1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Yuliya22 [10]
4 years ago
5

match each binomial expression with the set of coefficients of the terms obtained by expanding the expression (2x+y)^4, (x+2y)^4

, (x+3y)^4, (3x+2y)^4

Mathematics
1 answer:
adell [148]4 years ago
8 0
To solve this we are going to use the fourth row of Pascal's triangle. 
First, we are going to expand our binomials:
(2x+y)^4=16x^4+32x^3y+24x^2y^2+8xy^3+y^4

(x+2y)^4=x^4+8x^3y+24x^2y^2+32xy^3+16y^4

(x+3y)^4=x^4+12x^3y+54x^2y^2+108xy^3+81y^4

(3x+2y)^4=81x^4+216x^3y+216x^2y^2+96xy^3+16y^4

We can conclude that you should match each binomial expansion <span>with the set of coefficients of the terms obtained by expanding the expression</span> as follows:
(2x+y)^4----\ \textgreater \ (16,32,24,8,1)

(x+2y)^4----\ \textgreater \ (1,8,24,32,16)

(x+3y)^4----\ \textgreater \ (1,12,54,108,81)

(3x+2y)^4----\ \textgreater \ (81,216,216,96,16)

You might be interested in
PLEASE HELP THIS IS WORTH 20 POINTS PLUS ILL GIVE BRAINIEST​
torisob [31]
First you need to find the second base (b2) by plugging all the values into the area of a trapezoid formula.

A= 1/2 (b1+b2)h

After plugging it in you will find that b2= 21.3.

The questions is asking perimeter so you add all the sides together...
14.4+18.5+14.7+21.3= 68.9 ft

Hope it helps
7 0
3 years ago
A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now
olga_2 [115]

Answer:

1)  288.8 km due North

2)  144.9 km due East

3)  323.1 km

4)  207°

Step-by-step explanation:

<u>Bearing</u>: The angle (in degrees) measured clockwise from north.

<u>Trigonometric ratios</u>

\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}

where:

  • \theta is the angle
  • O is the side opposite the angle
  • A is the side adjacent the angle
  • H is the hypotenuse (the side opposite the right angle)

<u>Cosine rule</u>

c^2=a^2+b^2-2ab \cos C

where a, b and c are the sides and C is the angle opposite side c

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

Draw a diagram using the given information (see attached).

Create a right triangle (blue on attached diagram).

This right triangle can be used to calculate the additional vertical and horizontal distance the ship sailed after sailing north for 250 km.

<u>Question 1</u>

To find how far North the ship is now, find the measure of the short leg of the right triangle (labelled y on the attached diagram):

\implies \sf \cos(75^{\circ})=\dfrac{y}{150}

\implies \sf y=150\cos(75^{\circ})

\implies \sf y=38.92285677

Then add it to the first portion of the journey:

⇒ 250 + 38.92285677... = 288.8 km

Therefore, the ship is now 288.8 km due North.

<u>Question 2</u>

To find how far East the ship is now, find the measure of the long leg of the right triangle (labelled x on the attached diagram):

\implies \sf \sin(75^{\circ})=\dfrac{x}{150}

\implies \sf x=150\sin(75^{\circ})

\implies \sf x=144.8888739

Therefore, the ship is now 144.9 km due East.

<u>Question 3</u>

To find how far the ship is from its starting point (labelled in red as d on the attached diagram), use the cosine rule:

\sf \implies d^2=250^2+150^2-2(250)(150) \cos (180-75)

\implies \sf d=\sqrt{250^2+150^2-2(250)(150) \cos (180-75)}

\implies \sf d=323.1275729

Therefore, the ship is 323.1 km from its starting point.

<u>Question 4</u>

To find the bearing that the ship is now from its original position, find the angle labelled green on the attached diagram.

Use the answers from part 1 and 2 to find the angle that needs to be added to 180°:

\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{Total\:Eastern\:distance}{Total\:Northern\:distance}\right)

\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{150\sin(75^{\circ})}{250+150\cos(75^{\circ})}\right)

\implies \sf Bearing=180^{\circ}+26.64077...^{\circ}

\implies \sf Bearing=207^{\circ}

Therefore, as bearings are usually given as a three-figure bearings, the bearing of the ship from its original position is 207°

8 0
2 years ago
Read 2 more answers
If $2,500 is invested at 12% annual interest, which is compounded continuously, what is the account balance after 3 years, assum
Lelechka [254]

Answer: 3,583.32

Step-by-step explanation:

Use money chimp calculator

4 0
3 years ago
A train covers certain distance in two parts. Distance covered in first part is 200% more than the distance covered in second pa
Rama09 [41]

Answer:

Part A speed=96*2=192km/h

Step-by-step explanation:

Two parts, parts A and part B

Part A=200% more than B

Let part B=x

Part A=200 more than B

=2x

Speed ratio=2:1

Average speed of train=64km/h

Let Part A speed=2x

Part B speed=x

A:B=2:1

Total ratio =3

Speed in the part A can be calculated thus:

Totatl speed= ratio of speed in part A / total ratio

64=2x/3

192=2x

x=96km/h

Part A speed=96*2=192km/h

4 0
4 years ago
The sum of the series 20c0 - 20c1 + 20c2 - 20c3+........+20c10 is : ____
spin [16.1K]
20C0 - 20C1 + 20C2 - 20C3 + . . . + 20C10 = 1 - 20 + 190 - 1,140 + 4,845 - 15,504 + 38,760 - 77,520 + 125,970 - 167,960 + 184,756 = 92,378
4 0
4 years ago
Other questions:
  • What are two equivalent fractions for 54/168?
    13·1 answer
  • Solve for X<br><br> 4X-(9-3X)=8X-1
    8·1 answer
  • The temperature is increasing steadily at a rate of 3°F every five hours. By how much does the temperature change each hour? -°F
    13·1 answer
  • The shadow of a child is 2 feet and the height of the child is 4.5 feet. At the same time of day, the building shadow is 52 feet
    9·2 answers
  • Express the polynomial 7x + 3x + 5 in standard form
    15·1 answer
  • Which method do you prefer to use to divide polynomials? Long division or Synthetic division? Explain why.
    9·2 answers
  • Suppose A and B are dependent events. If P(A) =0.3 and P(B/A) = 0.9 What is P(A∩B)
    14·1 answer
  • Y= 4/3 x -8<br>4x- 3y=24<br>what is the nature of the solution. to linear equations.​
    12·1 answer
  • A fair coin is tossed 3 times, and a player wins $3 if 3 tails occur; wins $2 if 2 tails occur and loses $3 if no tails occur. I
    13·1 answer
  • Match the values associated with this data set to their correct descriptions. {6, 47, 49, 15, 43, 41, 7, 36}
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!