It would equal 1
Explanation
- keep (keep first fraction the same)
- change (change division to multiplication)
- flip (flip the last fraction)
Answer:
They are SIMILAR triangles
Step-by-step explanation:
Given triangles MNL and FHG with two of their angles to be 51° and 36° respectively. Their third angle is expressed as:
180°-(51°+36°)
= 180°-87°
= 93°
Their third angle is 93°
Since both triangles have the same angles in them, both triangles are considered as similar triangle even though the value of their sides are different due to angle differences in both.
Note that individual triangles are scalene triangles since their angles are not the same but the both triangles are similar triangles based on their relationship.
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
Answer:
Step-by-step explanation:
Multiplying 5 by itself is the same thing as raising it to the second power:
We multiply 2 by itself two seperate times. That's the same thing as raising 2 to the second power to the second power or in simpler terms raising two to the fourth power:
Since only multiplication is present, we can combine the two terms to get:
