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4 years ago

14

HELPP! Brainliest is rewarded!

2 answers:

maxonik [38]4 years ago

7 0

Answer:

DNE (does not exist)

Step-by-step explanation:

Our function is: $f(x)=\frac{x^2+2x}{x^4}$ . We want to find the limit of this as x approaches 0. The first thing we would want to do is to substitute 0 in for x. But when we do that, we get 0/0, which is undefined:

$\frac{0^2+2*0}{0^4}=\frac{0}{0}$

Let's divide the numerator and denominator both by x:

$\frac{x^2+2x}{x^4}=\frac{x+2}{x^3}$

Now substitute 0 in again:

$\frac{0+2}{0^3}=\frac{2}{0}$

Because we have a number divided by 0, this cannot exist. If we graph this function (see attachment), we'll also see that the graph diverges at x = 0, so the limit does not exist.

daser333 [38]4 years ago

6 0

Answer:

Limit doesn't exist

Step-by-step explanation:

As x --> 0,

(x² + 2x)/x⁴

From the LHS:

Let x = -0.1, (x² + 2x)/x⁴ = -1900

From the RHS:

Let x = 0.1, (x² + 2x)/x⁴ = 2100

Therefore the limit doesn't exist

You might be interested in

The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

Norma-Jean [14]

The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

So,

- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

3 0

3 years ago

What value represents the horizontal translation from the graph of the parent function f(x) = x2 to the graph of the function

kirza4 [7]

The negative 4 in the parenthesis caused the graph to move four units to the right.

7 0

4 years ago

Read 2 more answers

What is the surface area of the cylinder?

Ghella [55]

Answer:
$66\pi \: ft^{2}$

4 0

4 years ago

Read 2 more answers

Bob is saving up for a new phone. He has about $150 right now but she has a job and gets paid $15 an hour to babysit. If he work

guajiro [1.7K]

Current Balance: $150

Income rate: 15x

Equation: 15x+150

8 0

3 years ago

I’ve been stuck can someone explain

MariettaO [177]

Answer:

Option (3)

Step-by-step explanation:

From the figure attached,

AB and CD are two chords intersecting at O.

m∠AOD = 37°

m∠AOC + m∠AOD = 180° [Since these angles are supplementary angles]

m∠AOC = 180° - 37°

= 143°

By the theorem of intersecting chords,

Measure of angle formed is the half of the sum of measures of the arcs intercepted by the angle and vertical angle.

m∠AOC = $\frac{1}{2}(\widehat{AC}+\widehat{BD})$

143° = $\frac{1}{2}[(x+5)+(x-5)]$

143° = x

Therefore, Option (3) will be the answer.

3 0

3 years ago