A triangle has side lengths of x, x+11, and 3x+5. What is the perimeter?
2 answers:
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Answer:
Step-by-step explanation:
see picture
when doing questions like these it's all about breaking up the shape into digestible parts
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