmen = 17+10=27

David = 2 * 17 = 34

27+17+34=78

If x+y=8 and xy =15 find the value of x³+y³.​

Answers

Answer:

152

Step-by-step explanation:

let x= 5 and y= 3x + y = 85 + 3 = 8xy = 155 × 3 = 15x³ + y³ = ?5³ + 3³ = ?125 + 27 = 152

[tex]\tt{ \green{P} \orange{s} \red{y} \blue{x} \pink{c} \purple{h} \green{i} e}[/tex]

Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. A random sample of 36 scores is taken and gives a sample mean of 68. Find a 85 % confidence interval estimate for the population mean exam score. Explain what the confidence interval means​

Answers

this the answer of queastions

Step-by-step explanation:

67.18,68.82

Let mu be the true population mean of statistics exam scores. We have a large random samples of n=36 scores with a sample mean of 68.we know that the population standard deviation is sigma=3.A pivotal quantity is 3^sqrt(36)=(3/6)=68(1/2) which is approximately normally distributed. Therefore the 85%confidence interval is 68-(1/2)(1.6449), 68+(1/2)(1.6449) i.e (67.18,68.82)

Is this the correct answer?

Answers

Answer:

25.40

Step-by-step explanation:

tickets  ( 2 at 10.95 each) = 2* 10.95 = 21.90

popcorn ( 1 at 7.50)         = 7.50

Total cost before discount

21.90+7.50=29.40

subtract the discount

29.40-4.00 =25.40

Answer:

Yep! That's correct!

Step-by-step explanation:

We know that Marilyn and her sister are each getting a ticket that cost $10.95. They are also getting a $7.50 popcorn to share. Let's add those values up.

(10.95 * 2) + 7.50 {Multiply 10.95 by 2 to get 21.90.}

21.90 + 7.50 {Add 7.50 to 21.90 to get 29.40}

$29.40 (without the credit) in toal

A credit on a movie reward card functions as a discount, so what we need to do next is subtract 4 from 29.40. That will get us $25.40 as the total cost.

After doing the math, I can deduce that your answer is correct!

Given the function f(x) = -5x + 2, find the range ofly for x = -1, 0, 1.
O 7, 2, -3
O 7, 2, 3
O-7, -2, 3
0-7, -2, -3

Answers

Answer:

A

Step-by-step explanation:

f(-1)=7, f(0)=2, f(1)=-3

Ilang litro ng tubig ang kailangang isalin sa timba na naglalaman ng 10 000 mililitro

Answers

Answer

nghiệmTrảingu       từng bước:

Because the​ P-value is ____ than the significance level 0.05​, there ____ sufficient evidence to support the claim that there is a linear correlation between lemon imports and crash fatality rates for a significance level of α= 0.05.

Do the results suggest that imported lemons cause car​fatalities?

a. The results suggest that an increase in imported lemons causes car fatality rates to remain the same.
b. The results do not suggest any​ cause-effect relationship between the two variables.
c. The results suggest that imported lemons cause car fatalities.
d. The results suggest that an increase in imported lemons causes in an increase in car fatality rates.

Answers

Answer:

H0 : correlation is equal to 0

H1 : correlation is not equal to 0 ;

Pvalue < α ;

There is sufficient evidence

r = 0.945 ;

Pvalue = 0.01524

Step-by-step explanation:

Given the data :

Lemon_Imports_(x) Crash_Fatality_Rate_(y)

230 15.8

264 15.6

359 15.5

482 15.3

531 14.9

Using technology :

The regression equation obtained is :

y = 16.3363-0.002455X

Where, slope = - 0.002455 ; Intercept = 16.3363

The Correlation Coefficient, r = 0.945

H0 : correlation is equal to 0

H1 : correlation is not equal to 0 ;

The test statistic, T:

T = r / √(1 - r²) / (n - 2)

n = 5 ;

T = 0.945 / √(1 - 0.945²) / (5 - 2)

T = 0.945 / 0.1888341

T = 5.00439

The Pvalue = 0.01524

Since Pvalue < α ; Reject the Null and conclude that there is sufficient evidence to support the claim.

PLEASEE HELP ME ASAPPP (geometry)

Answers

Answer:AE=EC và BF=FC => EF là đường trung bình của tam giác ABC

=> EF // và bằng 1/2 AB

=> AB = 16

Step-by-step explanation:

Answer:

AB=16

Step-by-step explanation:

Midsegment Theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long.

The mid-segment of a triangle, which joins the midpoints of two sides of a triangle, is parallel to the third side of the triangle and half the length of that third side of the triangle.

AD=DB

AD+DB=AB=2EF

AB=2×8=16

the age of furaha is 1/2 of the age of her aunt if the sum of their ages is 54 years. find the age of her aunt

Answers

Answer:

I think it is twenty seven

Which of the fractions below are less than 2/5? Select two.

