3

Imagine that you need to compute e^0.4 but you have no calculator or other aid to enable you to compute it exactly, only paper a

labwork [276]

Answer:

0.0032

Step-by-step explanation:

We need to compute $e^{0.4}$ by the help of third-degree Taylor polynomial that is expanded around at x = 0.

Given :

$e^{0.4}$ < e < 3

Therefore, the Taylor's Error Bound formula is given by :

$|\text{Error}| \leq \frac{M}{(N+1)!} |x-a|^{N+1}$ , where $M=|F^{N+1}(x)|$

$\leq \frac{3}{(3+1)!} |-0.4|^4$

$\leq \frac{3}{24} \times (0.4)^4$

$\leq 0.0032$

Therefore, |Error| ≤ 0.0032

4 0

3 years ago

TJ-Maxx offered a 36% discount off the original price of a head board. The amount of the discount is $9. What is the regular pri

jok3333 [9.3K]

The amount if the original cost of the head board is $12.24

5 0

3 years ago

Find the sum of the arithmetic sequence 1,3,5,7, .. , 99.

Yakvenalex [24]

Answer:

Look at picture answer is 2500

Step-by-step explanation:

8 0

3 years ago

-4x-15y=-17<br>-x+5y=-13<br><img src="https://tex.z-dn.net/?f=%20-%204x%20-%2015y%20%3D%20%20-%2017%20%20%5C%5C%20%20-%20x%20%2B

salantis [7]

To solve this system of linear equations we need a little trick called elimination steps are below

we are given the following system of equations
$- 4x - 15y = - 17$
$- x + 5y = - 13$
if you look at the x variables in both equations you can see that we can easily eliminate them by multiplying the bottom equation by -4 as so
$- 4x - 15y = - 17$
$- 4( - x + 5y = - 13)$
simplify and we get
$- 4x - 15y = - 17$
$4x - 20y = 52$
now we can combine like terms on each side with the x's cancelling and get

$- 35y = 35$
now divide off -35
$y = - 1$
great now we can go back to our original system and pick a equation and substitute y back in to find x lets use

$- x + 5y = 13$
so now we substitute y and get
$- x + 5( - 1) = 13$
$- x - 5 = 13$
$- x = 17$
$x = - 17$

now we put x and y into one coordinate (x,y)
so now our FINAL ANSWER IS
(-17,-1)

6 0

3 years ago

For this function:

galina1969 [7]

Answer:

C)

Minimum of t = $-\infty$

Maximum of t = $+\infty$

Minimum of h(t) = $-\infty$

Maximum of h(t) = +20

D)

$D(-\infty,+\infty)$

$R(-\infty,+20]$

Step-by-step explanation:

C)

Here in this problem, we are given the function, which is

$h(t)=-5t^2+20$

We observe immediately the following:

- The function has no limitations on the value of t - in fact, it contains no square roots, no logarithms, and no fractions; therefore, every value of x is acceptable in this function. This means that it has the variable t has no minimum or maximum values.

- On the other hand, h(t) cannot take any value. In fact, we notice that this is a quadratic function with the second-degree term with a negative coefficient: this means that it is a downward parabola. So, it has no downward limit (it goes to $-\infty$ ), but it has a maximum, which corresponds to the vertex of the parabola.

The x-coordinate of the parabola is given by

$x_v = -\frac{b}{2a}=0$

because the coefficient of the 1st-degree term, b, is zero. So, the y-coordinate of the vertex is

$h(0)=-5\cdot 0^2+20 = 20$

So, the maximum of h(t) is 20. Therefore we have:

Minimum of t = $-\infty$

Maximum of t = $+\infty$

Minimum of h(t) = $-\infty$

Maximum of h(t) = +20

D)

The domain of a function is defined as the set of values of the independent function x for which the function has a valid value.

Therefore in this case, the domain of h(t) is the set of all values allowed for t: so, from part C, we can say that the domain is

$D(-\infty,+\infty)$

So, all real values.

The range of a function instead is the set of all values of the dependent variables, y.

So in this case, the range of h(t) is the set of all values that h(t) can take.

From part C, we know therefore that the range is

$R(-\infty,+20]$

Where +20 has the ] instead of ) because it is also an allowed value.

5 0

3 years ago