Suppose that a is a one-dimensional array of ints with a length of at least 2. Which of the following code fragments successfully exchange(s) the values of the first two elements of a?I a[ 0 ] = a[ 1 ];a[ 1 ] = a[ 0 ];II int t = a[ 0 ];a[ 0 ] = a[ 1 ];a[ 1 ] = t;III a[ 0 ] = a[ 0 ] - a[ 1 ];a[ 1 ] = a[ 0 ] + a[ 1 ];a[ 0 ] = a[ 1 ] - a[ 0 ];A) I onlyB) II onlyC) III onlyD) I and II onlyE) II and III only

Answer:

Solution: -4, 4

Step-by-step explanation:

You need to do 2 cases:

Case 1: [tex]y\geq 0[/tex]

[tex]When\ y\geq 0 => | y | = y[/tex]

y - 19 = -15

y = 19 - 15

y = 4

Case 2: [tex]y< 0[/tex]

[tex]When\ y < 0 => |y| = -y[/tex]

-y -19 = -15

-19 + 15 = y

-4 = y

Both are solutions of the equation.

Answer:

-4, 4 would be your answer

Step-by-step explanation:

Answer:

The answers are below

Step-by-step explanation:

1) p² - 6p + n =  p² - 6p + (3)² = (p-3)²         n = 9  

2) m² - 8m + n  = m² - 8m + (4)² = (m-4)²       n= 16

3) v² + 24v + n  =  v² + 24v + (12)² = (v + 12)²   n=144

4) y² - 40y + n  = y² - 40y + (20)² = (y - 20)²    n= 400

5) v² - 36w + n =  v² - 36w + (18)² = (v - 18)²  n = 324

Answer:  At least 3

Step-by-step explanation:

Lets solve this problem by the pidgeon hole principle. Suppose that Matt pulls out 2 socks, if he gets a matching pair then he is done and that's it. But now suppose that he gets one black sock and one white sock, if he then pulls out a third sock it will be either white or black, in either case he would have gotten with a 100% certainty a matching pair, since he already has a black and a white sock. So the minimum number to ensure he gets a matching pair is 3 socks.

Answer:

Step-by-step explanation:

Answer:

Malia's allowance changed from $10.00 to $([tex]10-2x[/tex]) after buying ice cream cones.

Step-by-step explanation:

We are given the following information:

Malia's allowance to spend at carnival = $10.00

Ice cream ordered by her = 2

Let x dollars be the cost of one ice cream cone.

Total money spent on ice creams = [tex]2x[/tex]

Formula:

[tex]\text{Change in Malia's allowance} = \text{Total allowance} - \text{Money spent on ice cream cones}\\= 10 - 2x[/tex]

Thus, Malia's allowance changed from $10.00 to $([tex]10-2x[/tex]) after buying ice cream cones.

Answer:

1 pound = 0.454 kilograms

1 ounce = 0.0296 liters

1 ounce = 0.0283 kilograms

Step-by-step explanation:

You need to convert the quantities to their corresponding unit.

Convert pounds to kilograms (For the entrecote)

1 pound = 0.454 kilograms

Thus, 1 pound of entrecote = 0.454 kilograms

Ounces to liters (for the Béarnaise, since it is a liquid)

1 ounce = 0.0296 liters

2 ounces of Béarnaise = 0.0592 liters

Ounces to kilograms (for the asparagus)

1 ounce = 0.0283 kilograms

4 ounces of asparagus =0.1132 kilograms

By using unit conversion and dimensional analysis, the 1 pound of entrecôte converts to roughly 0.4536 kg, 2 ounces of Béarnaise to around 56.7 grams, and 4 ounces of asparagus to approximately 113.4 grams.

To answer your question about Chef Andy and his need to convert his recipes to metric units for his European kitchen staff, we'll need to follow some simple unit conversion steps:

By unit conversion and dimensional analysis, these are the metric conversions for each of those items. It's an essential method in fields like science, math, and of course, international cooking, where one often needs to convert from one type of unit to another.

