Answer:
Step-by-step explanation:
f(x)=x²-5x+7
f(2)=2²-5.2+7=4-10+7=11-10=1
f(-1)=(-1)²-5.(-1)+7=1+5+7=13
f(1/3)=(1/3)²-5(1/3)+7=1/9-5/3+7=49/9
∴, f(2)-f(-1)+f(1/3)
=1-13+49/9
=58/9-13
=-59/9
mark be brainlest plss
When you have 3 choices for each of 6 spins, the number of possible "words" is
3^6 = 729
The number of permutations of 6 things that are 3 groups of 2 is
6!/(2!×2!×2!) = 720/8 = 90
A) The probability of a word containing two of each of the letters is 90/729 = 10/81
The number of permutations of 6 things from two groups of different sizes is
(2 and 4) : 6!/(2!×4!) = 15
(3 and 3) : 6!/(3!×3!) = 20
(4 and 2) : 15
(5 and 1) : 6
(6 and 0) : 1
B) The number of ways there can be at least 2 "a"s and no "b"s is
15 + 20 + 15 + 6 + 1 = 57
The probability of a word containing at least 2 "a"s and no "b"s is 57/729 = 19/243.
_____
These numbers were verified by listing all possibilities and actually counting the ones that met your requirements.
You have to get rid of 1/2 by multiplying it by its recriprocal which is like this:
1) 1/2 x 2/1 which equals to zero and you do the same to the other side.
2) -12 x 2/1 = -24x then multiply-24x with 7 which equals -168.
16-x= -24x -168
3) subtract 16 to the other side and subtract 24x with x
23x= -184
4) divide by 23 to isolate x and you get -8
5) x=-8
no. No results should be presented in a fair and objective manner.
<span>52 rounded to the nearest hundred </span>= 100
cause 52 > 50
hope it helps
