Can anyone help me out with these two problems please? Thanks!

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0

3 years ago

Use the net to compute the surface area of the three-dimensional figure.

Grace [21]

9514 1404 393

Answer:

382 square units

Step-by-step explanation:

The central four rectangles down the middle of the net are 9 units wide, and alternate between 8 and 7 units high. Then the area of those four rectangles is ...

9(8+7+8+7) = 270 . . . square units

The rectangles making up the two left and right "wings" of the net are 8 units high and 7 units wide, so have a total area of ...

2×(8)(7) = 112 . . . square units

Then the area of the figure computed from the net is ...

270 +112 = 382 . . . square units

__

Additional comment

You can reject the first two answer choices immediately, because they are odd. Each face will have an area that is the product of integers, so will be an integer. There are two faces of each size, so the total area of this figure must be an even number.

You may recognize that the dimensions are 8, 8+1, 8-1. Then the area is roughly that of a cube with dimensions of 8: 6×8² = 384. If you use these values (8, 8+1, 8-1) in the area formula, you find the area is actually 384-2 = 382. That area formula is A = 2(LW +H(L+W)).

3 0

3 years ago

In a class of 42 students, about 71% of them went trick or treating. About how many students in the class went trick or treating

Amanda [17]

About 29 students went trick or treating

6 0

3 years ago

Read 2 more answers

Write the inequality in slope-intercept form. 10x + 5y < 15

mestny [16]

Solve for y
minus 10x from both sides
5y<-10x+15
divide both sides by 5
y<-2x+3

7 0

3 years ago

Jerome burns 4 cal/min walking and 10 cal/min running. He walks between 10 and 20 min each day and runs between 30 and 45 min ea

erik [133]

Answer:

20 minutes should spend on walking and 30 minutes on running.

Step-by-step explanation:

Let x represents the minutes spent on walking and y represents the minutes spent on running,

∵ He burns 4 cal/min walking and 10 cal/min running,

So, the total calories burnt,

Z = 4x + 10y

Which has to be maximised.

Now, he walks between 10 and 20 min each day,

i.e. 10 < x < 20,

Also, he runs between 30 and 45 min each day,

i.e. 30 < y < 45,

By graphing 10 < x < 20 and 30 < y < 45,

We obtained the vertices of feasible region,

(10, 45), (20, 45), (10, 30) and (20, 30)

At (10, 45),

Z = 4(10) + 10(45) = 40 + 450 = 490,

At (20, 45),

Z = 4(20) + 10(45) = 80 + 450 = 530,

At (10, 30),

Z = 4(10) + 10(30) = 40 + 300 = 340,

At (20, 30)

Z = 4(20) + 10(30) = 80 + 300 = 380

Hence, maximum calories is 530 when 20 minutes spent on walking and 30 minutes spent on running.

5 0

3 years ago