Equivalent expressions are expressions of equal values
The equivalent expressions are 4x+ (y - 8y) + (2z-5z) +6 and 6x-3x-6x + (2y - 10y) + (4 - 8) + (z - 88z)
<h3>How to determine the equivalent expressions</h3>
The first expression has been solved.
So, we have the following expressions
4x−7y−5z+6 and -3x−8y−4−87z
<u>4x−7y−5z+6</u>
We have:
4x-7y-5z+6
Rewrite as:
4x+ (y - 8y) + (2z-5z) +6
<u>-3x−8y−4−87z</u>
We have:
-3x−8y−4−87z
Rewrite as:
3x-6x + (2y - 10y) + (4 - 8) + (z - 88z)
Hence, the equivalent expressions are 4x+ (y - 8y) + (2z-5z) +6 and 6x-3x-6x + (2y - 10y) + (4 - 8) + (z - 88z)
Read more about equivalent expressions at:
brainly.com/question/2972832
Answer:
how should I know?
Step-by-step explanation:
Answer:
84 degrees
Step-by-step explanation:
Applying the Cosine Rule (Choice 2 is the correct one) :-
12^2 = 11^2 + 6^2 - 2*11*6 cos Y
144 = 121 + 36 - 132 cos Y
cos Y = (121 + 36 - 144) / 132
cos Y = 0.09848
Y = 84.3 degrees
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:
Solution given:
<6+24=180[co- interior angle]
<6=180-24=156°
<6=156°