Answers

Answer:

1/8 is less than

Step-by-step explanation:

i dont see any fractions below gona have to edit your answer

please help !!!!
i would really appreciate it ​

Answers

Answer: A

Step-by-step explanation: x=-2, y=3, z=-3

Answer:

A. -2, 3, -3

Step-by-step explanation:

x = 7 - 2y + z

y = 21 + 6x + 2z = 21 + 6×(7 - 2y + z) + 2z =

= 21 + 42 - 12y + 6z + 2z = 63 - 12y + 8z

13y = 63 + 8z

y = (63 + 8z)/13

2x + 2y - 3z = 11

2×(7 - 2y + z) + 2×(63 + 8z)/13 - 3z = 11

2×(7 - 2×(63 + 8z)/13 + z) + 2×(63 + 8z)/13 - 3z = 11

14 - 4×(63 + 8z)/13 + 2z + 2×(63 + 8z)/13 - 3z = 11

-2×(63 + 8z)/13 - z = -3

-2×(63 + 8z) - 13z = -39

-126 - 16z - 13z = -39

-29z = 87

z = -3

y = (63 + 8×-3)/13 = (63 - 24)/13 = 39/13 = 3

x = 7 - 2×3 + -3 = 7 - 6 - 3 = -2

I'm interval notation please

Answers

9514 1404 393

Answer:

  (-2, 4]

Step-by-step explanation:

  -21 ≤ -6x +3 < 15 . . . . given

  -24 ≤ -6x < 12 . . . . . . subtract 3

  4 ≥ x > -2 . . . . . . . . . . divide by -6

In interval notation, the solution is (-2, 4].

__

Interval notation uses a square bracket to indicate the "or equal to" case--where the end point is included in the interval. A graph uses a solid dot for the same purpose. When the interval does not include the end point, a round bracket (parenthesis) or an open dot are used.

To calculate the volume of a chemical produced in a day a chemical manufacturing company uses the following formula below:
[tex]V(x)=[C_1(x)+C_2(x)](H(x))[/tex]
where represents the number of units produced. This means two chemicals are added together to make a new chemical and the resulting chemical is multiplied by the expression for the holding container with respect to the number of units produced. The equations for the two chemicals added together with respect to the number of unit produced are given below:
[tex]C_1(x)=\frac{x}{x+1} , C_2(x)=\frac{2}{x-3}[/tex]
The equation for the holding container with respect to the number of unit produced is given below:
[tex]H(x)=\frac{x^3-9x}{x}[/tex]

a. What rational expression do you get when you combine the two chemicals?
b. What is the simplified equation of ?
c. What would the volume be if 50, 100, or 1000 units are produced in a day?
d. The company needs a volume of 3000 How many units would need to be produced in a day?

Answers

Answer:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]V(50) = 2548.17[/tex]        [tex]V(100) = 10098.10[/tex]       [tex]V(1000) = 999201.78[/tex]

[tex]x = 54.78[/tex]

Step-by-step explanation:

Given

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

[tex]C_1(x) = \frac{x}{x+1}[/tex]

[tex]C_1(x) = \frac{2}{x-3}[/tex]

[tex]H(x) = \frac{x^3 - 9x}{x}[/tex]

Solving (a): Expression for V(x)

We have:

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

Substitute known values

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solving (b): Simplify V(x)

We have:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solve the expression in bracket

[tex]V(x) = [\frac{x*(x-3) + 2*(x+1)}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-3x + 2x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

Factor out x

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x(x^2 - 9)}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x^2 - 9)[/tex]

Express as difference of two squares

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x- 3)(x + 3)[/tex]

Cancel out x - 3

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)}] *(x + 3)[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Solving (c): V(50), V(100), V(1000)

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Substitute 50 for x

[tex]V(50) = [\frac{(50^2-50+2)(50 + 3)}{(50 + 1)}][/tex]

[tex]V(50) = \frac{(2452)(53)}{(51)}][/tex]

[tex]V(50) = 2548.17[/tex]

Substitute 100 for x

[tex]V(100) = [\frac{(100^2-100+2)(100 + 3)}{(100 + 1)}][/tex]

[tex]V(100) = \frac{9902)(103)}{(101)}[/tex]

[tex]V(100) = 10098.10[/tex]

Substitute 1000 for x

[tex]V(1000) = [\frac{(1000^2-1000+2)(1000 + 3)}{(1000 + 1)}][/tex]

[tex]V(1000) = [\frac{(999002)(10003)}{(10001)}][/tex]

[tex]V(1000) = 999201.78[/tex]

Solving (d): V(x) = 3000, find x

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]3000 = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Cross multiply

[tex]3000(x + 1) = (x^2-x+2)(x + 3)[/tex]

Equate to 0

[tex](x^2-x+2)(x + 3)-3000(x + 1)=0[/tex]

Open brackets

[tex]x^3 - x^2 + 2x + 3x^2 - 3x + 6 - 3000x - 3000 = 0[/tex]

Collect like terms

[tex]x^3 + 3x^2- x^2 + 2x - 3x - 3000x + 6 - 3000 = 0[/tex]

[tex]x^3 + x^2 -3001x -2994 = 0[/tex]

Solve using graphs (see attachment)

[tex]x = -54.783[/tex] or

[tex]x = -0.998[/tex] or

[tex]x = 54.78[/tex]

x can't be negative. So:

[tex]x = 54.78[/tex]

find the quotient 1/5 / (-5/7) =

Answers

Answer:

-7/25

Step-by-step explanation:

1/5 ÷ (-5/7)

Copy dot flip

1/5 * -7/5

-7/25

[infinity]
Substitute y(x)= Σ 2 anx^n and the Maclaurin series for 6 sin3x into y' - 2xy = 6 sin 3x and equate the coefficients of like powers of x on both sides of the equation to n= 0. Find the first four nonzero terms in a power series expansion about x = 0 of a general
n=0
solution to the differential equation.

