What is a transformation and 3 types of transformations.

Answers

Answer 1

a transformation happen when the shape changes.
There are four transformations
translation
rotation
reflection
dilation

translation, rotation and reflection do not change the shape or the size of the shape and are known as rigid motions

Answer 2

A transformation is the movement or change in size of a shape on a graph.

The four types:
Dilation: Change of size.
Rotation: A clockwise turn in each of the four quadrants in a graph.
Translation: Simply relocating the shape.
Reflection: Mirroring the shape across either the y or x axis.

Hope that helps!

Related Questions

How do you figure out what 1/10 of 1,7000.000 km squared is?

Answers

The first thing you would do is square 17,000, which is 298,000,000. Then you would multiply 298,000,000 by 0.10. (or divide the number by 10)

This would get you 28,900,000.

Suppose you have two credit cards. The first has a balance of $415 and a credit limit of $1,000. The second has a balance of $215 and a credit limit of $750. What is your overall credit utilization?

Answers

Compute for the total balance:

total balance = $415 + $215 = $630

Then we compute for the total credit limit:

total credit limit = $1,000 + $750 = $1,750

The credit utilization would simply be the percentage ratio of total balance over total credit limit. That is:

credit utilization = ($630 / $1,750) * 100%

credit utilization = 36%

Round 5836197 to the nearest hundred

Answers

5838200 is to the nearest hundred.

Answer:

5836200.

Step-by-step explanation:

Given : 5836197 .

To find : Round 5836197 to the nearest hundred.

Solution : We have given 5836197

Step 1 : First, we look for the rounding place which is the hundreds place.

Step 2 : Rounding place is 97.

Step 3 : 97 is greater than 50 then it would be rounded up mean next number to 97 would be increase to 1 and 97 become 00.

Step 4 : 5836200.

Therefore, 5836200.

Given sina=6/7 and cosb=-1/6, where a is in quadrant ii and b is in quadrant iii , find sin(a+b) , cos(a-b) and tan(a+b)

Answers

[tex]\bf sin(a)=\cfrac{\stackrel{opposite}{6}}{\stackrel{hypotenuse}{7}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{7^2-6^2}=a\implies \pm\sqrt{13}=a \\\\\\ \textit{now, angle "a" is in the II quadrant, where the adjacent is negative} \\\\\\ -\sqrt{13}=a\qquad \qquad \boxed{cos(a)=\cfrac{-\sqrt{13}}{7}}[/tex]

now, keep in mind that, the hypotenuse is just a radius unit, and thus is never negative, so if a fraction with it is negative, is the other unit. A good example of that is the second fraction here, -1/6, where the hypotenuse is 6, therefore the adjacent side is -1. Anyhow, let's find the opposite side to get the sin(b).

[tex]\bf cos(b)=\cfrac{\stackrel{adjacent}{-1}}{\stackrel{hypotenuse}{6}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a \\\\\\ \textit{now, angle "b" is in the III quadrant, where the opposite is negative} \\\\\\ -\sqrt{35}=b\qquad \qquad \boxed{sin(b)=\cfrac{-\sqrt{35}}{6}}[/tex]

now

[tex]\bf \textit{Sum and Difference Identities} \\ \quad \\ sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}}) \\ \quad \\ sin({{ \alpha}} - {{ \beta}})=sin({{ \alpha}})cos({{ \beta}})- cos({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} + {{ \beta}})= cos({{ \alpha}})cos({{ \beta}})- sin({{ \alpha}})sin({{ \beta}}) \\ \quad \\ cos({{ \alpha}} - {{ \beta}})= cos({{ \alpha}})cos({{ \beta}}) + sin({{ \alpha}})sin({{ \beta}}) \\ \quad \\ [/tex]

[tex]\bf tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\qquad tan({{ \alpha}} - {{ \beta}}) = \cfrac{tan({{ \alpha}})- tan({{ \beta}})}{1+ tan({{ \alpha}})tan({{ \beta}})}[/tex]

[tex]\bf sin(a+b)=\cfrac{6}{7}\cdot \cfrac{-1}{6}+\cfrac{-\sqrt{13}}{7}\cdot \cfrac{-\sqrt{35}}{6}\implies \cfrac{-1}{7}+\cfrac{\sqrt{455}}{42} \\\\\\ \cfrac{-6+\sqrt{455}}{42}\\\\ -------------------------------\\\\ cos(a-b)=\cfrac{-\sqrt{13}}{7}\cdot \cfrac{-1}{6}+\cfrac{6}{7}\cdot \cfrac{-\sqrt{35}}{6}\implies \cfrac{\sqrt{13}}{42}-\cfrac{\sqrt{35}}{7} \\\\\\ \cfrac{\sqrt{13}-6\sqrt{35}}{42}[/tex]

