Given a group of
n object. We want to make a selection of
k objects out of the n object. This can be done in
C(n, k) many ways, where
,
where k!=1*2*3*...(k-1)*k
Thus, we can do the selection of 3 cd's out of 5, in C(5,3) many ways,
where
Answer: 10
Answer:
x = -4
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
Step-by-step explanation:
<u>Step 1: Define</u>
-15x + 5 = -10x + 25
<u>Step 2: Solve for </u><em><u>x</u></em>
- Add 15x on both sides: 5 = 5x + 25
- Subtract 25 on both sides: -20 = 5x
- Divide 5 on both sides: -4 = x
- Rewrite: x = -4
I believe the answer is 2.849
f(n) = f(n+1) +3
f(1) = f(2) +3
f(2) =f(3) +3
f(3) = f(4) +3
So f(1) = f(n) + 3* (n-1)
Put n=4 here so you found
f(1) = f(4) + 3* 3
= 22 +9
=31 ans
hope it helps
Answer:
3
Step-by-step explanation:
Standard form is written in the order of the exponent <em>on</em><em> </em><em>a</em><em> </em><em>variable</em> from highest to lowest.
#1 exponents go 2, 1 , then zero. So, this is standard form.
#2 exponents go 2 then zero, so this is standard form.
#3 exponents go 1 then 2 then zero, so this is <em><u>not</u></em> in standard form.
#4 exponents go 3 to 2 to 1, so this is still in standard form.
hope this helps!