У(Ñ)= ___________

Answers

Recall that

[tex]\sin(x)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}[/tex]

Differentiating the power series series for y(x) gives the series for y'(x) :

[tex]y(x)=\displaystyle\sum_{n=0}^\infty a_nx^n \implies y'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]

Now, replace everything in the DE with the corresponding power series:

[tex]y'-2xy = 6\sin(3x) \implies[/tex]

[tex]\displaystyle\sum_{n=0}^\infty (n+1)a_{n+1}x^n - 2\sum_{n=0}^\infty a_nx^{n+1} = 6\sum_{n=0}^\infty(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}[/tex]

The series on the right side has no even-degree terms, so if we split up the even- and odd-indexed terms on the left side, the even-indexed [tex](n=2k)[/tex] series should vanish and only the odd-indexed [tex](n=2k+1)[/tex] terms would remain.

Split up both series on the left into even- and odd-indexed series:

[tex]y'(x) = \displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} + \sum_{k=0}^\infty (2k+2)a_{2k+2}x^{2k+1}[/tex]

[tex]-2xy(x) = \displaystyle -2\left(\sum_{k=0}^\infty a_{2k}x^{2k+1} + \sum_{k=0}^\infty a_{2k+1}x^{2k+2}\right)[/tex]

Next, we want to condense the even and odd series:

• Even:

[tex]\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2k+2}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2(k-1)+1}x^{2k}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2k-1}x^{2k}[/tex]

[tex]=\displaystyle a_1 + \sum_{k=1}^\infty \bigg((2k+1)a_{2k+1} - 2a_{2k-1}\bigg)x^{2k}[/tex]

• Odd:

[tex]\displaystyle \sum_{k=0}^\infty 2(k+1)a_{2(k+1)}x^{2k+1} - 2\sum_{k=0}^\infty a_{2k}x^{2k+1}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2(k+1)}-2a_{2k}\bigg)x^{2k+1}[/tex]

[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1}[/tex]

Notice that the right side of the DE is odd, so there is no 0-degree term, i.e. no constant term, so it follows that [tex]a_1=0[/tex].

The even series vanishes, so that

[tex](2k+1)a_{2k+1} - 2a_{2k-1} = 0[/tex]

for all integers k ≥ 1. But since [tex]a_1=0[/tex], we find

[tex]k=1 \implies 3a_3 - 2a_1 = 0 \implies a_3 = 0[/tex]

[tex]k=2 \implies 5a_5 - 2a_3 = 0 \implies a_5 = 0[/tex]

and so on, which means the odd-indexed coefficients all vanish, [tex]a_{2k+1}=0[/tex].

This leaves us with the odd series,

[tex]\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1} = 6\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}[/tex]

[tex]\implies 2(k+1)a_{2k+2} - 2a_{2k} = \dfrac{6(-1)^k}{(2k+1)!}[/tex]

We have

[tex]k=0 \implies 2a_2 - 2a_0 = 6[/tex]

[tex]k=1 \implies 4a_4-2a_2 = -1[/tex]

[tex]k=2 \implies 6a_6-2a_4 = \dfrac1{20}[/tex]

[tex]k=3 \implies 8a_8-2a_6 = -\dfrac1{840}[/tex]

So long as you're given an initial condition [tex]y(0)\neq0[/tex] (which corresponds to [tex]a_0[/tex]), you will have a non-zero series solution. Let [tex]a=a_0[/tex] with [tex]a_0\neq0[/tex]. Then

[tex]2a_2-2a_0=6 \implies a_2 = a+3[/tex]

[tex]4a_4-2a_2=-1 \implies a_4 = \dfrac{2a+5}4[/tex]

[tex]6a_6-2a_4=\dfrac1{20} \implies a_6 = \dfrac{20a+51}{120}[/tex]

and so the first four terms of series solution to the DE would be

[tex]\boxed{a + (a+3)x^2 + \dfrac{2a+5}4x^4 + \dfrac{20a+51}{120}x^6}[/tex]

190 of 7
6 7 8 9 10
-3
4
5
6
The slope of the line shown in the graph is
and the intercept of the line is

Answers

Answer:slope 2/3

Y-int 6

Step-by-step explanation:

i would like some help please i am stuck ​

Answers

Answer: -2(d) is the answer.

Step-by-step explanation:

x1 = 3

y1 = -5

x2 = -2

y2 = 5

slope (m) = rise/run = (y2 - y1)/(x2-x1)

                                =(5-(-5))/(-2-3)

                                 =   10/-5

                                 = -2

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