Is there a relationship between area or perimeter of a rectangle

Answers

Answer 1

Answer: Hello,

For a given perimeter (P) there are an infinity of Area (A)
Let's say x the length, and y the wide of the rectangle

P=2(x+y)
A=xy
k=x-y >=0

As (x+y)²-4xy=(x-y)²: A²-4P=k² or P=(A²-k²)/4
In primus, you will find a graph (abacus) giving P for a A and k given.
Negative Area or P are excluded.(just remind the first quadrant, A>=0 and P>=0)

Related Questions

Twice a number is equal to seven more than four times the number
Twenty five X squared minus one hundred and forty four
The average cost of tuition and room and board at a small private liberal arts college is reported to be $8,500 per term. A financial administrator believes that the average cost is higher. A study conducted using 350 small liberal arts colleges showed that the average cost per term is $8,745. The population standard deviation is $1,200. Let α
What is the area of the polygon below?
What is the middle term of the product of (5a-7)(2a-1)

A triangle has squares on its three sides as shown below what is the value of x

Answers

Answer:

5 cm

Step-by-step explanation:

Which of the following best defines density?A.
A measure of the volume of a given substance

B.
A number between 0 and 1 that defines how thick a substance is

C.
A measure of how much mass a given volume of a substance will have

D.
A measure of the mass of a given substance

Answers

Answer:

Density = mass/volume

Step-by-step explanation:

The mass per unit volume of a given substance is called its density. It can be represented by d or ρ. Mathematically, it is given by :

density=mass/volume

The SI unit of mass is kg and the SI unit of volume is m³. So, the SI unit of density is kg/m³.

So, the statement that defines density is "A measure of how much mass a given volume of a substance will have". Hence, the correct option is (c).

Density is mass per unit volume, therefore the correct answer is C. A measure of how much mass a given volume of a substance will have

M(5, 7) is the midpoint of mc140-1.jpg. The coordinates of S are (6, 9). What are the coordinates of R? (4, 5) (7, 11) (5.5, 8) (10, 14)

Answers

Answer:

(4,5)

Step-by-step explanation:

Given that M is the midpoint of the Line RS. Where the coordinates of M, R and S are

M(5,7)

R(X,Y)

S(6,9)

Here we have to find the valuer of X and Y

We will use the mid point formula which is given below.

$x=(x_1+x_2)/(2) \ny=(y_1+y_2)/(2)\n$

Let put the coordinated of M and S and find coordinates of R

$x=(x_1+x_2)/(2) \n5=(X+6)/(2) \n10=X+6\nX=4$

$y=(y_1+y_2)/(2)\n7=(Y+9)/(2) \n14=Y+9\ny=5$

Hence our coordinates are

(4,5)

the coordinate of R(4,5)

The data set represents the number of rings each person in a room is wearing.0, 2, 4, 0, 2, 3, 2, 8, 6

What is the interquartile range of the data?

2
3
4
6

Answers

Answer:

the interquartile range of the data is:

Step-by-step explanation:

We are given a data set as:

0, 2, 4, 0, 2, 3, 2, 8, 6

On arranging the data in the ascending i.e. increasing order is given by:

0 0 2 2 2 3 4 6 8

The minimum value of data set=0

Maximum value of data set is: 8

Range of data set= Maximum value-Minimum value

i.e. Range= 8-0

i.e. Range= 8

Also, Median of set is the central tendency of the data and is given by:

Median= 2

Lower set of data is:

0 0 2 2

Hence, The median of lower set of data is the lower quartile or first quartile.

i.e. $Q_1$

Hence, $Q_1=(0+2)/(2)\n\n\nQ_1=(2)/(2)\n\n\nQ_1=1$

Hence, Lower quartile=1

Similarly upper set of data is:

3 4 6 8

Hence, The median of upper set of data is the upper quartile or third quartile.

i.e. $Q_3$

Hence, $Q_3=(4+6)/(2)\n\n\nQ_3=(10)/(2)\n\n\nQ_3=5$

Hence, Upper quartile=5

Hence, the interquartile range(IQR) is given by:

IQR=Upper quartile-Lower quartile

IQR=5-1

IQR=4

If the data set represents the number of rings each person is wearing, being: 0,2,4,0,2,3,2,8,6, the interquartile range of the data is 2. Being, 4 as the Q1, 3 as the Q2 or median, and 6 as the Q3. Where the formula of getting the interquartile range is IQR= Q1-Q2.

A number x is no more than 9 units away from 8. Write and solve an absolute-value inequality to show the range of possible values of x.|x − 8| ≤ 9
−1 ≤ x ≤ 17

|x + 8| ≤ 9
−17 ≤ x ≤ 1

|x − 9| ≤ 8
1 ≤ x ≤ 17

|x + 9| ≤ 8
−17 ≤ x ≤ −1

Answers

Answer:

Option A is right

Step-by-step explanation:

Given that a number x is no more than 9 units away from 8

This implies that number can be either larger than 8 by less than 9 units or smaller than 8 by less than 9 units.

Difference between x and 8 is at most 9.

We represent in absolute value as

$|x-8|\leq 9\n-9\leq x-8\leq 9\n-1\leq x\leq 17$

Hence option A is right

The answer is a (|x - 8| ≤ 9 and -1 ≤ x ≤ 17) Because the difference between x and 8 can't be greater than 9, but it can be less than or equal to it, and because the range that x could be would be -1 to 17, since 8 + 9 = 17 and 8 - 9 = -1.

Which is a point 6units to the left of 4

Answers

-2 i believe, have a great day :)

Other Questions

What is the value of d? 11d = –44

Hi Krystal, when you submit this form, the owner will be able to see your name and email address,1. A is located ( 1,-1) and B is located (4,4) Find the coordinates of point p that partitionAB in the ratio 1:4. 03Enter your answerISubmitThis content is created by the owner of the form. The data you submit will be sent to the form owner. Never give out your password.Powered by Microsoft Forms Privacy and cookies | Terms of use

Domain definition in math

Solve the equation for the indicated variable. A = bh; h