Answer:
20,18,16,14,12,10,8,6,4,2
10,9,8,7,6,5,4,3,2,1
Step-by-step explanation:
Pattern A:
Rule : start with 20 and subtract 2
Pattern B:
Rule : Start with 10 and subtract 1
Pattern 1:
20 - 2 = 18
18 - 2 = 16
16 - 2 = 14
14 - 2 = 12
12 - 2 = 10
10 - 2 = 8
8 - 2 = 6
6 - 2 = 4
4 - 2 = 2
20,18,16,14,12,10,8,6,4,2
Pattern 2:
10 - 1 = 9
9 - 1 = 8
8 - 1 = 7
7 - 1 = 6
6 - 1 = 5
5 - 1 = 4
4 - 1 = 3
3 - 1 = 2
2 - 1 = 1
10,9,8,7,6,5,4,3,2,1
Answer:
3
Step-by-step explanation:
3x4 is 12 and then you subtract them from each 9ther
Answer:
Topics
2x-5y=4. 2x−5y=4. Add 5y to both sides. Add 5y to both sides.
2x=4+5y. 2x=4+5y. The equation is in standard form. The equation is in standard form.
2x=5y+4. 2x=5y+4. Divide both sides by 2. ...
\frac{2x}{2}=\frac{5y+4}{2} 22x=25y+4 Dividing by 2 undoes the multiplication by 2. ...
x=\frac{5y+4}{2} x=25y+4 Divide 4+5y by 2.
Step-by-step explanation: How do I determine how many solutions an equation has?
If solving an equation yields a statement that is true for a single value for the variable, like x = 3, then the equation has one solution. If solving an equation yields a statement that is always true, like 3 = 3, then the equation has infinitely many solutions.
Answer:
x = -10
Step-by-step explanation:
6 • (x - 2) - (8x + 8) = 0
Pulling out like terms :
3.1 Pull out like factors :
-2x - 20 = -2 • (x + 10)
Equation at the end of step 3 :
-2 • (x + 10) = 0
Equations which are never true :
4.1 Solve : -2 = 0
This equation has no solution.
A a non-zero constant never equals zero.
Solving a Single Variable Equation :
4.2 Solve : x+10 = 0
Subtract 10 from both sides of the equation :
x = -10
One solution was found :
x = -10
Processing ends successfully
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Answer:
Step-by-step explanation:
Given
Success
Failure
Consecutive trials until S happens
Number of trials
For every possible outcome, the last 3 must be SSS. So, we have:
The only possibility here is: 
because 3 trials implies that all outcomes must be S.
The only possibility here is:
because 4 trials implies that the first outcome must be F
The possibilities are:
because 5 trials implies that the first and the second outcomes must be FS or SF.
The possibilities are:
because 6 trials implies that the first three outcomes are: FFF, SSF and SFF
The possibilities are:
because 7 trials implies that the first three outcomes are: FFFF, FSSF and SSFF and FFSF (similar to (e))
