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VMariaS [17]
3 years ago
13

What is the diameter of the circle?

Mathematics
1 answer:
LuckyWell [14K]3 years ago
6 0
You need to include the image of the question it doesn’t make sense if
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Graph the function, y=-5x+3
Thepotemich [5.8K]
Graph a point at (0,2)

then graph a point at (-2,-1)
3 0
3 years ago
Solve 2x+6 ≤ 10 or 2x + 8 > 20.
Firlakuza [10]

Answer:

<h2>A. x ≤2 or x>6</h2>

Step-by-step explanation:

2x+6 ≤ 10  or  2x + 8 > 20

⇔ 2x ≤ 4  or  2x > 12

⇔ x ≤ 2  or  x > 6

6 0
2 years ago
Find all the real square roots of 0.0625
Fudgin [204]
Answer and workings in the attachment below.

8 0
3 years ago
20 cm
lawyer [7]

The area of the arrow given in the figure is 610 square cm

<h3>Area of composite figure</h3>

The given figure is made up of rectangle and triangle. The area is expressed as:

Area = Area of rectangle + area of triangle

Substitute the given parameters

Area of the arrow = (15*20) + 0.5(31 * 20)

Area of the arrow = 300 + 310
Area of the arrow = 610 square cm

Hence the area of the arrow given in the figure is 610 square cm

Learn more on area of composite figures here: brainly.com/question/21135654

#SPJ1

7 0
1 year ago
I need answer Immediately pls!!!!!!
Illusion [34]

Given:

Total number of students = 27

Students who play basketball = 7

Student who play baseball = 18

Students who play neither sports = 7

To find:

The probability the student chosen at randomly from the class plays both basketball and base ball.

Solution:

Let the following events,

A : Student plays basketball

B : Student plays baseball

U : Union set or all students.

Then according to given information,

n(U)=27

n(A)=7

n(B)=18

n(A'\cap B')=7

We know that,

n(A\cup B)=n(U)-n(A'\cap B')

n(A\cup B)=27-7

n(A\cup B)=20

Now,

n(A\cup B)=n(A)+n(B)-n(A\cap B)

20=7+18-n(A\cap B)

n(A\cap B)=7+18-20

n(A\cap B)=25-20

n(A\cap B)=5

It means, the number of students who play both sports is 5.

The probability the student chosen at randomly from the class plays both basketball and base ball is

\text{Probability}=\dfrac{\text{Number of students who play both sports}}{\text{Total number of students}}

\text{Probability}=\dfrac{5}{27}

Therefore, the required probability is \dfrac{5}{27}.

3 0
3 years ago
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