Answer:
Step-by-step explanation:
5
Answer:
- The sequence is an Arthemtic Progression
An=A1+(n-1)d
A1 is first term, An is nth term, n is number of term, and d is common difference
therefore
A4=35, A1= -17
A4=A1+(4-1)d
35= -17+3d
35+17=3d
52=3d
52/3=3d/3
14=d
common diffrence(d)=14
- The general solution is given by
An= -17+(n-1)14
An= -17+14n-14
An= -31+14n
<u>An= 14n-31</u>
A14 term, means n=14
From An=A1+(n-1)d
A14= -17+(14-1)14
= -17+(13×14)
= -17+182
= 165.
<u>Therfore, the 14th term is 165.</u>
2. A sequence has a CR of 4/5 and its eighth term (a8) is (393216/3125). What is its general equation? Its 3rd term?
<u>solution</u>
common ratio(r)=4/5
eighth term(G8)=393216/3125
From Gn= G1r^(n-1)
G8 means n=8
G8=G1r^(n-1)
393216/3125=G1(4/5)^(8-1)
393216/3125=G1(4/5)^7
G1=(393216/3125)/(4/5)^7
G1=600
<u>The first term is given by G1=600</u>
The General equation is given by
The General equation is given by Gn= 600(4/5)^(n-1)
3rd term (G3)
G3= G1(4/5)^(3-1) where n=3,
=600(4/5)^2
=600(16/25)
=384
<u>Therefore, the 3rd term is given by G3= </u><u>3</u><u>8</u><u>4</u><u>.</u>
<u>I</u><u> </u><u>h</u><u>a</u><u>v</u><u>e</u><u> </u><u>m</u><u>a</u><u>d</u><u>e</u><u> </u><u>s</u><u>o</u><u>m</u><u>e</u><u> </u><u>C</u><u>o</u><u>r</u><u>r</u><u>e</u><u>c</u><u>t</u><u>i</u><u>o</u><u>n</u><u>s</u><u> </u><u>i</u><u> </u><u>m</u><u>e</u><u>s</u><u>s</u><u>e</u><u>d</u><u> </u><u>u</u><u>p</u><u> </u><u>s</u><u>o</u><u>m</u><u>e</u><u>w</u><u>h</u><u>e</u><u>r</u><u>e</u><u>.</u>
Answer:
24 possible outcomes
Step-by-step explanation:
Combination has to do with selection. For example, if r object is selected from a pool of n objects, the number if possible ways can be expressed according to the combination formula:
nCr = n!/(n-r)!r!
Applying this in question, if each student receives one of 4 calculator models and one of 3 types of ruler, the number of ways this can be done is:
4C1 × 3C1
4C1 = 4!/(4-1)!1! {If a student gets one calculator)
4C1 = 4×3×2/3×2
4C1 = 4ways
3C1 = 3!/(3-2)!1! {If a student gets a ruler}
3C1 = 3×2/1
3C1 = 6ways
Total number of possible outcomes if a student gets one ruler and one calculator will be 4×6 = 24ways
12.6 because it has to be the same as ED
You can start by subtracting different equations from each other.
3x + 2y + 3z = 1
subtract
3x + 2y + z = 7
2z = -6
divide by 2
z = -3
add the following two expressions together:
3x + 2y + z = 7
3x + 2y + 3z =1
6x + 4y + 4z = 8
subtract the following two expressions:
6x + 4y + 4z = 8
5x + 5y + 4z = 3
x - y = 5
^multiply the whole equation above by 3
3x - 3y = 15
subtract the following two expressions:
3x - 3y = 15
3x + 2y = 10
-5y = 5
divide each side by -5
y=-1
take the following expression from earlier:
x - y = 5
substitute y value into above equation
x - - 1 = 5
2 negatives make a positive
x + 1 = 5
subtract 1 from each side
x = 4
Therefore x = 4, y = -1, z = -3
I checked these with all 3 equations and they worked :)
(it's quite complicated, comment if you don't understand anything) :)