[tex]\bf -------------------------------\\\\ tan(a)=\cfrac{\frac{6}{7}}{-\frac{\sqrt{13}}{7}}\implies -\cfrac{6}{\sqrt{13}}\implies -\cfrac{6\sqrt{13}}{13} \\\\\\ tan(b)=\cfrac{\frac{-\sqrt{35}}{6}}{\frac{-1}{6}}\implies -\sqrt{35}\\\\ -------------------------------\\\\[/tex]

[tex]\bf tan(a+b)=\cfrac{-\frac{6}{\sqrt{13}}-\sqrt{35}}{1-\left( -\frac{6}{\sqrt{13}} \right)\left( -\sqrt{35} \right)}\implies \cfrac{\frac{-6-\sqrt{455}}{\sqrt{13}}}{1-\frac{6\sqrt{35}}{\sqrt{13}}} \\\\\\ \cfrac{\frac{-6-\sqrt{455}}{\sqrt{13}}}{\frac{\sqrt{13}-6\sqrt{35}}{\sqrt{13}}}\implies \cfrac{-6-\sqrt{455}}{\sqrt{13}-6\sqrt{35}}[/tex]

and now, let's rationalize the denominator of that one, hmmm let's see

[tex]\bf \cfrac{-6-\sqrt{455}}{\sqrt{13}-6\sqrt{35}}\cdot \cfrac{\sqrt{13}+6\sqrt{35}}{\sqrt{13}+6\sqrt{35}} \\\\\\ \cfrac{-6\sqrt{13}-36\sqrt{35}-\sqrt{5915}-6\sqrt{15925}}{({\sqrt{13}-6\sqrt{35}})({\sqrt{13}+6\sqrt{35}})} \\\\\\ \cfrac{-6\sqrt{13}-36\sqrt{35}-13\sqrt{35}-210\sqrt{13}}{(\sqrt{13})^2-(6\sqrt{35})^2} \\\\\\ \cfrac{-216\sqrt{13}-49\sqrt{35}}{13-210}\implies \cfrac{-216\sqrt{13}-49\sqrt{35}}{-197} \\\\\\ \cfrac{216\sqrt{13}+49\sqrt{35}}{197}[/tex]

sin(a+b) = -1/7 +√455/42 = 0.8721804464845457

cos(a-b) = √13/42 - √35/7 = -0.7761476987942811

tan(a+b )= (6√13/13 + √35) / (1 - 6√455/13) = -0.525

Given sin(a) = 6/7 and cos(b) = -1/6, with a in quadrant II and b in quadrant III, we need to utilize trigonometric identities to find sin(a+b), cos(a-b), and tan(a+b).

Firstly, since a is in quadrant II, cos(a) is negative. We use the identity sin²(a) + cos²(a)=1 to find cos(a):

cos(a) = -√(1 - sin²(a)) = -√(1 - (6/7)²) = -√(1 - 36/49) = -√(13/49) = -√13/7

Similarly, since b is in quadrant III, sin(b) is also negative. We use the identity sin²(b) + cos²(b)=1 to find sin(b):

sin(b) = -√(1 - cos²(b)) = -√(1 - (-1/6)²) = -√(1 - 1/36) = -√(35/36) = -√35/6

Now we can use the angle addition and subtraction formulas:

1. sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

sin(a + b) = (6/7)(-1/6) + (-√13/7)(-√35/6) = -1/7 + √(13×35)/(7×6) = -1/7 + √455/42 = -1/7 +√455/42

2. cos(a - b) = cos(a)cos(b) + sin(a)sin(b)

cos(a - b) = (-√13/7)(-1/6) + (6/7)(-√35/6) = √13/(7×6) - (6√35)/(7×6) = √13/42 - √35/7

3. tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a)tan(b))

Using tan(a) = -sin(a)/cos(a) = -(6/7)/(-√13/7) = 6/√13 and tan(b) = sin(b)/cos(b) = (-√35/6)/(-1/6) = √35:tan(a + b) = (6/√13 + √35) / (1 - (6/√13)(√35)) = (6√13/13 + √35) / (1 - 6√455/13)

what is the answer of 9X23+3X39-28=n

Answers

9x23+3x39-28=n

n= 296

Answer: n=296

Step-by-step explanation: 9 x 23 + 3 x 39 - 28 + n = 296

Hope this helps! <3

Forty percent of the players on a soccer team are experienced players. The coach wants to increase this percent. There are 10 players on the team now. The expression below represents the percent of experienced players on the team after the coach adds x experienced players.

