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ICE Princess25 [194]
3 years ago
8

Does a limit exist if the left hand side and the right hand side limit are not the same? Especially when there is only a line to

the right side of the limit, but not a line to the left side.
Mathematics
2 answers:
In-s [12.5K]3 years ago
6 0
I agree with the answer below. The right side limit exists in this case, however, the left side limit doesn't exist, because there is not a line (think of it as a route) for the point to travel from left side of the limit to the point, so left side is basically DNE.
anzhelika [568]3 years ago
4 0
No, the limit does not exist if the left-sided limit does not equal to the right-sided limit. Furthermore, if the limit lets say is x --> 2, and there is a line from the 2+ direction but not a line from 2- direction, then the left-sided limit doesn't even exist.
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Franky puts £1000 in a bank account for six years with 2.5% rate of interest how much will be in the account at the end of that
yanalaym [24]

Answer:

£1159.69

Step-by-step explanation:

Assuming that the rate of interest is compound interest, you perform the equation:

New amount = Original amount × InterestRate^Number of Years

New amount = 1000 × 1.025⁶

New amount = 1159.693418

= £1159.69 (to the nearest penny)

8 0
3 years ago
For the function, find all critical numbers and then use the second-derivative test to determine whether the function has a rela
Serhud [2]

Answer:

relative maximum: x = 1

relative minimum: x = 7

Step-by-step explanation:

Critical points:

Values of x for which f'(x) = 0.

Second derivative test:

For a critical point, if f''(x) > 0, the critical point is a relative minimum.

Otherwise, if f''(x) < 0, the critical point is a relative maximum.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

f(x) = x^{3} - 12x^{2} + 21x - 8

Finding the critical points:

f'(x) = 3x^{2} - 24x + 21

3x^{2} - 24x + 21 = 0

Simplifying by 3

x^{2} - 8x + 7 = 0

So a = 1, b = -8, c = 7

\bigtriangleup = (-8)^{2} - 4*1*7 = 36

x_{1} = \frac{-(-8) + \sqrt{36}}{2} = 7

x_{2} = \frac{-(-8) - \sqrt{36}}{2} = 1

Second derivative test:

The critical points are x = 1 and x = 7.

The second derivative is:

f''(x) = 6x - 24

f''(1) = 6*1 - 24 = -18

Since f''(1) < 0, at x = 1 there is a relative maximum.

f''(7) = 6*7 - 24 = 18

Since f''(x) > 0, at x = 7 there is a relative minumum.

8 0
3 years ago
Suppose that X is a random variable with mean 30 and standard deviation 4. Also suppose that Y is a random variable with mean 50
Vladimir79 [104]

Answer:

a) Var[z] = 1600

    D[z] = 40

b) Var[z] = 2304

    D[z] = 48

c) Var[z] = 80

    D[z] = 8.94

d) Var[z] = 80

    D[z] = 8.94

e) Var[z] = 320

    D[z] = 17.88

Step-by-step explanation:

In general

V([x+y] = V[x] + V[y] +2Cov[xy]

how in this problem Cov[XY] = 0, then

V[x+y] = V[x] + V[y]

Also we must use this properti of the variance  

V[ax+b] = a^{2}V[x]

and remember that

standard desviation = \sqrt{Var[x]}

a) z = 35-10x

   Var[z] = 10^{2} Var[x] = 100*16 = 1600

   D[z] = \sqrt{1600} = 40

b) z = 12x -5

   Var[z] = 12^{2} Var[x] = 144*16 = 2304

   D[z] = \sqrt{2304} = 48

c) z = x + y

   Var[z] =  Var[x+y] = Var[x] + Var[y] = 16 + 64 = 80

   D[z] = \sqrt{80} = 8.94  

d) z = x - y

   Var[z] =  Var[x-y] = Var[x] + Var[y] = 16 + 64 = 80

   D[z] = \sqrt{80} = 8.94

e) z = -2x + 2y

   Var[z] = 4Var[x] + 4Var[y] = 4*16 + 4*64 = 320

  D[z] = \sqrt{320} = 17.88

   

7 0
3 years ago
Write an expression to describe the sequence below. Use n to represent the position of a term in the sequence, where n = 1 for t
ollegr [7]

Answer:

<u>The expression to describe this sequence is Aₙ =  (n - 1) * 1  - 6</u>

Step-by-step explanation:

1. Let's write an expression to describe the sequence which first four terms are as follows:

- 6, -5, -4, - 3.....

Aₙ = (n - 1) * 1  - 6; A₁ = - 6

A₁ = (n - 1) * 1  - 6 = (1 - 1) * 1  - 6 = 0 * 1 - 6 ⇒A₁ = - 6

A₂ = (n - 1) * 1  - 6 = (2 - 1) * 1  - 6 = 1 * 1 - 6⇒ A₁ = - 5

A₃ = (n - 1) * 1  - 6 = (3 - 1) * 1  - 6 = 2 * 1 - 6⇒ A₁ = - 4

A₄ = (n - 1) * 1  - 6 = (4 - 1) * 1  - 6 = 3 * 1 - 6⇒ A₁ = - 3

A₅ = (n - 1) * 1  - 6 = (5 - 1) * 1  - 6 = 4 * 1 - 6⇒ A₁ = - 2

A₆ = (n - 1) * 1  - 6 = (6 - 1) * 1  - 6 = 5 * 1 - 6⇒ A₁ = - 1

A₇ = (n - 1) * 1  - 6 = (7 - 1) * 1  - 6 = 6 * 1 - 6⇒ A₁ = 0

5 0
3 years ago
For each equation given complete the table and then use the points in your table to sketch the graph of the equation, Only #1
zloy xaker [14]

Answer:FOR THE FIRST TABLE ON THE LEFT

X____Y

-2      1/4

-1       1/2

0        1

1          2

2         4

FOR THE SECOND GRAPH TO THE RIGHT IS

X________Y

-2              1/9

-1          1/3

0        1

1            3

2           9

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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