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horsena [70]
3 years ago
11

In a certain Algebra 2 class of 24 students, 9 of them play basketball and 13 of them

Mathematics
1 answer:
marta [7]3 years ago
7 0

Answer:

20/24 or 5/6

for deltamath

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I'm too lazy to do the work by myself anyway, someone please help me thank you <3 !!
In-s [12.5K]

Answer:

98

Step-by-step explanation:

10-3=7

7x7=49

49 x2 = 98

Have a good day! uwu

7 0
3 years ago
Does (9,8) make the inequality y < 10×+8 true
ExtremeBDS [4]

Answer:

Yes, the point makes the inequality true

Step-by-step explanation:

y < 10x+8

Substitute x =9 and y=8 into the inequality

8 < 10*9+8

8<90+8

8<98

This is true so the point makes the inequality true

7 0
4 years ago
Read 2 more answers
I really need help with these 2 geometry questions tysm
qwelly [4]

Answer:

180 + 9 - 4 = 185

5x = 185

185/5= 37

A = 37 X 2 - 9 = 65

B = 37 X 2 + 4 = 78

C = 37

A + B + C = 180

65 + 78 + 37 = 180

A = 65 degrees

B = 78 degrees

7 0
3 years ago
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
3 years ago
What is the product of 48 and 0.36
BartSMP [9]

Answer:

17.28

Step-by-step explanation:

6 0
3 years ago
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