100(x+4)/x+10

What is:
a. Original ratio of experienced to total players
b. Final ratio of experienced to total players
c. Final total number of players

Choices

4/10
x+10
4
10
x+4/x+10

Answers

Part A:

Given that forty percent of the players on a soccer team are experienced players.

Then, the original ratio of experienced to total players is given by 40/100 = 4/10

Part B:

Given that the expression [tex] \frac{100(x+4)}{x+10} [/tex] represents the percent of experienced players on the team after the coach adds x experienced players.

Then, the final ratio of experienced to total players is given by:

[tex] \frac{\frac{100(x+4)}{x+10}}{100} =\frac{x+4}{x+10}[/tex]

Part C:

Given that there are 10 players on the team now, and that the coach adds x experienced player, then the number of players on the team now is given by:

10 + x.

D is the midpoint of CE.E has coordinates (-3,-2), and D has coordinates (2 1/2, 1). Find the coordinate of C.

Answers

midpoint formula : (x1 + x2) / 2, (y1 + y2) / 2
(-3,-2)....x1 = -3 and y1 = -2
(x,y)......x2 = x and y2 = y
now we sub
(-3 + x) / 2, (-2 + y) / 2

(-3 + x) / 2 = 2 1/2
-3 + x = 5/2 * 2
-3 + x = 5
x = 5 + 3
x = 8

(-2 + y) / 2 = 1
-2 + y = 1 * 2
-2 + y = 2
y = 2 + 2
y = 4

so the coordinates of point C are : (8,4)

The coordinate of point C on line CE will be (8, 4).

What is Coordinates?

A pair of numbers which describe the exact position of a point on a cartesian plane by using the horizontal and vertical lines is called the coordinates.

Given that;

D is the midpoint of CE.

E has coordinates (-3,-2), and D has coordinates (2 1/2, 1).

Now, By the definition of midpoint;

Let the coordinate of point C = (x, y)

Then,

((x + (-3))/2 , (y + (-2))/2) = (2 1/2, 1)

By comparison we get;

x + (-3) / 2= 2 1/2

x - 3 = 2 (5/2)

x - 3 = 5

x = 3 + 5

x = 8

And, (y + (-2))/2 = 1

y - 2 = 2

y = 2 + 2

y = 4

Thus, The coordinate of point C = (x, y) = (8, 4)

So, The coordinate of point C on line CE will be (8, 4).

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4r+3=r+21

How do I do this?

Answers

You first:
4r+3-3=r+21-3
4r=r+18
4r-r=r-r+18
3r=18
3r/3=18/3
r=6
remember that what you do in one side, do it in the other

simplify: 4r + 3 = r + 21, then reorder the terms: 3 + 4r = r + 21.
Reorder the terms 3 + 4r = 21 + r. Solve: 3 + 4r = 21 + r.
Move all terms containing r to the left, all other terms to the right.
Solving for variable 'r'.
Add '-1r' to each side of the equation. 3 + 4r + -1r = 21 + r + -1r.
Combine like terms: 4r + -1r = 3r3 + 3r = 21 + r + -1r
Combine like terms: r + -1r = 0 3 + 3r = 21 + 0 3 + 3r = 21
Add '-3' to each side of the equation. 3 + -3 + 3r = 21 + -3
Combine like terms: 3 + -3 = 0 0 + 3r = 21 + -3 3r = 21 + -3
Combine like terms: 21 + -3 = 18 3r = 18
Divide each side by '3'. r = 6 Simplifying r = 6

If the radius of a circle measures 2 inches, what is the measure of its diameter?

Answers

the diameter of the circle would be 4 inches

The diameter will equal 4 inches because the radius is half of the diameter...hope this helps..Good luck!

~~Alexis

write a fraction less than 1 with a denominator of 6 that is greater than 3/4

Answers

The answer would be 5/6.

Hope this helps!

Answer:= 5/6

Step-by-step explanation:hope this helps

"unable to determine because x can't take on the value 5.5"

Answers

k
f(x) = -------------------
x - (5.5)

is an example where there's no finite limit if x approaches 5.5.

