The marks obtained by students are 10, 13, 12, 15, 25, 30, 20, 35, 40, 18. Calculate the median mark.
1 answer:
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The 20th term of the arithmetic sequence is 122.
<h3>What is arithmetic sequence?</h3>nth term = a1 + d(n - 1) so here:
20th term = -30 + 8(20 - 1)
- 30 + 152
= 122.
An ordered group of numbers with a shared difference between each succeeding word is known as an arithmetic sequence. For instance, the common difference in the arithmetic series 3, 9, 15, 21, and 27 is 6.
Arithmetic sequences are sequences containing these patterns. The distance between succeeding terms in an arithmetic series is always the same. The difference between consecutive words is always two, hence the sequence 3, 5, 7, 9... is arithmetic.
An explicit formula that states a = d (n - 1) + c, where d is the common difference between succeeding words, and c = a1, can be used to establish an arithmetic sequence.
To learn more about arithmetic sequence from the given link:
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Find VA
simplify fraction
set deonmenator equal to zero
that value is VA
cannot cross VA
fidn HA
if the degree of the numerator is less than the degree of the denomenator, then the HA is y=0
if the degree is equal, then divide he leading coeficient of the numerator by the leading coeficient of the deonmenator
to find if the fn crosses the HA, set the HA equaal to the reduced fn and solve, if you get a false statement, then it does not cross
holes are found by where if you have the numberator and deonmentoar are the same degree and they have a factor of same multiplicity example
f(x)=
there is a hole at x=3 and to find the y coordinate, subsitute x=3 into reduced fraction
so
14.
make one fraction
f(x)=(2x-1)/(x-1)
x and y intercept
xint is f(x)=0
xintercept= 1/2 (1/2,0)
y intercept is when x=0 so set x=0
-1/-1=1
yintercept is y=1 aka (0,1)
VA set denom to zeero
x-1=0
x=1
VA at x=1
HA degree is same so divide leading coefs
2/1=2
HA at y=2
crosses HA?
2=(2x-1)/(x-1)
2x-1=x-1
x-1=-1
x=0
crosses ha at x=0 and
f(x)=(2(0)-1)/(0-1)=-1/-1=1
crosses HA at (0,1)
no holes
find where the fn is negative and positive
(2x-1) is zero at x=1/2
x-1 is zero at x=1
so in between, those, (1/2 and 1), the graph is negative (positive/negative=negative)
outside of that interval, the graph is positive (positive/positive=positive, negative/negative=negative) so graph is drawn on attachment
15. factor
f(x)=(x-2)/[(x-4)(x+1)]
VA set denom to zero
x-4=0
x+1=0
VAs at x=4 and -1
HA
degree of numberateor is smaller to HA is y=0
crosses HA?
0=(x-2)/(x^2-3x-4)
0=x-2
2=x
yes, at (2,0)
holes? no
find positive and negative
graph included
16.
VA=x is x=1
x=1
x-1=0
reduced denom is (x-1)
HA is y=2
degrees are same
2/1=2
(2x+something)/(x-1)
xint at -4,=
deonm when set to zero, etuals -4
2x+something=0 yeilds x=-4 so
2(-4)+something=0
-8+something=0
something=8
(2x+8)/(x-1)
hole at (6,4)
factored out bit is x=6
x=6
x-6=0
multiply whole equation by (x-6)/(x-6)
the function is
f(x)=
aka
f(x)=
17.
factor
f(x)=
x+5=0
x=-5
hole at x=-5
sub into reduced fn (f(x)=
)
4/7
hole at (-5,4/7)
degree is same so divide leading coeficients
1/1=1
HA=1
VA is set reduced denom to zero
(x-2) is reduced
x-2=0
x=2
VA is at x=2
simplify fraction
set deonmenator equal to zero
that value is VA
cannot cross VA
fidn HA
if the degree of the numerator is less than the degree of the denomenator, then the HA is y=0
if the degree is equal, then divide he leading coeficient of the numerator by the leading coeficient of the deonmenator
to find if the fn crosses the HA, set the HA equaal to the reduced fn and solve, if you get a false statement, then it does not cross
holes are found by where if you have the numberator and deonmentoar are the same degree and they have a factor of same multiplicity example
f(x)=
there is a hole at x=3 and to find the y coordinate, subsitute x=3 into reduced fraction
so
14.
make one fraction
f(x)=(2x-1)/(x-1)
x and y intercept
xint is f(x)=0
xintercept= 1/2 (1/2,0)
y intercept is when x=0 so set x=0
-1/-1=1
yintercept is y=1 aka (0,1)
VA set denom to zeero
x-1=0
x=1
VA at x=1
HA degree is same so divide leading coefs
2/1=2
HA at y=2
crosses HA?
2=(2x-1)/(x-1)
2x-1=x-1
x-1=-1
x=0
crosses ha at x=0 and
f(x)=(2(0)-1)/(0-1)=-1/-1=1
crosses HA at (0,1)
no holes
find where the fn is negative and positive
(2x-1) is zero at x=1/2
x-1 is zero at x=1
so in between, those, (1/2 and 1), the graph is negative (positive/negative=negative)
outside of that interval, the graph is positive (positive/positive=positive, negative/negative=negative) so graph is drawn on attachment
15. factor
f(x)=(x-2)/[(x-4)(x+1)]
VA set denom to zero
x-4=0
x+1=0
VAs at x=4 and -1
HA
degree of numberateor is smaller to HA is y=0
crosses HA?
0=(x-2)/(x^2-3x-4)
0=x-2
2=x
yes, at (2,0)
holes? no
find positive and negative
graph included
16.
VA=x is x=1
x=1
x-1=0
reduced denom is (x-1)
HA is y=2
degrees are same
2/1=2
(2x+something)/(x-1)
xint at -4,=
deonm when set to zero, etuals -4
2x+something=0 yeilds x=-4 so
2(-4)+something=0
-8+something=0
something=8
(2x+8)/(x-1)
hole at (6,4)
factored out bit is x=6
x=6
x-6=0
multiply whole equation by (x-6)/(x-6)
the function is
f(x)=
f(x)=
17.
factor
f(x)=
x+5=0
x=-5
hole at x=-5
sub into reduced fn (f(x)=
4/7
hole at (-5,4/7)
degree is same so divide leading coeficients
1/1=1
HA=1
VA is set reduced denom to zero
(x-2) is reduced
x-2=0
x=2
VA is at x=2