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faltersainse [42]
3 years ago
15

What type of angle is show​

Mathematics
1 answer:
9966 [12]3 years ago
6 0

A right angle is 90 degrees, an obtuse angle is larger than 90 degrees and an acute angle is less than 90 Degrees.

The angle shown is smaller than 90 degrees so it is c. Acute

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BRAINLIEST IF CORRECT. NO LINKS OR WILL REPORT
labwork [276]

Answer:

WHere is the question??

Step-by-step explanation:

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3 years ago
The radius of the circle is 7 inches what is the approximate length of AB (AB is 135 degrees
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We know that
circumference=2*pi*r
for r=7 in
circumference=2*pi*7------> 14*pi in

if 360° (full circle)  has a length of------------ 14*pi in
 135°-----------------------------> x
x=135*14*pi/360----------> x=5.25*pi in--------> 16.49 in

the answer is
5.25*pi in  or 16.49 in 
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What is the distance between (–6, 2) and (8, 10) on a coordinate grid?
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Write an inequality to show the remaining distance, d, in feet must she climb to reach the peak.
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Read 2 more answers
A computer programming team has 13 members. a. How many ways can a group of seven be chosen to work on a project? b. Suppose sev
Julli [10]

Answer:

1716 ;

700 ;

1715 ;

658 ;

1254 ;

792

Step-by-step explanation:

Given that :

Number of members (n) = 13

a. How many ways can a group of seven be chosen to work on a project?

13C7:

Recall :

nCr = n! ÷ (n-r)! r!

13C7 = 13! ÷ (13 - 7)!7!

= 13! ÷ 6! 7!

(13*12*11*10*9*8*7!) ÷ 7! (6*5*4*3*2*1)

1235520 / 720

= 1716

b. Suppose seven team members are women and six are men.

Men = 6 ; women = 7

(i) How many groups of seven can be chosen that contain four women and three men?

(7C4) * (6C3)

Using calculator :

7C4 = 35

6C3 = 20

(35 * 20) = 700

(ii) How many groups of seven can be chosen that contain at least one man?

13C7 - 7C7

7C7 = only women

13C7 = 1716

7C7 = 1

1716 - 1 = 1715

(iii) How many groups of seven can be chosen that contain at most three women?

(6C4 * 7C3) + (6C5 * 7C2) + (6C6 * 7C1)

Using calculator :

(15 * 35) + (6 * 21) + (1 * 7)

525 + 126 + 7

= 658

c. Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project?

(First in second out) + (second in first out) + (both out)

13 - 2 = 11

11C6 + 11C6 + 11C7

Using calculator :

462 + 462 + 330

= 1254

d. Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?

Number of ways with both in the group = 11C5

Number of ways with both out of the group = 11C7

11C5 + 11C7

462 + 330

= 792

8 0
3 years ago
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