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Answer:

[tex]2,050,000\ daily\ calls[/tex]

Step-by-step explanation:

The question in English is

An expanding mobile phone company handled eight hundred and fifty thousand calls a day during the quarter. In the next quarter it expects to reach one million and increase quarterly by the same amount over the next two years. How many daily calls do you expect to handle in two years?

Let

x ----> the number  of quarter

y ---> number of daily calls

we know that

we have the points

(0,850,000) and (1,1,000,000)

Find the slope m

[tex]m=(1,000,000-850,000)/(1-0)=150,000\ daily\ calls[/tex]

The linear equation that represent this situation is

[tex]y=150,000x+850,000[/tex]

How many daily calls do you expect to handle in two years?

In two years there are 8 quarterly

so

For x=8

substitute

[tex]y=150,000(8)+850,000=2,050,000\ daily\ calls[/tex]

Answer:

The answer to your question is: d = √73  

Step-by-step explanation:

data

Point E (-3, 4)

Point F (5, 1)

Formula

d = √((x2-x1)² + (y2-y1)²)

d = √(5-(-3))² + (1-4)²)          substitution

d= √ (5+3)² + (-3)²               simplify

d= √ 8² + -3²

d = √ 64 + 9

d = √73                               result

The distance between points E(5,1) and F(-3,4) in coordinate geometry is calculated by using the distance formula derived from the Pythagorean theorem. The computed distance between points E and F comes out to be √73 units.

The subject of this question is the distance formula in coordinate geometry, specifically between two points E(5, 1) and F(-3, 4). The formula is derived from the Pythagorean theorem and is represented as: D = √[(x₂ - x₁)² + (y₂ - y₁)²]. It is used to determine the simplest distance or 'path' between two points in a coordinate plane.

distance = √((x2 - x1)^2 + (y2 - y1)^2)

Substituting the coordinates of E and F, we have:

distance = √((-3 - 5)^2 + (4 - 1)^2)
distance = √((-8)^2 + (3)^2)

distance = √(64 + 9)

distance = √73

So, the exact distance between point E and point F is approximately √73 units.

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The complete question is given below:

What is the exact distance between point E and point F. The coordinates are E(5, 1) and F(-3, 4).

Answer: Jane started growing Sphere 3 hours ago

Step-by-step explanation:

Farm Rod starting population (Rsp) = 2

Farm Sphere starting population (Ssp) = 8

Let´s name "Rh" the quantity of hours since Rod started growing, and

"Sh" the quantity of hours since Sphere started growing.

And, let´s name "R" the population of farm Rod at 8 p.m. and "S" the population of farm Sphere at 8 p.m.

Population of Rod doubles every hour, therefore:

R = [tex]Rsp * 2^{Rh}[/tex]

R = [tex]2(2^{Rh})[/tex]

Population of Sphere is quadrupled every hour, therefore:

S = [tex]Ssp * 4^{Rh}[/tex]

S = [tex]8(4^{Rh})[/tex]

At 8 p.m. Jane found that R = S

Therefore, at 8 p.m:

[tex]2(2^{Rh})[/tex] = [tex]8(4^{Sh})[/tex]

dividing both sides by 2

[tex]2^{Rh} =4(4^{Sh})[/tex]

adding exponents

[tex]2^{Rh} = 4^{Sh+1}[/tex]

[tex]2^{Rh} =2^{2^{Sh+1} }[/tex]

the bases are the same; exponents must be the same

Rh = 2Sh + 2    (equation 1)

And we also know that Jane started growing Rod five hours before Sphere:

Rh = Sh + 5    (equation 2)

Replacing equation 2 into equation 1:

(Sh + 5) = 2Sh + 2

5 - 2 = 2Sh - Sh

3 = Sh, or

Sh = 3

Jane started growing Sphere 3 hours ago.  