The diagram represents a reduction of a triangle by using a scale factor of 0.8.

What is the height of the reduced triangle?
4.0 inches
4.8 inches
5.2 inches
7.5 inches

Answers

The reduction factor of 0.8 means the lengths in the reduced triangle are 0.8 times those of the original.

Then the original 6 inch length is reduced to 0.8×6 inches = 4.8 inches in the reduced triangle.

Answer:

4.8 inches

Step-by-step explanation:

The scale factors are used to convert a figure into another one with similar characteristics but different lengths, in this example is a triangle, and in order to calculate the measure of the height you just have to multiply the original height by the scale factor:

6 inches * scale factor

6 inches* 0.8= 4.8 inches

So the resultant triangle will have a height of 4.8 inches.

What is the expanded notation for 2.918?

Answers

2 + 0.9+ 0.01+ 0.008

I hope this helps.

Every two years, a manufacturer's total yearly sales decline 20%. She sells 400 total units in her first year. Rounded to the nearest unit, how many total units will she sell her ninth year?

Answers

[tex]\bf \textit{Periodic Exponential Decay}\\\\ A=I(1 - r)^{\frac{t}{p}}\qquad \begin{cases} A=\textit{accumulated amount}\\ I=\textit{initial amount}\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\\ p=period \end{cases}[/tex]

we know, the first year she sold 400 units, thus year 0, 400 units, t = 0, A = 400.

[tex]\bf \textit{Periodic Exponential Decay}\\\\ A=I(1 - r)^{\frac{t}{p}}\qquad \begin{cases} A=\textit{accumulated amount}\to &400\\ I=\textit{initial amount}\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\to &0\\ p=period \end{cases} \\\\\\ 400=I(1-r)^{\frac{0}{p}}\implies 400=I\cdot 1\implies 400=I[/tex]

therefore then

[tex]\bf \textit{Periodic Exponential Decay}\\\\ A=I(1 - r)^{\frac{t}{p}}\qquad \begin{cases} A=\textit{accumulated amount}\\ I=\textit{initial amount}\to &400\\ r=rate\to 20\%\to \frac{20}{100}\to &0.20\\ t=\textit{elapsed time}\\ p=period\to &2 \end{cases} \\\\\\ A=400(1-0.2)^{\frac{t}{2}}[/tex]

now, how many units will it had decreased by the 9th year? t = 9

[tex]\bf A=400(1-0.2)^{\frac{9}{2}}[/tex]

146.54 units are the total units will she sell her ninth year.

Here, we have,

periodic exponential decay is:

[tex]A = I(1-r)^{t/p}[/tex]

where,

A = amount

I = initial amount

r = rate

t = time

p = period

we know, the first year she sold 400 units,

thus year 0, 400 units, t = 0, A = 400.

so, we get,

periodic exponential decay is:

[tex]A = I(1-r)^{t/p}[/tex]

=> 400 = I × 1

=> 400 = I

therefore then, when, rate = 20 %

we get,

periodic exponential decay is:

[tex]A = I(1-r)^{t/p}[/tex]

[tex]= > A = 400(1-0.2)^{t/2}[/tex]

now, how many units will it had decreased by the 9th year? t = 9

so, we get,

[tex]= > A = 400(1-0.2)^{9/2}[/tex]

=> A = 146.54

Learn more about exponential function here:

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605 mi in 11 hours at the same rate how many miles would he drive in 13 hours

Answers

First you need to find the unit rate

605/ 11= 55

55*13=715

In 13 hours, 715 miles will be traveled

What is the completely factored form of x3 – 64x? x(x – 8)(x – 8) (x-4)(x2+4x+16) x(x – 8)(x + 8) (x – 4)(x + 4)(x + 4)

Answers

After factoring x from both terms, you can factor the difference of two squares.

= x(x² -64)

= x(x -8)(x +8)

_____

It is worth remembering the "speciall form" that is the difference of two squares:

... a² - b² = (a -b)(a +b)

Answer:

x(x -8)(x +8)

Step-by-step explanation:

What is the answer to the problem below?

Solve the system of equations graphically.

x=3
y=4

Answers

Hi there!

They already give us the answer, which is (3,4). So all we have to do is graph it on the graph. 3 is for x and 4 is for y.

Check the picture below.

