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siniylev [52]
2 years ago
8

Find two points on the line to be graphed. y-5=0

Mathematics
1 answer:
Nonamiya [84]2 years ago
3 0
Any point with the y of 5 could be the answer
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Factor the expression

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4 19/30

Step-by-step explanation:

1  3/10 + 3  1/3 = 139/30 or 4.63333...

Simplyfiy  139/30  to  4 19/30

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True or False. The mapping diagram shows a function.
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Which statements are true given the histogram? Check all that apply.
vova2212 [387]

Answer:

(a)The histogram is symmetrical. FALSE  

(b) The histogram is evenly distributed.  FALSE

(c) The cluster from 4 p.m.-10 p.m. means that most of  the texts were sent in the afternoon and evening. TRUE  

(d) More texts were sent from 6 p.m.-8 p.m. than from 8  p.m.-10 p.m.  FALSE.  

(e)The peak from 4 p.m.-6 p.m. means that the highest  number of texts was sent during this interval. TRUE

Step-by-step explanation:

(a)The histogram is symmetrical.  

FALSE, as the histogram has no line of symmetry.

(b) The histogram is evenly distributed.  

FALSE,  the histogram is not evenly distributed as there is no data in the  

10: AM to 3:00 PM slot.

(c) The cluster from 4 p.m.-10 p.m. means that most of  the texts were sent in the afternoon and evening.

TRUE, the data shows that the most texts were sent in afternoon and evening as it has the highest peaks in the given slot.

(d) More texts were sent from 6 p.m.-8 p.m. than from 8  p.m.-10 p.m.

The number of texts sent from  6 p.m.-8 p.m = 19

The number of texts sent  from 8  p.m.-10 p.m = 29

But, 19 < 29

Hence, the given statement is FALSE.

(e)The peak from 4 p.m.-6 p.m. means that the highest  number of texts was sent during this interval.  TRUE, the highest peak indicates the highest number of texts sent in the given interval.

8 0
2 years ago
Read 2 more answers
Quadrilateral PEST has vertices (-1, -5), (8, 2), (11, 13), and (2, 6), respectively. Classify the quadrilateral as a square, rh
Alchen [17]

Answer:

The figure PEST is a rhombus

Step-by-step explanation:

* Lets talk about the difference between all these shapes

- At first to prove the shape is a parallelogram you must have one

 of these conditions

# Each two opposite sides are parallel OR

# Each two opposite sides are equal in length OR

# Its two diagonals bisect each other

- After that to prove the parallelogram is:

* A rectangle you must have one of these conditions

# Two adjacent sides are perpendicular to each other OR

# Its two diagonals are equal in length

* A rhombus you must have one of these conditions

# Two adjacent sides are equal in length OR

# Its two diagonals perpendicular to each other OR

# Its diagonals bisect its vertices angles

* A square you must have two of these conditions

# Its diagonals are equal and perpendicular OR

# Two adjacent sides are equal and perpendicular

* Now lets solve the problem

∵ The vertices of the quadrilateral PEST are

   P (-1 , -5) , E (8 , 2) , S (11 , 13) , T (2 , 6)

- Lets find the slope from each two points using this rule :

 m = (y2 - y1)/(x2 - x1), where m is the slope and (x1 , y1) , (x2 , y2)

 are two points on the line

- Let (x1 , y1) is (-1 , -5) and (x2 , y2) is (8 , 2)

∴ m of PE = (2 - -5)/(8 - -1) = 7/9

- Let (x1 , y1) is (8 , 2) and (x2 , y2) is (11 , 13)

∴ m of ES = (13 - 2)/(11 - 8) = 11/3  

- Let (x1 , y1) is (11 , 13) and (x2 , y2) is (2 , 6)

∴ m of ST = (6 - 13)/(2 - 11) = -7/-9 = 7/9

- Let (x1 , y1) is (2 , 6) and (x2 , y2) is (-1 , -5)

∴ m of TP = (-5 - 6)/(-1 - 2) = -11/-3 = 11/3

∵ m PE = m ST = 7/9

∴ PE // ST ⇒ opposite sides

∵ m ES = m TP = 11/3

∴ ES // TP ⇒ opposite sides

- Each two opposite sides are parallel

∴ PEST is a parallelogram

- Lets check if the parallelogram can be rectangle or rhombus or

 square by one of the condition above

∵ If two line perpendicular , then the product of their slops = -1

- Lets check the slopes of two adjacent sides (PE an ES)

∵ m PE = 7/9

∵ m ES = 11/3

∵ m PE × m ES = 7/9 × 11/3 = 77/27 ≠ -1

∴ PE and ES are not perpendicular

∴ PEST not a rectangle or a square (the sides of the rectangle and

  the square are perpendicular to each other)

- Now lets check the length of two adjacent side by using the rule

 of distance between two points (x1 , y1) and (x2 , y2)

 d = √[(x2 - x1)² + (y2 - y1)²]

- Let (x1 , y1) is (-1 , -5) and (x2 , y2) is (8 , 2)

∴ PE = √[(8 - -1)² + (2 - -5)²] = √[9² + 7²] = √[81 + 49] = √130 units

- Let (x1 , y1) is (8 , 2) and (x2 , y2) is (11 , 13)

∴ ES = √[(11 - 8)² + (13 - 2)²] = √[3² + 11²] = √[9 + 121] = √130 units

∴ PE = ES ⇒ two adjacent sides in parallelogram

∴ The four sides are equal

* The figure PEST is a rhombus

4 0
2 years ago
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