Answer:

1/5 is the answer

Step-by-step explanation:

1/5 had blue

2/5 had purple

1/5 had gray

add those together you get 4/5

leaving you with 1/5 who chose different backgrounds.

Answer:

264.

Step-by-step explanation:

Let x represent the number of trout in the lake.

We have been given that a group of naturalists catch, tag and release 121 trout into a lake. The next day they catch and release 48 trout, of which 22 had been tagged.

Using proportions, we will get:

[tex]\frac{\text{Total trouts}}{\text{Tagged trouts}}=\frac{48}{22}[/tex]

[tex]\frac{x}{121}=\frac{48}{22}[/tex]

[tex]\frac{x}{121}*121=\frac{48}{22}*121[/tex]

[tex]x=\frac{24}{11}*121[/tex]

[tex]x=24*11[/tex]

[tex]x=264[/tex]

Therefore, there would be approximately 264 trout in the lake.

264 trouts are estimated to be in the lake.

Using proportion, we write: Tagged trout/Total Trout

This becomes 22/48.

In order to use cross multiplication, we write another fraction using x, the total number of trout in the lake: 121/x.

Now we have 121/x=22/48.

We cross multiply.

121*48=22x.

(121*48)/22 becomes:

264.

Answer:

1). Option E

2). Option H

3). Option C

4). Option J

5). Option C

Step-by-step explanation:

1). (5x + 3xy + 4y) + (4x - 2xy - 2y)

= (5x + 4x) + (3xy - 2xy) + (4y - 2y)

= 9x + xy + 2y

Option E. is the answer.

2). 5y + 2z + 3x²- (2y - 2z + 4x)

= 5y - 2y - 4x + 2z + 2z + 3x²

= 3y - 4x + 4z + 3x²

Option H is the answer.

3). (x - 1)(x + 1)(x - 1)

= (x² - 1)(x - 1)    [Since (a - b)(a + b) = a² - b²]

= x²(x - 1) - 1(x - 1)

= x³ - x² - x + 1

Option C is the answer.

4). (2a + 6)²

= 4a² + 24a + 36

[Since (a + b)² = a² - 2ab + b²]

Option J. is the answer.

5). For subtraction,

(6 - 1)³ - (4 - 1)²

= 5³ - 3²

= 125 - 9

= 116

But the result is 200 so operation subtraction is not the answer.

For Multiplication,

(6 × 1)³ - (4 × 1)²

= 6³ - 4²

= 216 - 16

= 200

For division,

(6 ÷ 1)³ - (4 ÷ 1)²

= 6³ - 4²

= 216 - 16

= 200

Therefore, Option C. is the answer.

The two region under whose we have to find area is

 f(x)=x

 [tex]g(x)=x^{\frac{1}{n}}\\\\x \geq 0\\\\n\geq 2[/tex]

The Point of Intersection of two curves is always , x=0 and x=1.

Area of the Region

    =Area under the line - Area Under the curve g(x), when n take different value, that is ≥2.

[tex]\rightarrow[- \int\limits^1_0 {x} \, dx + \int\limits^1_0 {x^{\frac{1}{2}} \, dx]+[ -\int\limits^1_0 {x} \, dx + \int\limits^1_0 {x^{\frac{1}{3}} \, dx]+[ -\int\limits^1_0 {x} \, dx + \int\limits^1_0 {x^{\frac{1}{4}} \, dx]+[ -\int\limits^1_0 {x} \, dx + \int\limits^1_0 {x^{\frac{1}{5}} \, dx]+......[/tex]

[tex]=\int\limits^1_0({x^{\frac{1}{2}}+x^{\frac{1}{3}}+x^{\frac{1}{4}}+x^{\frac{1}{5}}+.......+x^{\frac{1}{200}}}) \, dx -\int\limits^1_0 ({x}+{x}+{x}+{x}...........+200\text{times}) \, dx[/tex]

When, n=200, the first quadrant is completely occupied by the curve

       [tex]g(x)=x^{\frac{1}{n}},x\geq 0\\\\2\leq n \leq 200[/tex]