Here we're given the solution on a silver platter:

(3,4)

You could solve this graphically by graphing the vertical line x = 3 and the horiz. line y = 4. A visual check would show that these two lines intersect at (3,4).

what 5x(4+gy) if x= 1 g= 20 Y=14
please show work

Answers

To solve this one, we'll plug in the given x, g, and y values:
[tex]5x(4+gy)[/tex]
[tex]5(1)(4+(20)(14))=5(4+280)=5(284)=1420[/tex]

5•1(4+20•14) =1420 bc 20 • 14 is 280 + 4 = 284 • 5 is 1420

Can I have help on number 16 and please explain how you got the answer

Answers

The area of the rectangle is defined as its length multiplied by its width. This rectangle has a length of 3 ft and a width of (c + 4) ft, so its area would be [tex]3(c+4)[/tex] feet. Using the distributive property, we can simplify this to:

[tex]3c+3(4)\\ 3c+12[/tex]

So the area of this particular rectangle would be 3c + 12 sq. ft

how can I adjust a quotient to solve a division problem

Answers

Ask them to first estimate the quotient and then to find the actual

The five-number summary for scores on a statistics test is 11, 35, 61, 70, 79. in all, 380 students took the test. about how many scored between 35 and 61

Answers

Answer: There are 95 students who scored between 35 and 61.

Step-by-step explanation:

Since we have given that

The following data : 11,35,61,70,79.

So, the median of this data would be = 61

First two data belongs to "First Quartile " i.e. Q₁

and the second quartile is the median i.e. 61.

The last two quartile belongs to "Third Quartile" i.e. Q₃

And we know that each quartile is the 25th percentile.

And we need "Number of students who scored between 35 and 61."

So, between 35 and 61 is 25% of total number of students.

So, Number of students who scored between 35 and 61 is given by

[tex]\dfrac{25}{100}\times 380\\\\=\dfrac{1}{4}\times 380\\\\=95[/tex]

Hence, There are 95 students who scored between 35 and 61.

The number of students who scored between 35 and 61 is 95

The 5 number summary is the value of the ;

Minimum Lower quartile Median Upper quartile and Maximum values of a distribution.

The total Number of students = 380

The lower quartile (Lower 25%) = 35

The median (50%) = 61

The Number of students who scored between 35 and 61 : 50% - 25% = 25%

This means that 25% of the total students scored between 35 and 61.

25% of 380 = 0.25 × 380 = 95

Hence, 95 students scored between 35 and 61.

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$185 DVD player 6% markup

Answers

$196.10 dbdhdbrbhtbt

after the markup it will be 196.1$

i think.....

if it is right dont forget to give me brainliest

What is the unit rate for 822.6 km in 18 hours? Enter your answer, as a decimal, in the box. Please Help

Answers

unit rate = 822.6 / 18 = 45.7

answer
unit rate = 45.7 km per hour

Find the unit rate by dividing 822.6 km by 18 hours:

822.6 km
-------------- = 45.7 km/hr
18 hrs

Note: this is approx 45.7 km/hr 0.625 mile
--------------- * ----------------- = 28.6 mph
1 1 km

What's 24.67 to one significant figure?

Answers

24.67 is just barely before 25.00, so we should be able to round 20.00, which gives us only one significant digit.

Find a solution x = x(t) of the equation x′ + 2x = t2 + 4t + 7 in the form of a quadratic function of t, that is, of the form x(t) = at2 + bt + c, where a, b, and c are to be determined.

Answers

The particular quadratic solution to the ODE is found as follows:

[tex]x=at^2+bt+c[/tex]
[tex]x'=2at+b[/tex]

[tex](2at+b)+2(at^2+bt+c)=t^2+4t+7[/tex]
[tex]2at^2+(2a+2b)t+(b+2c)=t^2+4t+7[/tex]

[tex]\begin{cases}2a=1\\2(a+b)=4\\b+2c=7\end{cases}\implies a=\dfrac12,b=\dfrac32,c=\dfrac{11}4[/tex]

Note that there's also the fundamental solution to account for, which is obtained from the characteristic equation for the ODE:

[tex]x'+2x=0\implies r+2=0\implies r=-2[/tex]

so that [tex]x_c=Ce^{-2t}[/tex] is a characteristic solution to the ODE, and the general solution would be

[tex]x=Ce^{-2t}+\dfrac{t^2}2+\dfrac{3t}2+\dfrac{11}4[/tex]

To find a, b, c for the solution:

