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olganol [36]
4 years ago
15

In a couple of sentences and in your own words describe in what ways horizontal, vertical, and oblique asymptotes can be identif

ied.
Mathematics
1 answer:
drek231 [11]4 years ago
5 0
A horizontal asymptote is one in which y has a limit as x approaches positive or negative infinite. It is usually due to both the denominator and the numerator having the same highest degree term, and the coefficient created by their proportion serves as the asymptote. For example, (2x^2 + 1) / (3x^2) would have a horizontal asymptote of 2/3

A vertical asymptote is an x value at which y approaches infinite. One example includes when the denominator of the function approaches zero at a certain point. For example, (x^2 + 3) / (x + 1) has a vertical asymptote at x=-1, since the denominator approaches zero as x approaches this point. 

For an oblique asymptote, y generally takes the form of a linear function as x approaches infinite. This is the case when the highest term in numerator is one degree higher than the highest degree term in the denominator. 

Examples include (5x^2 + 2) over 2x, where the oblique asymptote is (5/2)x, and even the linear function 2x+3 has an oblique asymptote of 2x
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Please answer this correctly
lana66690 [7]

Answer:

  698 cm²

Step-by-step explanation:

The volume is given by ...

  V = LWH

Filling in the given values, we have ...

  1020 = (17)(5)y . . . . . . . using L=17, W=5, H=y

  y = 1020/(17·5) = 12

The surface area is given by ...

  A = 2(LW +H(L+W))

  A = 2(17·5 +12(17+5)) = 2(85 +264) . . . . . . . using L=17, W=5, H=y=12

  A = 698 . . . . square centimeters

6 0
3 years ago
Which point best approximates √3?<br><br> A<br> B<br> C<br> D
noname [10]

Answer:

B

Step-by-step explanation:

Square root of 3 is between 1 and 2

1^2 =1

(sqrt(3)) ^2 =3

2^2 = 4

So it must be point B

7 0
4 years ago
Find the value of x if log3 (x+6) =2​
Vanyuwa [196]

Answer:

3(x+6) =2

3x+18 =2

3x=2-18

3x=16

x=16/3

x=5.33333333

8 0
3 years ago
Read 2 more answers
5-4. Solve each of the following equations for the indicated variable. Use your Equation Mat if it is helpful. Write down each o
balu736 [363]
QUESTION 1

The given equation is :

2(y - 3) = 4
We want to solve for y. This means we have to isolate y on one side of the equation while all other constants are also on the other side of the equation,

We first divide both sides by 2 to obtain,

\frac{2(y - 3)}{2} = \frac{4}{2}

We cancel out the common factors to get,

y - 3 = 2

We now group like terms to get,

y = 3 + 2

We simplify to obtain,

y = 5

QUESTION 2

The given equation is

2x + 5y = 10
We want to solve for x. This means that we need to isolate x on one side of the equation, while all other variables and constant are also on the other side of the equation.

This implies that,

2x= 10 - 5y

We now divide through by 2, to obtain,

\frac{2x}{2} = \frac{10}{2} - \frac{5y}{2}

This will now give us,

x = 5 - \frac{5}{2} y

QUESTION 3

The given equation is
6x + 3y = 4y + 11

We want to solve for y. This means that we need to isolate y on one side of the equation, while all other variables and constants are also on the other side of the equation.

We first of all group all the y terms on one side of the equation to obtain,

6x - 11 = 4y - 3y

We simplify to get,

6x - 11 = y

This implies that,

y = 6x - 11

QUESTION 4

The given equation is,

3(2x + 4) = 2 + 6x + 10

We want to solve for x. This means that we need to isolate x on one side of the equation, while all other variables and constant are also on the other side of the equation.

Let us first expand the brackets to get,

6x + 12 = 2 + 6x + 10
This implies that

6x + 12 = 6x + 12

We group like terms to get,

6x - 6x = 12 - 12

We now simplify both sides to get,

0x = 0

We don't want x to vanish, so let us try to divide both sides by zero to get,

x = \frac{0}{0}

This is an indeterminate form, which implies that, x has infinitely many solutions. This means that all the real numbers are solution to the equation.

\therefore \: x \in \: R

QUESTION 5

The given equation is,

y = - 3x + 6

We want to solve for x, so we add the additive inverse of 6, which is -6 to both sides of the equation to get,

y - 6 = - 3x

We rewrite this to obtain,

- 3x = y - 6

We divide through by -3 to get,

\frac{ - 3x}{ - 3} = \frac{y}{ - 3} - \frac{6}{ - 3}

This will give us,

x = - \frac{y}{3} + 2

or

x= 2 - \frac{1}{3} y
5 0
3 years ago
5+6(2x-1) <br> i need this like now pls
Crank

Answer:

Step-by-step explanation:

5 + 6 = 11 x 2 - 1 = 21 therefore its 21

5 0
3 years ago
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