[tex]=\int\limits^1_0{x^{\frac{1}{n}} \, dx=\frac{n\times x^{(\frac{1}{n}+1)}}{n+1}\left \{ {{x=1} \atop {x=0}} \right.}\\\\= \frac{n}{n+1}\\\\\int\limits^1_0{x^{n} \, dx=\frac{x^{n+1}}{n+1}\left \{ {{x=1} \atop {x=0}} \right.}\\\\=\frac{1}{n+1}[/tex]

[tex]=\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+...........+\frac{200}{201}-199 \times \frac{1}{2}\\\\=1-\frac{1}{3}+1-\frac{1}{4}+1-\frac{1}{5}+...........+1-\frac{1}{201}-99.5\\\\=199-99.5+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...........+\frac{1}{201}\\\\=99.5+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...........+\frac{1}{201}-1-\frac{1}{2}\\\\=98+\frac{1}{d} \times\ln(\frac{2a+(2n-1)d}{2a-d})\\\\=98+\ln(\frac{2 \times 1+(2\times 201-1)\times 1}{2\times 1-1})\\\\=98+\ln 403\\\\=98+6({\text{approx}})\\\\=104 \text{square units}[/tex]

Sum of n terms of Harmonic Progression is

     [tex]=\frac{1}{a}+\frac{1}{a+d}+\frac{1}{a+2d}+\frac{1}{a+3d}+\frac{1}{a+4d}.....+\frac{1}{a+(n-1)d}\\\\=\frac{1}{d} \times \ln(\frac{2a+(2n-1)d}{2a-d})[/tex]    

Answer:

The total change in the stock market from the beginning of the day to the end of the day is [tex]-89\frac{3}{4}[/tex] or the stock market goes down [tex]89\frac{3}{4}[/tex]

Step-by-step explanation:

At the beginning of the day the stock market goes up [tex]30\frac{1}{2}[/tex] points. This means, we have to add [tex]30\frac{1}{2}[/tex]

At the end of the day the stock market goes down [tex]120\frac{1}{4}[/tex] points. This means, we have to subtract [tex]120\frac{1}{4}[/tex]

Change [tex]=+30\frac{1}{2}-120\frac{1}{4}=(30-120)+\left(\frac{1}{2}-\frac{1}{4}\right)=-90+\frac{1}{4}=-89\frac{3}{4}[/tex]

So, the total change in the stock market from the beginning of the day to the end of the day is [tex]-89\frac{3}{4}[/tex] or the stock market goes down [tex]89\frac{3}{4}[/tex]

Answer 89 3/4

I dont know what to do help me

Answer:

Step-by-step explanation:

As the problem statement suggests, look up the rules if you have to.

In the case of trailing zeros, the decimal point makes them significant.

__

In a number without a decimal point, such as 300, the zeros are sometimes significant. (Different authors may tell you the zeros are never significant in such a number. The problem comes when, for example, you're trying to represent 296 rounded to the nearest 10. That is 30. tens--with 2 significant digits, just as 286 rounded to the nearest 10 would be 290--with 2 significant digits. In some numbers, you cannot tell the difference between a 0 that is a placeholder and a 0 that is present because the value of the number cannot be represented exactly any other way.)

Answer:

The drawer contains 40 nickels, 20 dimes and 14 quarters.

Step-by-step explanation:

Let d, q and n represent number of dimes, quarters and nickels respectively.

We have been given that there are twice as many nickels as dimes.

[tex]n=2d...(1)[/tex]

Further, the number of dimes and quarters sum to 34.

[tex]d+q=34...(2)[/tex]

As the change drawer contains $7.50 made up entirely of quarters, nickels, and dimes.

