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3 years ago

Give the equation of the horizontal asymptote shown below.

Mathematics

2 answers:

Kazeer [188]3 years ago

5 0

Answer:

the other person is right, the answer is y = 7

Step-by-step explanation:

Oxana [17]3 years ago

4 0

Polynomial division yields

$\dfrac{7x^2}{x^2+5}=7-\dfrac{35}{x^2+5}$

As $x\to\pm\infty$ , you have the remainder term approaching $0$ , so the horizontal asymptote is the line $y=7$ .

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On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

Answer:

a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

7 0

3 years ago

Please answer within the next 7 min, giving 38 points of a ligament answer!

frosja888 [35]

The answer is choice C

6 0

2 years ago

YALL PLEASEE HELP MEE i am very confused and I need to pass this

aliya0001 [1]

Answer: Go to
Math-way!! I think it’s the third option though❤️ (give me Brainly pls)

6 0

3 years ago

Read 2 more answers

Evaluate the expression \dfrac{5\cdot x^3}{x^2} x 2 5⋅x 3 start fraction, 5, dot, x, cubed, divided by, x, squared, end fracti

Fantom [35]

Answer:

Step-by-step explanation:

Given the expression:

(5 * x³) / x² ; for x = 2

Simplifying (5 * x³) / x²

5 * x^(3-2)

5 * x^1

= 5x

For x = 2

5x = 5*2 = 10

8 0

2 years ago

Solve the equation:<br><br> 3x + 34 when x = 4<br><br> Show work!

balandron [24]

3x + 34

x = 4

Substitute x:

3 times 4 = 12

12 + 34 = 46

The answer is 46.

Hope this helped

4 0

3 years ago

Read 2 more answers