Let's start by writing down the expression for the function x(t) and its derivative:

We have:

x(t) = at² + bt + c
and
x'(t) = 2at + b

Using x' and x into the differential equation x′ + 2x = t² + 4t + 7 gives us:

2at + b + 2*(at² + bt + c) = t² + 4t + 7
Expanding this gives:
2at² + 2bt + b + 4at + 2c = t² + 4t + 7

By equating the coefficients of equivalent powers of t on both sides, we get three equations:

For t² :
2a = 1
So, a = 1/2

For t:
2b + 4a = 4
Substitute a = 1/2 into the equation gives b = 1 - 2 = -1

For the constant term:
b + 2c = 7
Substituting b = -1 gives c = 4.

So the solution is a = 1/2, b = -1, c = 4.

So the specific solution of this differential equation is given by x(t) = (1/2)t² - t + 4.

Simplify each expression (combine like terms)

17x+12-13x

-35+17-3w+12w

Answers

the first one is 4x+17 and the second one is 9w-18

For the first one the answer is 4x+12

And for the second one - 18+9w

Rochelle found the quotient of an integer, x, and 13. Is her quotient a rational number

Answers

It helps, in this case, to ask what exactly a rational number is, and how they come up. The various number systems we work with in math come up in response to different kinds of algebraic problems. The natural numbers (sometimes abbreviated N) are the most basic, and they include all of the "counting numbers" you learn in early gradeschool - 1, 2, 3, 4, etc. - and occasionally 0. If we add or multiply any two natural numbers together, we get another natural number, so we don't need to worry about creating any new numbers when we're just working with those operations. Subtraction poses a few issues, though. If we take an expression like 3 - 4, our answer is going to be outside of N. Rather than throwing our hands up and calling it a day at this point, we create a new set of numbers to deal with this situation.

The integers (abbreviated Z) contain all of the natural numbers, and add negative numbers, too (-1, -2, -3, etc.). Multiplication, addition, and subtraction work fine now, and no matter which integers we use, those operations will always give us back another integer. It's not all perfect, though; division gives us a bit of a hassle now. Expressions like 6/3 are fine, since they just produce another integer (2, in this case), but there's no integer result for something like 2/5 or 3/8, so again, we have to create a new set of numbers to compensate.

That set of numbers is what we refer to as the rational numbers (abbreviated Q), named after the fact that all of them are expressed as ratios. By expanding our number system with these, we gain the ability to express every integer as a rational number by putting it over 1 (6 = 6/1, 5 = 5/1, etc.).

So, getting back to the original question, "is the quotient of an integer x and 13 a rational number?" Yes. The quotient of two integers can always be represented as a rational number.

A, B, and C are polynomials, where A = n, B = 2n + 6, and C = n2 – 1. What is AB – C in simplest form? A=–n2 + 3n + 5 B=n2 + 6n + 1 C=2n2 + 6n – 1 D=3n2 + 5

Answers

A, B, and C are polynomials, where A = n, B = 2n + 6, and C = n2 – 1. What is AB – C in simplest form? We want (n)(2n+6) - (n^2-1). (Please note: n^2 is correct, while n2 is not.)

Multiplying out the first term: 2n^2 + 6n
Subtracting n^2-1: - (n^2 -1)
----------------------
n^2 + 6n + 1 (answer)

Answer:

Step-by-step explanation:

What can be used as a reason in a two-column proof?
Select each correct answer.
conjecture
postulate
definition
premise

Answers

Two column proofs are organized into statement and reason columns. Each statement must be justified in the reason column. The reason column will typically include "given", vocabulary definitions, and theorems.

Therefore, what can be used as a reason in a two-column proof are:

Postulates

Definitions

Answer:

Therefore, what can be used as a reason in a two-column proof are:

Postulates

Definitions

Determine whether the function f : z × z → z is onto if
a.f(m,n)=m+n. b)f(m,n)=m2+n2.
c.f(m,n)=m.
d.f(m,n) = |n|.
e.f(m,n)=m−n.

Answers

a. Yes; [tex]\mathbb Z[/tex] is closed under addition
b. No; [tex]m^2+n^2\ge0[/tex] for any integers [tex]m,n[/tex]
c. Yes; self-evident
d. No; similar to (b), because [tex]|n|\ge0[/tex] for any [tex]n\in\mathbb Z[/tex]
e. Yes; [tex]\mathbb Z[/tex] is closed under subtraction

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