[tex]0.10d+0.25q+0.05n=7.50...(3)[/tex]  

From equation (2), we will get:

[tex]q=34-d[/tex]

Substituting equation (1) and equation (2) in equation (3), we will get:

[tex]0.10d+0.25(34-d)+0.05(2d)=7.50[/tex]  

[tex]0.10d+8.50-0.25d+0.10d=7.50[/tex]  

[tex]-0.05d+8.50=7.50[/tex]  

[tex]-0.05d+8.50-8.50=7.50-8.50[/tex]  

[tex]-0.05d=-1[/tex]  

[tex]\frac{-0.05d}{-0.05}=\frac{-1}{-0.05}[/tex]  

[tex]d=20[/tex]  

Therefore, drawer contains 20 dimes.

Substitute [tex]d=20[/tex] in equation (1):

[tex]n=2d[/tex]

[tex]n=2(20)[/tex]

[tex]n=40[/tex]

Therefore, drawer contains 40 nickels.

Substitute [tex]d=20[/tex] in equation (2):

[tex]20+q=34[/tex]

[tex]20-20+q=34-20[/tex]

[tex]q=14[/tex]

Therefore, drawer contains 14 quarters.

Answer:

There are 40 nickels, 14 quarters and 20 dimes.

Step-by-step explanation:

A change drawer contains $7.50

Let the number of quarters in the drawer = q

let the number of nickels in the drawer = n

and number of dimes = d

So, (0.25q + 0.05n + 0.10d) = 7.5

By dividing the equation by 0.50

5q + n + 2d = 150 ---------(1)

Now statement says " There are twice as nickels as dimes"

n = 2d -------(2)

And "the number of dimes and quarters sum to 34"

d + q = 34 -------(3)

We replace n = 2d from equation (2) in equation (1)

5q + 2d + 2d = 150

5q + 4d = 150 ---------(4)

Multiply equation (3) by 4 and subtract it from equation (4)

(5q + 4d) - 4(d + q) = 150 - 34×4

5q - 4q + 4d - 4d = 150 - 136

q = 14

We plug in the value of q in equation (3)

d + 14 = 34

d = 34 - 14

d = 20

Since n = 2d

So n = 2×20

n = 40

Therefore, there are 40 nickels, 14 quarters and 20 dimes.

Answer:

A.A line segment consist of two points and all points in between .

Step-by-step explanation:

Definition of  a line segment : It is a portion of a line .It has two end points and all point contain between them.It is shortest distance between two points.

A.A line segment consist of two points and all points in between .

By definition of  a line is is true.

Hence, option A is true.

B.A line segment is a set of a points that extend infinitely far in one direction.

No, line segment can not be extend infinitely in any direction.

Therefore, option is false.

C.A line segment is a set of points that extend far in both directions.

No, by definition of line segment , it is false.

D.A line segments is the set of points in a plane that are equidistant from a given point.

By definition of line segment , it is false.

Answer:

6 mph

Step-by-step explanation:

g(x) = x^4 − 4x^2 + 7x − 8

when x = 2,

g(2) = (2)^4 - 4(2^2) + 7 (2) - 8

= 16 - 16 + 14 - 8

= 6 mph

Hey!

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

Solution:

We know that 2 = x

g(x) = (2)^4 - 4(2)^2 + 7(2) - 8

g(x) = 16 - 16 + 14 - 8

g(x) = 0 + 14 - 8

g(x) = 14 - 8

g(x) = 6

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

Answer:

6 miles per hour!

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

Hope This Helped! Good Luck!

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Answer:

At the maxium point

Step-by-step explanation:

A maxium point is a point where the function gets the higher value. For example, in the basic sine curve, we can find a maxium point when [tex]x=\frac{\pi }{2}[/tex]. We won't find any maxium point when x=0.

As you can see on the graph I attached, the zeros of a basic sine curve (or, said in other words, the points were x=0), are at the beginning, ath the middle and at the end of the function's graphic.

Answer:

6

Step-by-step explanation:

In a square, all diagonals, vertices, sides, and angles are congruent, and since 36 is its area, take the square root of 36, giving you 6.

* When we are talking about length, we want the NON-NEGATIVE root.

I am joyous to assist you anytime.

