I don't know number 10

Find the horizontal and vertical asymptotes of f(x). f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the

miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* Lets the meaning of vertical and horizontal asymptotes

- Vertical asymptotes are vertical lines which correspond to the zeroes

of the denominator of a rational function

- A horizontal asymptote is a y-value on a graph which a function

approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

then there is a horizontal asymptote at y = leading coefficient of the

numerator ÷ leading coefficient of the denominator

* Lets solve the problem

∵ $f(x)=\frac{6x}{x+2}$

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ The horizontal asymptote can be described by the line y = 6

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ The vertical asymptote can be described by the line x = -2

6 0

3 years ago

Read 2 more answers

Sample Work Help Needed! Please answer these questions and show the steps used to solve them. NO DECIMAL ANSWERS (except on thre

anygoal [31]

1. Using the exponent rule (a^b)·(a^c) = a^(b+c) ...

$\left(-2x^3y^4\right)\left(5x^9y^{-2}\right)=(-2)(5)\left(x^{(3+9)}y^{(4-2)}\right)\\\\=-10x^{12}y^{2}$

Simplify. Write in Scientific Notation

2. You know that 256 = 2.56·100 = 2.56·10². After that, we use the same rule for exponents as above.

$8 \left(32\times 10^{11}\right)=(8)(32)\times 10^{11}=256\times 10^{11}=2.56\times 10^{13}$

3. The distributive property is useful for this.

(3x – 1)(5x + 4) = (3x)(5x + 4) – 1(5x + 4)

... = 15x² +12x – 5x –4

... = 15x² +7x -4

4. Look for factors of 8·(-3) = -24 that add to give 2, the x-coefficient.

-24 = -1×24 = -2×12 = -3×8 = -4×6

The last pair of factors adds to give 2. Now we can write

... (8x -4)(8x +6)/8 . . . . . where each of the instances of 8 is an instance of the coefficient of x² in the original expression. Factoring 4 from the first factor and 2 from the second factor gives

... (2x -1)(4x +3) . . . . . the factorization you require

8 0

3 years ago

All sacks of sugar have the same weight. All sacks of flour also have the same weight, but not necessarily the same as the weigh

maksim [4K]

Sugar: x pounds
flour: 2x + 5 pounds
2x + 3(2x + 5) = 40
2x + 6x + 15 = 40
8x = 25
x = 3.125
-> a sack of flour is 3.125 + 5 = 8.125 pounds
I hope it‘s coorect

4 0

3 years ago

Maths easy class 8 whoever gets it I will give u brainlist

asambeis [7]

Answer:

424 cm²

Step-by-step explanation:

The figure is composed of a square and a trapezium on top

A of square = 18² = 324 cm²

A of trapezium = $\frac{1}{2}$ h (b₁ + b₂ )

where h is the perpendicular height and b₁, b₂ the parallel bases

Here h = 8, b₁ = 18 and b₂ = 7 , then

A = 0.5 × 8 × (18 + 7) = 4 × 25 = 100 cm²

Area of hexagonal park = 324 + 100 = 424 cm²

8 0

3 years ago

5. If position of object x = 3 sinΘ – 7 cosΘ then motion of object is bounded between position.

lesya692 [45]

9514 1404 393

Answer:

±√58 ≈ ±7.616

Step-by-step explanation:

The linear combination of sine and cosine functions will have an amplitude that is the root of the sum of the squares of the individual amplitudes.

|x| = √(3² +7²) = √58

The motion is bounded between positions ±√58.

_____

Here's a way to get to the relation used above.

The sine of the sum of angles is given by ...

sin(θ+c) = sin(θ)cos(c) +cos(θ)sin(c)

If this is multiplied by some amplitude A, then we have ...

A·sin(θ+c) = A·sin(θ)cos(c) +A·cos(θ)sin(c)

Comparing this to the given expression, we find ...

A·cos(c) = 3 and A·sin(c) = -7

We know that sin²+cos² = 1, so the sum of the squares of these values is ...

(A·cos(c))² +(A·sin(c))² = A²(cos(c)² +sin(c)²) = A²(1) = A²

That is, A² = (3)² +(-7)² = 9+49 = 58. This tells us the position function can be written as ...

x = A·sin(θ +c) . . . . for some angle c

x = (√58)sin(θ +c)

This has the bounds ±√58.

3 0

3 years ago