The number of students that are on the track team are 18.
The number of students that are on the baseball team are 15.
<h3>What are the linear equations that represent the question?</h3>
a + b = 33 equation 1
a - b = 3 equation 2
Where:
- a = number of students that are on the track team
- b = number of students that are on the baseball team
<h3>How many
students that are on the
baseball team?</h3>
Subtract equation 2 from equation 1
2b = 30
Divide both sides by 2
b = 30/2 = 15
<h3>How many
students that are on the track
team?</h3>
Subtract 15 from 33: 33 - 15 = 18
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Answer:
Below in bold
Step-by-step explanation:
The sequence is:
3, x, y, 18
If this is an A P then
x - 3 = y - x
2x - y = 3 (A) and
y - x = 18 - y
2y - x = 18 (B)
Multiply (A) by 2:
4x - 2y = 6 (C)
Adding B and C:
3x = 24
x = 8.
and
2y - 8 = 18
2y = 26
y = 13.
So x = 8 and y = 13.
b) ar + ar^2 = 6ar^3 where a = first term and r = common ratio
Divide by a:
r + r^2 = 6r^3
6r^3 - r^2 - r = 0
r(6r^2 - r - 1) = 0
r(3r + 1)(2r - 1) = 0
So the 2 possible values of r
= -1/3 and 1/2.
i) The common ratio is positive so it must be 1/2.
Second term ar = 8
1/2 a = 8
a = 16.
So the first 6 terms are:
16, 8, 4, 2, 1, 1/2.
X = 4 ; x = 3 + i ; x = 3 - i
(If you get a zero that is adding or subtracting, you always need to write it twice but change the sign do they cancel out)
f(x) = (x-4)(x-3-i)(x-3+i)
Distributing the last two parenthesis first is always the best way to start off
(x-3-i)(x-3+i) has (x-3) in common so it can be separated to
(x-3)^2 + (-i)(+i)
(x^2 - 6x + 9) ; (-i)(+i) is always +1
(x^2 - 6x + 9) + 1
(x^2 - 6x + 10)
Now multiply this with (x-4)
x^3 - 6x^2 + 10x
- 4x^2 + 24x - 40
x^3 - 10x^2 + 34x - 40 = f(x)
3 but in most cases it is 2 real roots and one root has a multiplicity of 2