Initial population size is 188 and after 9 years we get 1259 fishes.

We have given that the function

[tex]P(t)=\frac{1500}{1+7e^(-0.04t)}[/tex]

Where t is the number of years from the time the species was added to the lake.

We have to find the initial population size of the species and the population size after 9 years.

where p(0) is the population size of species and p(9) is the population size after 9 years.

Therefore we have the given function is at t=0

[tex]P(0)=\frac{1500}{1+7e^(-0.04(0)))}=\frac{1500}{1+07e^0} \\=\frac{1500}{1+7} \\=\frac{1500}{8} \\=187.5\\=188[/tex]

at t=9 we have

[tex]P(9)=\frac{1500}{1+7ex^{-0.4(9)} }\\=\frac{1500}{1+7e^{-3.6} } \\=\frac{1500}{1.191}\\ =1259.16\\=1259 fishes[/tex]

Therefore after 9 years we get 1259 fishes.

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The initial population size of the species is approximately 188. After 9 years, the population size is approximately 1260.

To find the initial population size of the species, we need to evaluate the population function ( P(t) ) at ( t = 0 ).

Given the population function:

[tex]\[ P(t) = \frac{1500}{1 + 7e^{-0.4t}} \][/tex]

1. **Initial population size (( P(0) )):**

  Substitute ( t = 0 ) into the function:

[tex]\[ P(0) = \frac{1500}{1 + 7e^{-0.4 \cdot 0}} \][/tex]

[tex]\[ P(0) = \frac{1500}{1 + 7e^0} \][/tex]

 [tex]\[ P(0) = \frac{1500}{1 + 7 \cdot 1} \][/tex]

[tex]\[ P(0) = \frac{1500}{1 + 7} \][/tex]

[tex]\[ P(0) = \frac{1500}{8} \][/tex]

[tex]\[ P(0) = 187.5 \][/tex]

  Round to the nearest whole number:

[tex]\[ P(0) \approx 188 \][/tex]

  So, the initial population size is[tex]\( \boxed{188} \).[/tex]

2. **Population size after 9 years (( P(9) )):**

  Substitute ( t = 9 ) into the function:

[tex]\[ P(9) = \frac{1500}{1 + 7e^{-0.4 \cdot 9}} \][/tex]

[tex]\[ P(9) = \frac{1500}{1 + 7e^{-3.6}} \][/tex]

  Calculate [tex]\( e^{-3.6} \):[/tex]

[tex]\[ e^{-3.6} \approx 0.02732 \][/tex]

  Substitute this value back into the equation:

[tex]\[ P(9) = \frac{1500}{1 + 7 \cdot 0.02732} \][/tex]

[tex]\[ P(9) = \frac{1500}{1 + 0.19124} \][/tex]

 [tex]\[ P(9) = \frac{1500}{1.19124} \][/tex]

[tex]\[ P(9) \approx 1260 \][/tex]

  Round to the nearest whole number:

[tex]\[ P(9) \approx \boxed{1260} \][/tex]

  So, the population size after 9 years is [tex]\( \boxed{1260} \).[/tex]

Answer:

y = - [tex]\frac{3}{4}[/tex] x + 3

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (0, 3) and (x₂, y₂ ) = (4, 0) ← 2points on the line

m = [tex]\frac{0-3}{4-0}[/tex] = - [tex]\frac{3}{4}[/tex]

Note the line crosses the y- axis at (0, 3 ) ⇒ c = 3

y = - [tex]\frac{3}{4}[/tex] x + 3 ← equation of line

The cheapest cost to fill the Paolo's tank = $33.75

We have Paolo who bought a used car for $3,500 and wants to fill it up with gasoline. His gas tank holds 15 gallons and is currently empty. Paolo has $100. One gas station sells gas for $2.50 per gallon. The station across the street sells gas for $2.25 per gallon.

We have to determine the cheapest cost to fill up Paolo's tank.

