This is Algebra 2. help fast!
this is Algebra 2. help fast! - 1

Answers

Answer 1
Answer: 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)=((x+2)(x-3))/((x-3)(x-5))
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)=((2x+8)(x-6))/((x-1)(x-6)) aka
f(x)=(2x^2-4x-48)/(x^2-7x+6)








17.
factor
f(x)=((x+1)(x+5))/((x-2)(x+5))
x+5=0
x=-5
hole at x=-5
sub into reduced fn (f(x)=(x+1)/(x-2))
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

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I don't get 7/8n-3=5/8n+12??

(5xy+x-2) + (x-2xy+9)

Answers

The answer would be 3xy+2x+7 in simplest form.

Round 7.456 to the nearest hundredth

Answers

I think it's going to be 7.46 hoped i helped
7.46  hundredths are the second digit after the decimal

What is the answer for -2= - (n-8)​

Answers

Answer:

n=10

Step-by-step explanation:

-2=-(n-8)

-n+8=-2

-n=-2-8

-n=-10

n=10

Simplest form of 4/8

Answers

Since 4 is half of the number 8, the simplest form of that would be 1/2
~1/2 is the same as 4/8
1/2, 4 is half of 8 so it is 1/2

What is the quotient of 43.8 and 12

Answers

Answer:

The quotient of our given numbers is 3.65.

Step-by-step explanation:

We are asked to find the quotient of 43.8 and 12.

To find the quotient of 43.8 and 12, we need to divide 43.8 by 12 as shown below:

\text{Quotient}=(43.8)/(12)

\text{Quotient}=3.65

Therefore, the quotient of our given numbers would be 3.65.

When you divide 43.8 / 12 you get 3.65.

The table shows the data of a random sample of fish that were collected from and later released into a lake.Type of Fish Bass Eel Gar
Number of Fish 92 56 52


How many gars are estimated to be present if a random sample of 1,000 fish is collected from the lake?
104
108
260
320

Answers

Answer:

Gars estimated to be present if a random sample of 1,000 fish is collected from the lake are:

260

Step-by-step explanation:

According to the given sample there was a total of 200(92+56+52) fish collected from a lake.

Of those, 52 were gar.

52/200 = 0.26 = 26%

i.e. 26% of the total fishes collected were jar

So, if 1000 fish are collected

26% of 1000 = (26)/(100)* 1000

                     = 260

i.e. 260 fishes were jar

Hence, gars estimated to be present if a random sample of 1,000 fish is collected from the lake are:

260

there was a total of 200 fish collected and released from the pond...of those, 52 were gar......52/200 = 0.26 = 26%

so if 1000 fish are collected....26% of 1000 = 0.26(1000) = 260 gar <==