The price estimation is important in mathematics as it helps us to create mathematical relations among the various factors on which the price depends in order to extract accurate results.

According to question, we have -

Cost of Car = $3500.

Car's gasoline capacity = 15 gallons.

Amount of money with Paolo = $100.

Cost of gas at station A = $2.25.

Cost of gas at station B = $2.50.

Case 1 - If Paolo goes to Station A

Total cost to fill the tank = 2.25 x 15 = $33.75

Case 2 - If Paolo goes to Station B

Total cost to fill the tank = 2.50 x 15 = $37.50

Hence, the cheapest cost to fill the Paolo's tank = $33.75

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Answer: All data will be sent in 89.39 hours

Step-by-step explanation:

Data transfer rate: 100 Mbps=100 [tex]\frac{Mb}{s}[/tex] ×[tex]\frac{1Tb}{1024^{2} Mb}[/tex]×[tex]\frac{3600s}{h}[/tex]=0.3433 [tex]\frac{Tb}{h}[/tex]

The pigeon can fly 1000 km/day and it needs to fly 400 km (round trip).

Pigeon rate: [tex]\frac{1000km}{24h}[/tex]=41.67 [tex]\frac{km}{h}[/tex]

Time of round trip: 400 km÷[tex]\frac{1000km}{24h}[/tex]=9.6 h

So the pigeon can send 1 Tb every 9.6 hours.

If we sum the rates we can get the time for sending all the data:

40 Tb= (0.3433 Tb/h + 1 Tb/9.6h)×t

T= 89.39 hours

Answer:

The magnitude of A is 17.46 m and B is 1.50 m

Step-by-step explanation:

If the vector sum A+B+C =0, then the sum of the projection of the vector in axes x- is zero and the sum of the projection of the vector in the axes y- is also zero.

Ax+Bx+Cx = 0

Ay+By+Cy = 0

|Ax| = cos 41.9 * |A|

|Ay| = sin 41.9 * |A|

|Bx| = cos 28.2 * |B|

|By| = sin 28.2 * |B|

|Cx| = 0

|Cy| = 22.2

Ax+Bx+Cx = 0

|Ax|-|Bx|+0 =0

the vector Ax is in the positive direction of the x- axes and Bx in the negative direction and C do not have a component in the x- axes

cos 41.9 * |A| - cos 28.2 * |B| = 0 (I)

Ay+By+Cy = 0

|Ay|+|By|-|Cy|=0

the vector Ay and By are the positive direction of the y- axes and Cy in the negative direction

sin 41.9 * |A| + sin 28.2 * |B| - 22 =0 (II)

Now we have a system of 2 (I and II) equations and 2 variables (|A| and |B|)

cos 41.9 * |A| - cos 28.2 * |B| = 0

sin 41.9 * |A| + sin 28.2 * |B| = 22

cos 41.9 * |A| = cos 28.2 * |B|

|A| = cos 28.2 * |B| / cos 41.9

sin 41.9 * |A| + sin 28.2 * |B| = 22

sin 41.9 *  cos 28.2 * |B| / cos 41.9 + sin 28.2 * |B| = 22

tg 41.9 * cos 28.2 * |B| + sin 28.2 * |B| = 22

(tg 41.9 * cos 28.2 + sin 28.2) * |B| = 22

|B| = 22 / (tg 41.9 * cos 28.2 + sin 28.2)

|B| = 17.46

|A| = 1.50

The magnitude of vector A and the magnitude of vector B is 20.6198 and 17.4146 respectively.

The quantity which has magnitude, direction and follows the law of vector addition is called a vector.

Given

Vector C has a magnitude of 22.2 m and points in the negative y‑direction.

Vectors A and B both have positive y‑components and make angles of α=41.9° and β=28.2° with the positive and negative x-axis.

Let the vectors A, B, and C be concurrent.

Then vectors can be resolved in x-direction and y-direction.

Vectors in y-direction

[tex]\rm A\ sin 41.9^o + B \ sin28.2^o = C\\0.6678\ A\ \ +\ 0.4726\ B \ = 22[/tex].....eq(1)

Vectors in x-direction

[tex]\rm A \ cos41.9^o = B \ cos 28.2^o\\0.74431 \ \ A = B \ \ 0.8813[/tex].....eq(2)

From equations 1 and 2, we get

A = 20.6199 and B = 17.4147

Thus, the magnitude of vector A and the magnitude of vector B is 20.6198 and 17.4146 respectively.

More about the vector link is given below.

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Answer:

  -$5,500

Step-by-step explanation:

  The change in the one year from 2009 to 2010 was ...

  18,500 -24,000 = -5,500

The annual rate of change in the boat's value was -$5,500.

Thus, the annual rate of change of the boat's value is $5,500 per year.

To find the annual rate of change of the boat's value, we need to determine how much the value decreased from 2009 to 2010 and then calculate the yearly decrease. The boat's value in 2009 was $24,000; in 2010, it was $18,500.

First, calculate the decrease in the boat's value from 2009 to 2010:

Decrease = Initial Value - Final Value

Decrease = $24,000 - $18,500 = $5,500

Since this decrease occurred over one year, the annual rate of change in the boat's value is $5,500 per year.

Therefore, the annual rate of change of the boat's value is $5,500 per year.

-200,-100,-25,0,30

Step-by-step explanation:

negatives will always be less than positives with negatives you put greatest to least and then 0 and then positives from least to greatest

In order to get the answer you have to remember that bigger negatives are actually less then smaller negatives so -200 would be less then -25.

[tex]-100, -200, 0, -25, 30[/tex]

[tex]-200 < -100[/tex]

[tex]-100 <-25[/tex]

[tex]-25 < 0[/tex]

[tex]0 <30[/tex]

[tex]= -200,-100,-25,0,30[/tex]

Therefore your answer is "-200,-100,-25,0,30."

Hope this helps.

Answer:

It can be any digit between 7, 8 and 9.

Step-by-step explanation:

The given data set is: 2​, 4​, 5​, 7​, 7​, 10

Here mode is 7, So we have to make our median equal to 7.

If one more data point will be included in the data set then we will have an odd number of observations.

For getting a median equal to 7 we have to add a number between 7 and 10 or it can be equal to 7.

Now, if we add any data points between 7, 8 and 9 then our median and mode will equal to 7.

Hence, the any one number between 7, 8 and 9 can be included in given data set to get both median and mean equal.

Further, the Median is the middle observation of the given data. It can be found by following steps:

Arranging data in ascending or descending order.

Taking the average of the middle two values if the total number of observations is even, and this average is our median.

or, if we odd number of observation then the most middle value is our median.

The mode is the observation which has a high number of repetitions (frequency).

The entry that should be added to the data set 2, 4, 5, 7, 7, and 10 to make the median and mode equal is 7. This is because the current mode is 7, and to have the median also equal to 7, a 7 must be added to the data set.

The goal is to add an entry between 1 and 10 to your data set (which currently includes 2​, 4​, 5​, 7​, 7​, and 10) that will make the median and mode equal. To start, let's understand what the median and mode are. The median is the middle value in an ordered data set. If there's an even number of values in the set, the median is the average of the two middle values. The mode, on the other hand, is the most frequent value or values in a data set.

In your original data set, the mode is 7, as it appears twice. Currently, with six numbers in the data set, the median would be the average of the 3rd and 4th numbers when ordered (5 and 7), so the median is 6. To match the mode (7), a 7 needs to be added to the data set. Now the median is (7 + 7) / 2 = 7, which matches the mode of 7. So, the number you need to add to your dataset to make the median and mode both equal 7 is 7.

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Answer:

ST = 20

Step-by-step explanation:

RT is the sum of RS and ST

Replacing with length you get:

17 + x + 6 = 3x - 56

17 + 6 + 5 = 3x - x

28 = 2x

14 = x

ST = x + 6 = 14 + 6 = 20